a, \(16x^2-9\left(x+1\right)^2=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)

b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)

c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)

1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)

\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)

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

3: Ta có: \(5x^3-20x=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)

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

Rối mắt , loạn thần kinh toàn là x không

Nhiều quá bạn ơi 

a) 2x(x - 5) - x(3 + 2x) = 26

    2x² - 10x - 3x - 2x² = 26

    -10x - 3x = 26

    -13x = 26 => x = -2

b) 5x(x - 1) = x - 1

    5x(x - 1) - (x - 1) = 0

    (x - 1)(5x - 1) = 0

    x - 1 = 0 => x = 1

    5x - 1 = 0 => x = \(\frac{1}{5}\)

Vậy x = 0; x = \(\frac{1}{5}\)

\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)

\(\left(x^3+y^3\right)-\left(x+y\right)+3xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)+3xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

\(8x^3+12x^2+6x+1=0.\)

\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)

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

\(\Leftrightarrow2x+1=0\)

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

\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)

\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

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

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

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

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

\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)

\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)

\(=25x^2+45x+15+8+10x-40x-50x^2\)

\(=-25x^2+15x+23\)

\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)

\(=\left(x-y\right)^3-\left(x-y\right)^3+\left(x+y\right)^3-3x^2y-3xy^2\)

\(=\left(x+y\right)^3-3x^2y-3xy^2\)

\(=x^3+3x^2y+3xy^2+y^3-3xy^2-3x^2y\)

\(=x^3+y^3\)

\(2x^2-x+6x-3=0\)

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

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

....

a) \(5x\left(x-3\right)+x-3=0\)

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

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

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

câu b) bạn xem kĩ lại đề đi

sorry nha mik viết nhầm

b)x⁴-4x³+4x²-9=0