a) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

b) \(\Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\end{matrix}\right.\)

c) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\end{matrix}\right.\)

d) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

e) \(\Rightarrow x^2=\dfrac{16}{25}\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

f) \(\Rightarrow x^2=\dfrac{7}{36}\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{7}}{6}\\x=-\dfrac{\sqrt{7}}{6}\end{matrix}\right.\)

g) \(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) ĐKXĐ: \(n\ne-\dfrac{1}{3}\left(đúng.do.n\in Z\right)\)

 \(\dfrac{6n-2}{3n+1}=2-\dfrac{4}{3n+1}\in Z\)

\(\Rightarrow\left(3n+1\right)\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

Do \(n\in Z\Rightarrow n\in\left\{-1;0;1\right\}\)

c) ĐKXĐ: \(n\ne2\)

\(\dfrac{n^2-3n+5}{n-2}=\dfrac{\left(n-2\right)\left(n-1\right)+3}{n-2}=n-1+\dfrac{3}{n-2}\in Z\)

\(\Rightarrow\left(n-2\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow n\in\left\{-1;1;3;5\right\}\)

D

\(AB=\dfrac{5}{3}BC,AC=\dfrac{7}{5}AB\Rightarrow\left\{{}\begin{matrix}BC=\dfrac{3}{5}AB\\AC=\dfrac{7}{5}AB\end{matrix}\right.\)

\(AB+AC+BC=210:2\)

\(\Rightarrow AB+\dfrac{7}{5}AB+\dfrac{3}{5}AB=105\Rightarrow AB=35\left(cm\right)\)

\(\Rightarrow\left\{{}\begin{matrix}AC=\dfrac{7}{5}AB=49\left(cm\right)\\BC=\dfrac{3}{5}AB=21\left(cm\right)\end{matrix}\right.\)

a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)

\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

a) \(A=\dfrac{\left(x-1\right)\left(x-2\right)+6x-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)

b) \(\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)

c) \(A=\dfrac{x}{x-2}=1+\dfrac{2}{x-2}\in Z\Rightarrow\left(x-2\right)\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Kết hợp ĐKXĐ: 

\(\Rightarrow x\in\left\{0;1;3;4\right\}\)

\(\Rightarrow\left(2x+1\right)\inƯ\left(45\right)=\left\{-45;-15;-9;-5;-3;-1;1;3;5;9;15;45\right\}\)

\(\Rightarrow2x\in\left\{-46;-16;-10;-6;-4;-2;0;2;4;8;14;44\right\}\)

\(\Rightarrow x\in\left\{-23;-8;-5-3;-2-1;0;1;2;4;7;22\right\}\)

5B  6D

Tổng số phần bằng nhau: \(1+6=7\)(phần)

Giá trị một phần: \(42:7=6\)

Tuổi con là: \(6\times1=6\left(tuổi\right)\)

Tuổi bố là: \(6\times6=36\left(tuổi\right)\)

Tuổi ông là: \(36\times2=72\left(tuổi\right)\)

Bài 97:

\(\Leftrightarrow x-68=\left(\dfrac{43}{11}-4\right):\left(\dfrac{21}{22}-1\right):\left(\dfrac{33}{34}-1\right)=-68\Leftrightarrow x=0\)

Bài 98:

a) \(A=\dfrac{2002}{\left|x\right|+2003}\le\dfrac{2002}{0+2003}=\dfrac{2002}{2003}\)

\(maxA=2003\Leftrightarrow x=0\)

b) \(B=\dfrac{\left|x\right|+2002}{-2003}\le\dfrac{0+2002}{-2003}=-\dfrac{2002}{2003}\)

\(maxB=-\dfrac{2002}{2003}\Leftrightarrow x=0\)