`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`

1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)

\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)

\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)

\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)

`2)(6x+5)^2(3x+2)(x+1)=35`
`<=>12(6x+5)^2(3x+2)(x+1)=420`
`<=>(6x+5)^2+(6x+4)(6x+6)=420`
Đặt `6x+5=a` 
`pt<=>a^2(a+1)(a-1)=420`
`<=>a^2(a^2-1)-420=0`
`<=>a^4-a^2-420=0`
`<=>` $\left[ \begin{array}{l}a^2=-20(False)\\a^2=21(True)\end{array} \right.$
`<=>` $\left[ \begin{array}{l}a=\sqrt{20}\\a=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}6x+5=\sqrt{20}\\6x+5=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\dfrac{\sqrt{20}-5}{6}\\x=\dfrac{-\sqrt{20}-5}{6}\end{array} \right.$
Vậy `S={\frac{\sqrt{20}-5}{6},\frac{-\sqrt{20}-5}{6}}`

\(\frac{3-\frac{1}{2}+\frac{1}{4}}{\frac{2}{3}-\frac{5}{6}-\frac{3}{4}}=\frac{\left(3-\frac{1}{2}+\frac{1}{4}\right)\left(3.4\right)}{\left(\frac{2}{3}-\frac{5}{6}-\frac{3}{4}\right)\left(3.4\right)}=\frac{36-6+3}{8-10-9}=\frac{33}{-11}=-3\)

Xin lỗi m.n nhé gửi nhầm tí

1) 1/3 x 1/2 x 3/7 = 3/42 = 1/14

2) 5/4 x 1/3 +1/7 = 5/12 + 1/7 = 35/84 + 12/84 = 47/84

3) 8 x ( 8/9 - 2/3 ) = 8 x 2/9 = 16/9

4) 5/6 x 48/20 x 1/2 = 240/240 = 1

5) ( 2/5 + 3/4 ) + 8 = 23/20 + 8 = 23//20 + 160/20 = 183/20

6) 10 x ( 1/2 - 1/5 ) = 10 x 3/10 = 10/1 x 3/10 = 30/10 = 3

c: \(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2+2x+1}{x^2-x+1}\)

a: Để A là số nguyên thì \(13⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{2;0;14;-12\right\}\)

b. Ta có \(B=\dfrac{x+3}{x-2}=\dfrac{x-2+3+2}{x-2}=1+\dfrac{5}{x-2}\)

Để \(B\) nhận giá trị nguyên thì\(5⋮\left(x-2\right)\Rightarrow\left(x-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=5\\x-2=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\\\sqrt{x}=7\\\sqrt{x}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=1\\x=49\end{matrix}\right.\)

Vậy tất cả các x thỏa mãn ycbt là x=9; x=1 hoặc x=49

b: Để \(\dfrac{x+3}{x-2}\) là số nguyên thì \(x+3⋮x-2\)

\(\Leftrightarrow5⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{3;1;7;-3\right\}\)