`@` `\text {Ans}`

`\downarrow`

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

`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{75}{100}-\dfrac{3}{2}x=0\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}2x=-1\cdot3\\x=\dfrac{75}{100}\div\dfrac{3}{2}\end{matrix}\right.\)

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

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

Vậy, `x={-3/2; 1/2}.`

https://lazi.vn/edu/exercise/giai-phuong-trinh-x-1-x-22-x-1-x-4-32x-4-x-42-0-1

chỉ tiềm thấy  cái này thôi ~ vì mk k thể giải đc nên nhờ mạng nên thông cảm cho nha

a/

\(\left(x-1\right)^2-\left(x+1\right)^2=2x-6\\ x^2-2x+1-\left(x^2+2x+1\right)=2x-6\\ \)

\(\Leftrightarrow x^2-2x+1-x^2-2x-1-2x+6=0\)

\(\Leftrightarrow6-6x=0\)

=> x=1

b, \(4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4\)

\(\Leftrightarrow12x+9+48=-4x+4\Leftrightarrow16x=-53\Leftrightarrow x=-\dfrac{53}{16}\)

c, \(\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\Leftrightarrow x=3;x=\dfrac{1}{4}\)

d, \(x^2\left(x+4\right)-9\left(x+4\right)=0\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\Leftrightarrow x=-3;3;-4\)

a) Đặt  \(A=4x-x^2-5\)

\(-A=x^2-4x+5\)

\(-A=\left(x^2-4x+4\right)+1\)

\(-A=\left(x-2\right)^2+1\)

Mà  \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge1\)

\(\Leftrightarrow A\le-1< 0\left(đpcm\right)\)

b) Đặt  \(B=x^2-2x+5\)

\(B=\left(x^2-2x+1\right)+4\)

\(B=\left(x-1\right)^2+4\)

Mà  \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow B\ge4>0\left(đpcm\right)\)

a)4x-x²-5 = -(x²-4x+4)-1= -(x-2)^2 -1 < 0 với mọi x (đpcm)

b) x² -2x+5= (x²-2x+1)+4=(x-1)^2 +4 >0  với mọi x (đpcm)

rút gọn à banj

đúng rồi á

`@` `\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}.`

1/ \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)

\(\left(2x-1\right)^2\left(1-3\right)=0\)

\(\left(2x-1\right)^2\cdot\left(-2\right)=0\)

\(\Rightarrow\text{ }\left(2x-1\right)^2=0\)

\(2x-1=0\)

\(2x=0+1=1\)

\(x=\frac{1}{2}\)

1) \(\left(2x-1\right)^2-3\left(2x-1\right)^2=0\)

=> \(\left(2x-1\right)^2\left(1-3\right)=0\)

=> \(\left(2x-1\right)^2.\left(-2\right)=0\)

=> \(\left(2x-1\right)^2=0\)

=> \(2x-1=0\)

=> \(2x=1\)

=> \(x=1:2=\frac{1}{2}\)

2/ \(\left(x-1\right)^2\left(x+1\right)=x+1\)

\(\left(x-1\right)^2\left(x+1\right)\text{ : }\left(x+1\right)=1\)

\(\left(x-1\right)^2\left(x+1\right)\cdot\frac{1}{\left(x+1\right)}=1\)

\(\left(x-1\right)^2\left[\left(x+1\right)\cdot\frac{1}{\left(x+1\right)}\right]=1\)

\(\left(x-1\right)^2\cdot1=1\)

\(\Rightarrow\text{ }\left(x-1\right)^2=1\)

\(x-1=\pm1\)

\(\Rightarrow\orbr{\begin{cases}x=-1+1\\x=1+1\end{cases}}\)                     \(\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

                             \(\Rightarrow\text{ }x\in\left\{0\text{ ; }2\right\}\)

a)      \(2\left(x+5\right)-x^2-5x=0\)

  \(\Leftrightarrow2x+10-x^2-5x=0\)

 \(\Leftrightarrow-x^2-3x+10=0\)

\(\Leftrightarrow x^2+3x-10=0\)

 \(\Leftrightarrow x^2-2x+5x-10=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)

b) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)^3=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

c)\(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)

d) \(x^3+x=0\)

\(\Leftrightarrow x^2\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

e)\(x^2-2x-3=0\)

\(\Leftrightarrow x^2+x-3x-3=0\)

\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

Cảm ơn bạn nha