`@` `\text {Ans}`

`\downarrow`

`a)`

\(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\) phải k bạn nhỉ? `11/8` k có bậc lũy thừa nào `=5` á.

`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

`=>`\(\left(\dfrac{x}{2}-1\right)^3=-\dfrac{27}{8}\)

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

`=>`\(\dfrac{x}{2}-1=-\dfrac{3}{2}\)

`=>`\(\dfrac{x}{2}=-\dfrac{3}{2}+1\)

`=>`\(\dfrac{x}{2}=-\dfrac{1}{2}\)

`=> x=1`

Vậy, `x=1`

`b)`

\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\0,75-1\dfrac{1}{2}x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\-\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=-3\\-3x\cdot100=2\cdot75\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\-3x\cdot100=150\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy, `x={-3/2; -1/2}.`

a, 2\(xy\) - 2\(x\) + 3\(y\) = -9

(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12

2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12

(\(y-1\))(2\(x\) + 3) = -12

Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}

Lập bảng ta có:

Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)

b, (\(x+1\))²(\(y\) - 3) = -4 

    Ư(4) = {-4; -2; -1; 1; 2; 4}

Lập bảng ta có: 

Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là: 

(\(x;y\)) = (0; -1); (-3; 2); (1; 2)

c) \(\left(x+3\right)^2+\left(2y-1\right)^2< 44\)

\(\Leftrightarrow\left(x+3\right)^2< 44-\left(2y-1\right)^2< 44\) (do \(-\left(2y-1\right)^2\le0\)) (1) 

mà (x + 3)² là số chính phương 

Kết hợp (1) ta được \(\left(x+3\right)^2\le36\)

\(\Leftrightarrow\left(x+3\right)^2\le6^2\Leftrightarrow\left(x+3\right)^2\in\left\{0;1;4;9;25;36\right\}\)

Với (x + 3)² \(\in\left\{0;1;4\right\}\) ta được (2y - 1)² \(\in\left\{0;1;4;9;25;36\right\}\) 

Với (x + 3)² \(\in\left\{9;16\right\}\) ta được (2y - 1)² \(\in\left\{0;1;4;9;25\right\}\) 

Với (x + 3)² = 25 ta được (2y - 1)² \(\in\left\{0;1;4;9;16\right\}\)

Với (x + 3)² = 36 ta được (2y - 1)² \(\in\left\{0;1;4;9\right\}\)

`a)2x^2+3(x-1)(x+1)=5x(x+1)`

`<=>2x^2+3x^2-3=5x^2+5x`

`<=>5x=-3`

`<=>x=-3/5`

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`b)(x-3)^3+3-x=0` nhỉ?

`<=>(x-3)^3-(x-3)=0`

`<=>(x-3)(x^2-1)=0`

`<=>[(x=3),(x^2=1<=>x=+-1):}`

__________________________________________

`c)5x(x-2000)-x+2000=0`

`<=>5x(x-2000)-(x-2000)=0`

`<=>(x-2000)(5x-1)=0`

`<=>[(x=2000),(x=1/5):}`

__________________________________________

`d)3(2x-3)+2(2-x)=-3`

`<=>6x-9+4-2x=-3`

`<=>4x=2`

`<=>x=1/2`

__________________________________________

`e)x+6x^2=0`

`<=>x(1+6x)=0`

`<=>[(x=0),(x=-1/6):}`

a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)

\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)

\(\Rightarrow2x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)

i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)

\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)

\(\Rightarrow\dfrac{x}{2}-1=-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{x}{2}=-\dfrac{3}{2}+1\)

\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)

k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{3}{4}-1\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)

= -1

giải thì tự xử lí viết ra dài  nản

`x-1/9 =8/3`

`=>x=8/3 +1/9`

`=> x= 24/9 +1/9`

`=>x= 25/9`

Vậy `x=25/9`

__

`x-2/20=-5/2-x`

`=>x+x=-5/2 +2/20`

`=> 2x= -50/20 +2/20`

`=> 2x= -48/20`

`=> x= -12/5:2`

`=>x=-12/5 xx1/2`

`=>x= -12/10`

`=>x= -6/5`

Vậy `x=-6/5`

`@` `\text {Ans}`

`\downarrow`

`a)`

Đề là \(\dfrac{x-1}{9}=\dfrac{8}{3}\) phải hongg bạn?

\(\dfrac{x-1}{9}=\dfrac{8}{3}\)

`=>` `(x-1)3 = 8*9`

`=> (x-1)3=72`

`=> x-1=72 \div 3`

`=> x-1=24`

`=> x=25`

`b)`

\(\dfrac{x-2}{20}=\dfrac{-5}{2-x}\)

`=>` `(x-2)(2-x)=20*(-5)`

`=> (x-2)(2-x)=-100`

`=> -[(x-2)(x-2)]=-100`

`=> -(x-2)^2 = -100`

`=> (x-2)^2 = 100`

`=> (x-2)^2 = (+-10)^2`

`=>`\(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

Giải:

a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\) 

              \(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\) 

              \(\dfrac{25}{x+1}=\dfrac{1}{3}\) 

\(\Rightarrow1.\left(x+1\right)=25.3\)  

\(\Rightarrow x+1=75\) 

\(\Rightarrow x=75-1\) 

\(\Rightarrow x=74\) 

b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) 

              \(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\) 

             \(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\) 

Chúc bạn học tốt!