a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

a) 22 + (2x -13) = 83 => 2x -13 = 61 => x = 37.

b) 51 - (-12 + 3x) = 27 => 63 - 3x = 27 => x = 12.

c) - (2x + 2) + 21 = - 23 => 2x + 2 = 44 => x = 21.

d) 25 - (25 - x) = 0 => 25 - 25 + x = 0 => x = 0.

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

                 \(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

                \(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

           \(=>2x+\frac{3}{5}=\frac{3}{5}\)

                                    \(2x=\frac{3}{5}-\frac{3}{5}\)

                                    \(2x=0\)

                                      \(x=0:2\)

                                      \(x=0\)

b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)

=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.

\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)

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

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

\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

b) \(\left(3x-1\right)\left(-\frac{1}{2}x+5\right)=0\)

\(\left(3x-1\right)\left(-\frac{x}{2}+5\right)=0\)

\(\left(3x-1\right)\left(5-\frac{x}{2}\right)=0\)

\(\orbr{\begin{cases}3x-1=0\\5-\frac{x}{2}=0\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

a ) \(9x^2-49=9\)

\(\Leftrightarrow9x^2=58\)

\(\Leftrightarrow x^2=29\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)

Vậy ......................

b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)

\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)

\(\Leftrightarrow x^3+27-x^3+x-27=0\)

\(\Leftrightarrow x=0\)

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

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

\(\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)

Vây .....................

d với e thì tách hết ra, tự triệt tiêu là ra kết quả, dễ mà :) @La  Thị Thu Phượng

a) \(9x^2-49=0\)

\(\left(3x\right)^2-7^2=0\)

\(\Rightarrow\left(3x+7\right)\left(3x-7\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x+7=0\\3x-7=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{7}{3}\\x=\frac{7}{3}\end{array}\right.\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)

\(x^3+27-x\left(x^2-1\right)-27=0\)

\(x^3+27-x^3+x-27=0\)

\(x=0\)

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

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

\(x^2+4=0\)

\(x^2=-4\) (loại vì \(x^2\ge0\) với mọi x)

d) \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)

\(3x^2+2x+x^2+2x+1-4x^2+25=0\)

\(4x+26=0\)

\(4x=-26\)

\(x=-\frac{13}{2}\)

e) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)\left(2x+1\right)=7\)

\(4x^2-8x+x-2-4x^2-2x+6x+3=7\)

\(-3x+1=7\)

\(-3x=6\)

\(x=-2\)

a, \(3x^2-3x+2x^3-2x^2=0\)

\(\Rightarrow3x.\left(x-1\right)+2x^2.\left(x-1\right)=0\)

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

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

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

Vậy......

Câu b tương tự!!

a, \(3x^2-3x+2x^3-2x^2=0\)

\(\Leftrightarrow3x\left(x-1\right)+2x^2\left(x-1\right)=0\)

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

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

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

Vậy...