a) \(\left|x\right|=10\)

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

b) \(\left|x+7\right|=-5\)

Mà: \(\left|x+7\right|\ge0\forall x\)

\(\Rightarrow\) Không tìm được giá trị nào của \(x\) thoả mãn yêu cầu đề bài.

\(5x\left(x-3\right)=x-3\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

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

\(A=27x^3-8-27x^3+3=-5\\ B=8x^3+y^3+8x^3-y^3-16x^3=0\)

ta có: \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{x}.\)

\(A=1+\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{x}\)

\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2.2}+...+\frac{1}{x:2}\)

\(\Rightarrow2A-A=2-\frac{1}{x}\)

\(A=2-\frac{1}{x}=\frac{4095}{2048}\)

=> 1/x = 1/2048

=> x = 2048 ( 2048 = 2¹¹ )

a: \(x\left(1-2x\right)+2x^2=14\)

=>\(x-2x^2+2x^2=14\)

=>x=14

b: \(x\left(x-5\right)+3x-15=0\)

=>\(\left(x-5\right)\left(x+3\right)=0\)

=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

\(2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{2}{x}\)

=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{2}{x}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{2}{x}+\frac{1}{x}\right)\)

=> \(A=2-\frac{1}{x}\)

Giải phương trình:

 \(2-\frac{1}{x}=\frac{4095}{2048}\)

            \(\frac{1}{x}=2-\frac{4095}{2048}\)

              \(\frac{1}{x}=\frac{1}{2048}\)

                x=2048

a)\(2x\left(x-2016\right)-2x+4032=0\)

\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)

\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)

b)\(5x\left(x-3\right)=x-3\)

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)

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

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

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

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

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

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

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

d) Ta có: \(x^2-6x-16=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

e) Ta có: \(x^4-6x^2-7=0\)

\(\Leftrightarrow\left(x^2-7\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)

120 - 4x(y - 3) = 136

=> 4x(y - 3)     = 136 - 120

=> 4x(y - 3)     = 16

=>   x(y - 3)     = 16 : 4

=>   x(y - 3)     = 4

Mà 4 = 1.4 = 2.2 = (-1).(-4) = (-2).(-2)

Lập bảng xét tất cả 6 trường hợp : 

Vậy các cặp (x,y) thỏa mãn là : (1;7) ; (2;5) ; (4;4) ; (-2 ; 1) ; (-1 ; -1) ; (- 4;2)

120 - 4(y - 3) = 136

<=> -4(y - 3) = 136 - 120

<=> -4(y - 3) = 16

<=> y - 3 = 16 : (-4)

<=> y - 3 = -4

<=> y = -4 + 3

<=> y = -1

=> y = -1

120 - 4 x (y - 3) = 136

=>     4 x (y - 3) = 120 - 136

=>     4 x (y - 3) = - 16

=>            y - 3   = -16 : 4

=>            y - 3   = - 4

=>            y         = - 4 + 3

=>            y         = - 1