\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Ta có: x=2 

nên x-1=1

Ta có: \(B=\left(x+1\right)\left(x^7-x^6+x^5-x^4+x^3-x^2+x-1\right)\)

\(=\left(x+1\right)\left[x^6\left(x-1\right)+x^4\left(x-1\right)+x^2\left(x-1\right)+\left(x-1\right)\right]\)

\(=\left(x+1\right)\left(x^6+x^4+x^2+1\right)\)

\(=\left(x+1\right)\left(x+1\right)\left(x^4+1\right)\)

\(=\left(2^4+1\right)\left(2+1\right)^2=17\cdot9=153\)

a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

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

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)

Tại sao ý b có dấu - trước ngoặc đâu mà đổi dấu mong bn giải đáp

\(pt\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\\ \Leftrightarrow6x+20=0\Leftrightarrow x=-\dfrac{20}{6}=\dfrac{-10}{3}\)

Vậy ........

\(pt\text{⇔}x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\text{⇔}6x+20=0\text{⇔}x=-\dfrac{10}{3}\)

a. (3x - 1)² - (x + 3)² = 0

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

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

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

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

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

\(a,=\left(x+3\right)^3=\left(7+3\right)^3=10^3=1000\\ b,=\left(4-x\right)^3=\left(4-24\right)^3=\left(-20\right)^3=-8000\\ c,=\left(x-1\right)^3=\left(11-1\right)^3=10^3=1000\)