\(\left(x+3\right)\left(2x-1\right)-\left(x-4\right)\left(2x+1\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-\left(2x^2+x-8x-4\right)=10\)

\(\Leftrightarrow2x^2-x+6x-3-2x^2-x+8x+4=10\)

\(\Leftrightarrow12x+1=10\)

\(\Leftrightarrow12x=9\)

\(\Leftrightarrow x=\frac{9}{12}\)

\(\Rightarrow x=\frac{3}{4}\)

Vậy   \(x=\frac{3}{4}\)

a) Đặt x^2+2x+2=t

\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)

\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)

Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Để ( 2x - 1 ).( y - 3 ) = 10 <=> 2x - 1 và y - 3 ∈ Ư ( 10 ) = { - 10 ; - 5 ; - 2 ; - 1 ; 1 ; 2 ; 5 ; 10 }

Nếu 2x + 1 = 10 thì y - 3 = 1 => x = 9/2 ; y = 4 ( loại )

Nếu 2x + 1 = 5 thì y - 3 = 2 => x = 2 ; y = 5 ( chọn )

...................

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)

Hãy tích cho tui đi

vì câu này dễ mặc dù tui ko biết làm 

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Tui sẽ ko tích lại bạn đâu

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b, x thuộc rỗng vì giá trị tuyệt đối ko âm

c. | x +1| thuộc +-2

=> th1 : x+1=2=> x=1

Th2: x+1=-2=> x= -3

a. |3-x| thuộc +-1/2

=> th1: 3-x=1/2=> x= 5/2

Th2: 3-x=-1/2=> x= 7/2

Vậy...

x=0 ; y = 13

x=0 ; y =5