河南省南陽市六校2025-2025學(xué)年高二上學(xué)期期末聯(lián)考數(shù)學(xué)試題

河南省南陽市六校2025-2025學(xué)年高二上學(xué)期期末聯(lián)考數(shù)學(xué)試題考試時(shí)間：90分鐘?總分：150分?年級/班級：高二一、選擇題〔每題5分，共20分〕要求：在以下各題中，每題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)為哪一項(xiàng)符合題目要求的。1.知曉函數(shù)$f(x)=x^3-3x+2$，那么函數(shù)的對稱中心為〔〕A.$(1,-2)$B.$(-1,-2)$C.$(0,2)$D.$(0,-2)$2.設(shè)$a>0$，$b>0$，$a+b=1$，那么$2a^2+3b^2$的最小值為〔〕A.1B.2C.$\frac{7}{2}$D.33.假設(shè)直線$y=kx+1$與圓$(x-1)^2+y^2=1$相切，那么$k$的取值范圍為〔〕A.$[-\sqrt{2},\sqrt{2}]$B.$(-\infty,-\sqrt{2})\cup(\sqrt{2},+\infty)$C.$(-\sqrt{2},\sqrt{2}]$D.$(-\infty,\sqrt{2}]\cup[-\sqrt{2},+\infty)$4.知曉等差數(shù)列$\{a_n\}$的前$n$項(xiàng)和為$S_n$，假設(shè)$a_1=2$，$S_5=20$，那么公差$d$的值為〔〕A.2B.3C.4D.55.知曉復(fù)數(shù)$z$滿足$|z-1|=|z+1|$，那么$z$的取值范圍為〔〕A.$y\leq0$B.$y\geq0$C.$x^2+y^2=1$D.$x^2+y^2=4$6.設(shè)$a>0$，$b>0$，$a+b=1$，那么$\sqrt{a}+\sqrt$的最大值為〔〕A.$\sqrt{2}$B.$2\sqrt{2}$C.$\sqrt{\frac{5}{2}}$D.$2\sqrt{\frac{5}{2}}$二、填空題〔每題5分，共20分〕要求：把答案填在題目的橫線上。1.知曉等差數(shù)列$\{a_n\}$中，$a_1=3$，$a_4=7$，那么公差$d=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\三、解答題〔每題15分，共30分〕要求：解答以下各題。3.知曉函數(shù)$f(x)=e^x-\ln(x)$，求$f(x)$的導(dǎo)數(shù)$f'(x)$，并說明$f(x)$的單調(diào)區(qū)間。四、證明題〔每題15分，共30分〕要求：證明以下各題。4.證明：對于任意正實(shí)數(shù)$a$和$b$，不等式$a^2+b^2\geq2ab$成立。五、計(jì)算題〔每題15分，共30分〕要求：計(jì)算以下各題。5.計(jì)算定積分$\int_0^{\frac{\pi}{2}}\frac{\sin^2x}{\cos^2x+1}\,dx$。六、應(yīng)用題〔每題15分，共30分〕要求：解答以下各題。6.一輛汽車從靜止開始以勻加速直線運(yùn)動(dòng)，加速度為$a$，求汽車在時(shí)間$t$內(nèi)所行駛的距離$s$，并求出汽車從靜止加速到速度$v$所需的時(shí)間$t_0$。本次試卷答案如下：一、選擇題〔每題5分，共20分〕1.A解析：函數(shù)$f(x)=x^3-3x+2$的導(dǎo)數(shù)$f'(x)=3x^2-3$，令$f'(x)=0$，解得$x=\pm1$。由于$f(x)$在$x=1$處取得極小值，故對稱中心為$(1,f(1))=(1,-2)$。2.C解析：由均值不等式得$\sqrt{2a^2+3b^2}\geq\sqrt{2\sqrt{2a^2}\cdot\sqrt{3\sqrt{b^2}}}=\sqrt{2\sqrt{2}\cdot\sqrt{3}}=\sqrt{6}$，等號成立當(dāng)且僅當(dāng)$2a^2=3b^2$。由$a+b=1$，解得$a=\frac{\sqrt{6}}{2}$，$b=\frac{2-\sqrt{6}}{2}$。因此，$2a^2+3b^2=3+2\sqrt{3}$，最小值為$\frac{7}{2}$。3.B解析：圓的方程$(x-1)^2+y^2=1$表示圓心在$(1,0)$，半徑為1的圓。直線$y=kx+1$到圓心的距離為$\frac{|k\cdot1+1|}{\sqrt{k^2+1}}=1$，解得$k=\pm\sqrt{2}$。因此，$k$的取值范圍為$(-\infty,-\sqrt{2})\cup(\sqrt{2},+\infty)$。4.B解析：等差數(shù)列的前$n$項(xiàng)和公式為$S_n=\frac{n(a_1+a_n)}{2}$，代入$a_1=2$，$S_5=20$，得$20=\frac{5(2+a_5)}{2}$，解得$a_5=14$。由$a_5=a_1+4d$，解得$d=3$。5.A解析：復(fù)數(shù)$z=x+yi$，那么$|z-1|=|(x-1)+yi|$，$|z+1|=|(x+1)+yi|$。由$|z-1|=|z+1|$，得$(x-1)^2+y^2=(x+1)^2+y^2$，化簡得$x^2-2x+1=x^2+2x+1$，解得$x=0$。因此，$z$的取值范圍為$y\leq0$。6.D解析：由$a>0$，$b>0$，$a+b=1$，得$a^2+b^2=(a+b)^2-2ab=1-2ab$。由均值不等式得$1-2ab\geq2\sqrt{2ab}-2ab$，等號成立當(dāng)且僅當(dāng)$a=b$。因此，$\sqrt{a}+\sqrt$的最大值為$2\sqrt{\frac{5}{2}}$。二、填空題〔每題5分，共20分〕1.$d=2$解析：由等差數(shù)列的通項(xiàng)公式$a_n=a_1+(n-1)d$，代入$a_1=3$，$a_4=7$，解得$d=2$。三、解答題〔每題15分，共30分〕3.解析：函數(shù)$f(x)=e^x-\ln(x)$的導(dǎo)數(shù)$f'(x)=e^x-\frac{1}{x}$。當(dāng)$x>0$時(shí)，$e^x>\frac{1}{x}$，因此$f'(x)>0$，$f(x)$在$(0,+\infty)$上單調(diào)遞增；當(dāng)$x<0$時(shí)，$e^x<\frac{1}{x}$，因此$f'(x)<0$，$f(x)$在$(-\infty,0)$上單調(diào)遞減。四、證明題〔每題15分，共30分〕4.解析：由均值不等式得$\frac{a^2+b^2}{2}\geq\sqrt{a^2b^2}$，即$a^2+b^2\geq2ab$。等號成立當(dāng)且僅當(dāng)$a=b$。五、計(jì)算題〔每題15分，共30分〕5.解析：$\int_0^{\frac{\pi}{2}}\frac{\sin^2x}{\cos^2x+1}\,dx=\int_0^{\frac{\pi}{2}}\frac{1-\cos^2x}{\cos^2x+1}\,dx=\int_0^{\frac{\pi}{2}}\left(1-\frac{\cos^2x}{\cos^2x+1}\right)\,dx=\int_0^{\frac{\pi}{2}}\left(1-\frac{1}{\sec^2x}\right)\,dx=\int_0^{\frac{\pi}{2}}\left(\sec^2x-1\right)\,dx=\tanx\bigg|_0^{\frac{\pi}{2}}=+\infty-0=+\infty$。六、應(yīng)用題〔每題15分，共30分〕6.解析：由勻加速直線運(yùn)動(dòng)的位移公式$s=\frac{1}{2}at^2$，代入$a$和$t$，得$s=\frac{1}{2}at^2$。要求出汽車從靜止加速到速度$v$所需的時(shí)間$t_0$，由$v=at_0$，得$t_0=\frac{v}{a}$。四、證明題〔每題15分，共30分〕要求：證明以下各題。4.證明：對于任意正實(shí)數(shù)$a$和$b$，不等式$a^2+b^2\geq2ab$成立。答案：4.證明：對于任意正實(shí)數(shù)$a$和$b$，不等式$a^2+b^2\geq2ab$成立。解析：由均值不等式，我們有$(a-b)^2\geq0$，因?yàn)槠椒娇偸欠秦?fù)的。展開得到$a^2-2ab+b^2\geq0$。將$-2ab$移到