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

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

\(a,\Leftrightarrow x^8-2x^4-8=0\\ \Leftrightarrow x^8+2x^4-4x^4-8=0\\ \Leftrightarrow\left(x^4+2\right)\left(x^4-4\right)=0\\ \Leftrightarrow\left(x^4+2\right)\left(x^2-2\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\left(x^2+2>0;x^4+2>0\right)\\ b,\Leftrightarrow x^2+4x+3=0\\ \Leftrightarrow\left(x+1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 40x = 15 <=> x = 3/8

a) <=> (4x - 4x + 5)(4x + 4x - 5) = 15 <=> 5(8x-5) = 15

<=> 40x = 40 <=> x = 1

Cái này mới chuẩn

b) (2x+1)(1-2x)+(1-2x)²=18 <=> 1 - 4x² + 4x² - 4x + 1 = 18

<=> -4x = 16 <=> x = -4

a: ta có: \(\left(x+3\right)\left(x-1\right)-x\left(x-5\right)=11\)

\(\Leftrightarrow x^2+2x-3-x^2+5x=11\)

\(\Leftrightarrow x=2\)

b: Ta có: \(\left(x+4\right)\left(x^2-4x+16\right)-x\left(x+1\right)\left(x+2\right)+3x^2=0\)

\(\Leftrightarrow x^3+64-x^3-3x^2-2x+3x^2=0\)

\(\Leftrightarrow2x=64\)

hay x=32

a: Ta có: \(\left(8x^2-4x\right):\left(-4x\right)-\left(x+2\right)=8\)

\(\Leftrightarrow-2x+1-x-2=8\)

\(\Leftrightarrow-3x=9\)

hay x=-3

b: Ta có: \(\left(2x^4-3x^3+x^2\right):\left(-\dfrac{1}{2}x^2\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow-4x^2+6x-2+4x^2-8x+4=0\)

\(\Leftrightarrow-2x=-2\)

hay x=1

a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

=>\(13\sqrt{2x}=28\)

=>căn 2x=28/13

=>2x=784/169

=>x=392/169

b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>2*căn x-5=4

=>căn x-5=2

=>x-5=4

=>x=9

c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=-1 hoặc x=2

a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)

c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)

   \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)

a)\(\Leftrightarrow\)\(x^2-4x-21=0\)

\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)

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

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

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

b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)

\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)

\(\Leftrightarrow\)\((-2x+1)(-2x+9)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)

c)\((x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15\)

\(\Leftrightarrow\)\(x^3-9x^2+27x-27-x^3+27+9x^2+18x+9-15=0\)

\(\Leftrightarrow\)\(45x-6=0\)

\(\Leftrightarrow\)\(x=\dfrac{2}{15}\)

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

\(2\left(x+3\right)+x\left(3+x\right)=0\)

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

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

<=> (x+3)(x+2)=0

TH1 x+3=0 <=> x=-3

TH2 x+2=0 <=> x=-2

Vậy.... 

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) Ta có: \(6x\left(x-5\right)+3x\left(7-2x\right)=18\)

\(\Leftrightarrow6x^2-30x+21x-6x^2=18\)

\(\Leftrightarrow-9x=18\)

hay x=-2

Vậy: S={-2}

b) Ta có: \(2x\left(3x+1\right)+\left(4-2x\right)\cdot3x=7\)

\(\Leftrightarrow6x^2+2x+12x-6x^2=7\)

\(\Leftrightarrow14x=7\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

c) Ta có: \(0.5x\left(0.4-4x\right)+\left(2x+5\right)\cdot x=-6.5\)

\(\Leftrightarrow0.2x-2x^2+2x^2+5x=-6.5\)

\(\Leftrightarrow5.2x=-6.5\)

hay \(x=-\dfrac{5}{4}\)

Vậy: \(S=\left\{-\dfrac{5}{4}\right\}\)

d) Ta có: \(\left(x+3\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

Vậy: S={-5}

e) Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

Vậy: S={0}

a,(2x-5^2)-4x(x-3)=0

=> 2x-25-4x²+12x=0

=>-4x²+14x-25=0

đề bài ý a sai nha

b, 6x²-7x=0

=>x(6x-7)=0

=>x=0 và 6x-7=0

=>x=0 và x=7/6

vậy x=0 và x=7/6