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\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(3\left(x-1\right)^2\cdot3x\left(x-5\right)=0\)
\(\Rightarrow9x\left(x-1\right)^2\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=5\end{matrix}\right.\)
b) \(\left(x+3\right)^2-5x-15=0\)
\(\Rightarrow\left(x+3\right)^2-5\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x+3-5\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
c) \(2x^5-4x^3+2x=0\)
\(\Rightarrow2x\left(x^4-2x^2+1\right)=0\)
\(\Rightarrow2x\left[\left(x^2\right)^2-2\cdot x^2\cdot1+1^2\right]=0\)
\(\Rightarrow2x\left(x^2-1\right)^2=0\)
\(\Rightarrow2x\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
\(\text{#}Toru\)
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) x² - 4 = 0
x² = 4
x = 2 hoặc x = -2
b) 2x(x + 5) - 3(5 + x) = 0
(x + 5)(2x - 3) = 0
X + 5 = 0 hoặc 2x - 3 = 0
*) x + 5 = 0
x = -5
*) 2x - 3 = 0
2x = 3
x = 3/2
c) x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x² - 2x - 3x + 6) = 0
(x - 1)[(x² - 2x) - (3x - 6)] = 0
(x - 1)[x(x - 2) - 3(x - 2)] = 0
(x - 1)(x - 2)(x - 3) = 0
x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x - 3 = 0
x = 3
Vậy x = 1; x = 2; x = 3
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
a: (2x+1)(3-x)(4-2x)=0
=>(2x+1)(x-3)(x-2)=0
hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)
b: 2x(x-3)+5(x-3)=0
=>(x-3)(2x+5)=0
=>x=3 hoặc x=-5/2
c: =>(x-2)(x+2)+(x-2)(2x-3)=0
=>(x-2)(x+2+2x-3)=0
=>(x-2)(3x-1)=0
=>x=2 hoặc x=1/3
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
e: =>(2x+5+x+2)(2x+5-x-2)=0
=>(3x+7)(x+3)=0
=>x=-7/3 hoặc x=-3
f: \(\Leftrightarrow2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
a: \(x^2\left(2x-3\right)+8x-12=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x^2+4\right)=0\)
=>2x-3=0
hay x=3/2
b: \(\Leftrightarrow\left(2x-5\right)\left(2x+10\right)-\left(2x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+10-x+1\right)=0\)
=>(2x-5)(x+11)=0
=>x=5/2 hoặc x=-11
c: \(\Leftrightarrow2x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{0;4;-4\right\}\)