a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)

b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)

c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)

                              \(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)

d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)

        \(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)

        \(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)

a) \(A=\frac{a^{\frac{5}{2}}\left(a^{\frac{1}{2}}-a^{\frac{-3}{2}}\right)}{a^{\frac{1}{2}}\left(a^{\frac{-1}{2}}-a^{\frac{3}{2}}\right)}=\frac{a^3-a}{1-a^2}=-a\)

Do đó : \(A=-\left(\pi-3\sqrt{2}\right)=3\sqrt{2}-\pi\)

b) Rút gọn B ta có :

\(B=\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)\left[\left(a^{\frac{1}{3}}\right)^2+\left(b^{\frac{1}{3}}\right)^2\right]=\left(a^{\frac{1}{3}}\right)^3+\left(b^{\frac{1}{3}}\right)^3=a+b\)

Do đó :

\(B=\left(7-\sqrt{2}\right)+\left(\sqrt{2}+3\right)=10\)

1/ \(f'\left(x\right)=\frac{3\sqrt{x^2+1}-\frac{x\left(3x+1\right)}{\sqrt{x^2+1}}}{x^2+1}=\frac{3\left(x^2+1\right)-3x^2-x}{\left(x^2+1\right)\sqrt{x^2+1}}=\frac{3-x}{\left(x^2+1\right)\sqrt{x^2+1}}\)

Hàm số đồng biến trên \(\left(-\infty;3\right)\) nghịch biến trên \(\left(3;+\infty\right)\)

\(\Rightarrow f\left(x\right)\) đạt GTLN tại \(x=3\)

\(f\left(x\right)_{max}=f\left(3\right)=\frac{10}{\sqrt{10}}=\sqrt{10}\)

2/ \(y'=\frac{\sqrt{x^2+2}-\frac{\left(x-1\right)x}{\sqrt{x^2+2}}}{x^2+2}=\frac{x^2+2-x^2+x}{\left(x^2+2\right)\sqrt{x^2+2}}=\frac{x+2}{\left(x^2+2\right)\sqrt{x^2+2}}\)

\(f'\left(x\right)=0\Rightarrow x=-2\in\left[-3;0\right]\)

\(y\left(-3\right)=-\frac{4\sqrt{11}}{11}\) ; \(y\left(-2\right)=-\frac{\sqrt{6}}{2}\) ; \(y\left(0\right)=-\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}M=-\frac{\sqrt{2}}{2}\\N=-\frac{\sqrt{6}}{2}\end{matrix}\right.\) \(\Rightarrow MN=\frac{\sqrt{12}}{4}=\frac{\sqrt{3}}{2}\)

Tất cả các đáp án đều sai

3/ \(\left\{{}\begin{matrix}\left|x-3\right|\ge0\\\sqrt{x+1}>0\end{matrix}\right.\) \(\Rightarrow f\left(x\right)\ge0\) \(\forall x\Rightarrow N=0\) khi \(x=3\)

- Với \(0\le x< 3\Rightarrow f\left(x\right)=\left(3-x\right)\sqrt{x+1}\)

\(\Rightarrow f'\left(x\right)=-\sqrt{x+1}+\frac{\left(3-x\right)}{2\sqrt{x+1}}=\frac{-2\left(x+1\right)+3-x}{2\sqrt{x+1}}=\frac{-3x+1}{2\sqrt{x+1}}\)

\(f'\left(x\right)=0\Rightarrow x=\frac{1}{3}\)

- Với \(3< x\le4\Rightarrow f\left(x\right)=\left(x-3\right)\sqrt{x+1}\)

\(\Rightarrow f'\left(x\right)=\sqrt{x+1}+\frac{x-3}{2\sqrt{x+1}}=\frac{2\left(x+1\right)+x-3}{2\sqrt{x+1}}=\frac{3x-1}{2\sqrt{x+1}}>0\) \(\forall x>3\)

Ta có: \(f\left(0\right)=3\) ; \(f\left(\frac{1}{3}\right)=\frac{16\sqrt{3}}{9}\) ; \(f\left(4\right)=\sqrt{5}\)

\(\Rightarrow M=\frac{16\sqrt{3}}{9}\Rightarrow M+2N=\frac{16\sqrt{3}}{9}\)

Câu 2 hình như câu B mà người ta nói đạt GTLN . GTNN tại M , N nên là 0 x -2 =0

5.

Gọi M là trung điểm BC \(\Rightarrow AM\perp BC\)

\(\Rightarrow BC\perp\left(A'AM\right)\)

\(\Rightarrow\widehat{A'MA}\) là góc giữa (A'BC) và (ABC)

\(\Rightarrow\widehat{A'MA}=60^0\)

\(AM=\frac{a\sqrt{3}}{2}\Rightarrow A'A=AM.tan60^0=\frac{3a}{2}\)

\(B=\frac{a^2\sqrt{3}}{4}\Rightarrow V=B.A'A=\frac{3\sqrt{3}}{8}a^3\)

1.

\(V=Bh\)

2.

\(B=\frac{a^2\sqrt{3}}{4}\Rightarrow V=Bh=\frac{a^2\sqrt{3}}{4}.a\sqrt{6}=\frac{3\sqrt{2}}{4}a^3\)

3.

\(B=\frac{1}{2}\left(a\sqrt{2}\right)^2=a^2\Rightarrow V=Bh=a^2.5a=5a^3\)

4.

\(h=\sqrt{\left(2a\right)^2-\left(a\sqrt{3}\right)^2}=a\)

\(B=\frac{\left(a\sqrt{3}\right)^2\sqrt{3}}{4}=\frac{3\sqrt{3}}{4}a^2\)

\(V=Bh=\frac{3\sqrt{3}}{4}a^3\)

a) \(A=\log_{5^{-2}}5^{\frac{5}{4}}=-\frac{1}{2}.\frac{5}{4}.\log_55=-\frac{5}{8}\)

b) \(B=9^{\frac{1}{2}\log_22-2\log_{27}3}=3^{\log_32-\frac{3}{4}\log_33}=\frac{2}{3^{\frac{3}{4}}}=\frac{2}{3\sqrt[3]{3}}\)

c) \(C=\log_3\log_29=\log_3\log_22^3=\log_33=1\)

d) Ta có \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)

                   \(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)

                   \(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)

Đặt \(x=\frac{\sqrt{2}}{2}sint\Rightarrow dx=\frac{\sqrt{2}}{2}cost.dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=\frac{1}{2}\Rightarrow t=\frac{\pi}{4}\end{matrix}\right.\)

\(\int\limits^{\frac{1}{2}}_0f\left(\sqrt{1-2x^2}\right)dx=\frac{\sqrt{2}}{2}\int\limits^{\frac{\pi}{4}}_0f\left(cost\right).costdt=\frac{\sqrt{2}}{2}\int\limits^{\frac{\pi}{4}}_0f\left(cosx\right)cosxdx=\frac{7}{6}\)

\(\Rightarrow J=\int\limits^{\frac{\pi}{4}}_0f\left(cosx\right).cosx.dx=\frac{7\sqrt{2}}{6}\)

Đặt \(\left\{{}\begin{matrix}u=f\left(cosx\right)\\dv=cosx.dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-sinx.f'\left(cosx\right)dx\\v=sinx\end{matrix}\right.\)

\(\Rightarrow J=sinx.f\left(cosx\right)|^{\frac{\pi}{4}}_0+\int\limits^{\frac{\pi}{4}}_0f'\left(cosx\right)sin^2x.dx=\frac{\sqrt{2}}{2}+I\)

\(\Rightarrow I=\frac{7\sqrt{2}}{6}-\frac{\sqrt{2}}{2}=\frac{2\sqrt{2}}{3}\)

Đặt hệ trục \(Oxyz\) vào lăng trụ với \(O\equiv A;\) \(AB\equiv Ox\); \(AC\equiv Oy\); \(AA'\equiv Oz\), quy ước \(a\) bằng 1 đơn vị độ dài

\(\Rightarrow A'\left(0;0;1\right);B\left(1;0;0\right);C\left(0;\sqrt{3};0\right);B'\left(1;0;1\right)\)

\(\Rightarrow\overrightarrow{A'B}=\left(1;0;-1\right);\overrightarrow{BC}=\left(-1;\sqrt{3};0\right);\overrightarrow{BB'}=\left(0;0;1\right)\)

\(\Rightarrow\overrightarrow{n_{\left(A'BC\right)}}=\left[\overrightarrow{A'B};\overrightarrow{BC}\right]=\left(\sqrt{3};1;\sqrt{3}\right)\)

\(\overrightarrow{n_{\left(BCC'B'\right)}}=\left[\overrightarrow{BC};\overrightarrow{BB'}\right]=\left(\sqrt{3};1;0\right)\)

\(\Rightarrow cos\alpha=\frac{\left|3+1+0\right|}{\sqrt{3+1+3}\sqrt{3+1+0}}=\frac{2\sqrt{7}}{7}\)

// Giải theo kiểu 11:

Kẻ \(A'M\perp B'C'\); \(AN\perp BC\Rightarrow\left\{{}\begin{matrix}A'M\perp\left(BCC'B'\right)\\BC\perp\left(A'MN\right)\end{matrix}\right.\)

\(\Rightarrow\widehat{A'NM}\) là góc giữa \(\left(A'BC\right)\) và \(\left(BCC'B'\right)\)

\(\frac{1}{A'M^2}=\frac{1}{A'B'^2}+\frac{1}{A'C'^2}\Rightarrow A'M=\frac{a\sqrt{3}}{2}\)

\(\Rightarrow MN=AA'=a\Rightarrow A'N=\sqrt{A'M^2+MN^2}=\frac{a\sqrt{7}}{2}\)

\(\Rightarrow cos\alpha=\frac{MN}{A'N}=\frac{2\sqrt{7}}{7}\)

Hoặc là đáp án sai, hoặc là bạn đưa ra số liệu sai, cạnh bên lăng trụ có đúng bằng a ko bạn?