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Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
a: (3x-2)(4x+5)=0
=>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: (2,3x-6,9)(0,1x+2)=0
=>2,3x-6,9=0 hoặc 0,1x+2=0
=>x=3 hoặc x=-20
c: =>(x-3)(2x+5)=0
=>x-3=0 hoặc 2x+5=0
=>x=3 hoặc x=-5/2
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
c/ 
d/ 
)(x-
) = 0 g/ (3x-1)(2x-3)(x+5) = 0 
h)
m)
= 8 – x k) 
= – 4x +7
q) 
6:
k: =>x^2-9<x^2+2x+3
=>2x+3>-9
=>2x>-12
=>x>-6
1:
h: =>x(x-1)=0
=>x=0; x=1
i: =>x(x-3)=0
=>x=0; x=3
\(a,x^2-10x=-25\)
\(< =>x^2-10x+25=0\)
\(< =>\left(x-5\right)^2=0< =>x=5\)
b, \(4x^2-4x=-1\)
\(< =>4x^2-4x+1=0\)
\(< =>\left(2x-1\right)^2=0< =>x=\frac{1}{2}\)
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
bạn đăng tách cho mn cùng giúp nhé
Bài 1 :
a, \(\Leftrightarrow11-x=12-8x\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
b, \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow x=-2\)
c, \(\Leftrightarrow3-2x=-x-4\Leftrightarrow x=7\)
d, \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
\(\Leftrightarrow3x^2+12x-9=3x^2+3x+1\Leftrightarrow x=\dfrac{10}{9}\)
e, \(\Leftrightarrow2x^2-x-3=2x^2+9x-5\Leftrightarrow x=5\)
f, \(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)
\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
\(a,\left(2x+1\right)\left(x^2+2\right)=0\)
\(\left[{}\begin{matrix}2x=-1\\x^2=-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\pm\sqrt{2}\end{matrix}\right.\)
\(b,\left(x^2+x+1\right)\left(6-2x\right)=0\)
\(6-2x=0\Leftrightarrow2x=6\Leftrightarrow x=3\)
\(c,\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0\)
\(\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\2x=3\\3x=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=-\frac{4}{3}\end{matrix}\right.\)
\(d,\left(x^2+4\right)\left(7x-3\right)=0\)
\(\left[{}\begin{matrix}x^2+4=0\\7x-3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x^2=-4\\7x=3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\pm2\left(voli\right)\\x=\frac{3}{7}\end{matrix}\right.\)
\(e,\left(8x-4\right)=\left(x^2+x+2\right)\)
\(8x-4=x^2+x+2\)
\(8x-4-x^2-x-2=0\)
\(7x-6-x^2=0\)
\(\left(x-6\right)\left(x-1\right)=0\)
\(\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
\(f,\left(2x-1\right)\left(3x+2\right)\left(5-x\right)\)
đề thiếu hay là rút gọn vậy bn