`a)5x(x-1)-(x+2)(5x-7)=6`

`<=>5x^2-5x-(5x^2-7x+10x-14)=6`

`<=>5x^2-5x-(5x^2+3x-14)=6`

`<=>-8x+14=6`

`<=>8x=8<=>x=1`

Vậy `x=1`

`b)(x+2)^2-(x^2-4)=0`

`<=>x^2+4x+4-x^2+4=0`

`<=>4x+8=0`

`<=>4x=-8`

`<=>x=-2`

Vậy `x=-2`

a)x=5/2

b)x=-2

a) \(\Rightarrow\left(x-1\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b) \(\Rightarrow25x^2+10x+1-25x^2+9=30\)

\(\Rightarrow10x=20\Rightarrow x=2\)

a. x² - 2x + 1 = 25

<=> x² - 2x - 24 = 0

<=> x² - 6x + 4x - 24 = 0

<=> x(x - 6) + 4(x - 6) = 0

<=> (x + 4)(x - 6) = 0

<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)

b. (5x + 1)² - (5x - 3)(5x + 3) = 30

<=> 25x² + 10x + 1 - 25x² + 9 = 30

<=> 25x² - 25x² + 10x = 30 - 1 - 9

<=> 10x = 20

<=> x = 2

Lời giải:

a. $x^2-100x=0$

$\Leftrightarrow x(x-100)=0$

$\Rightarrow x=0$ hoặc $x-100=0$

$\Leftrightarrow x=0$ hoặc $x=100$

b.

$x^2+5x+6=0$

$\Leftrightarrow (x^2+2x)+(3x+6)=0$

$\Leftrightarrow x(x+2)+3(x+2)=0$

$\Leftrightarrow (x+2)(x+3)=0$

$\Leftrightarrow x+2=0$ hoặc $x+3=0$

$\Leftrightarrow x=-2$ hoặc $x=-3$

a) x = 0 :))

a) Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-1\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)

\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

b,\(< =>25x^2+10x+1-25x^2+9-30=0\)

\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)

c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)

\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)

\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)

\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)

a: Ta có: \(x^2-2x+1=25\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)

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

\(x^2+6x+9-x^2+9=0\)

\(6x+18=0\)

\(6x=-18\)

\(x=-3\)

Vậy x=-3

\(b,5x^3+20x=0\)

\(5x\left(x^2+4\right)=0\)

\(Th1:5x=0=>x=0\)

\(Th2:x^2+4=0\)

\(x^2=-4\)(vô lý)

Vậy x=0

\(X^2=49\\ Mà:7^2=49;\left(-7\right)^2=49\\ \Rightarrow X=7.hoặc.x=-7\\ ----\\ b,\left(5x+1\right)^2=121=11^2=\left(-11\right)^2\\ Nên:5x+1=11.hoặc.5x+1=-11\\ Nên:5x=10.hoặc.5x=-12\\ Vậy:x=2.hoặc.x=-\dfrac{12}{5}\\ ---\\ 3x+36=-7x-64\\ \Rightarrow3x+7x=-64-36\\ \Rightarrow10x=-100\\ \Rightarrow x=-\dfrac{100}{10}=-10\\ ---\\ -5x-1178=14x+145\\ \Rightarrow14x+5x=-1178-145\\ \Rightarrow19x=-1323\\ \Rightarrow x=\dfrac{-1323}{19}\)

b: \(5x^2+3x-2-4x^2+x+5=0\)

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

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

a) \(\Rightarrow x^8-2x^4-8=0\Rightarrow\left(x^4-4\right)\left(x^4+2\right)=0\)

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

\(\Rightarrow x^2=2\Rightarrow x=\pm\sqrt{2}\)(do \(x^2+2\ge2>0,x^4+2\ge2>0\))

b) \(\Rightarrow x^2+4x+3=0\Rightarrow\left(x+1\right)\left(x+3\right)=0\)

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