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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,x^2 +4x-5=0`
`<=> x^2-x+5x-5=0`
`<=> x(x-1)+5(x-1)=0`
`<=>(x-1)(x+5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
`b, x^2 -x-12=0`
`<=> x^2 +3x-4x-12=0`
`<=>(x^2+3x)-(4x+12)=0`
`<=>x(x+3)-4(x+3)=0`
`<=>(x+3)(x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
`c, (2x-7)^2 - 6(2x-7)(x-3)=0`
`<=>(2x-7)(2x-7 -6x+18)=0`
`<=>(2x-7) ( -4x+11)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\-4x+11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\-4x=-11\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{11}{4}\end{matrix}\right.\)
a: =>(x+5)(x-1)=0
=>x=1 hoặc x=-5
b: =>(x-4)(x+3)=0
=>x=4 hoặc x=-3
c: =>(2x-7)(2x-7-6x+18)=0
=>(2x-7)(-4x+11)=0
=>x=11/4 hoặc x=7/2
\(a.\left(x^2-2x+1\right)-4=0\\\Leftrightarrow \left(x-1\right)^2-2^2=0\\\Leftrightarrow \left(x-1-2\right)\left(x-1+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{3;-1\right\}\)
\(b.x^2-x=-2x+2\\\Leftrightarrow x^2-x+2x-2=0\\\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\\Leftrightarrow \left(x+2\right)\left(x-1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2;1\right\}\)
\(c.4x^2+4x+1=x^2\\ \Leftrightarrow4\left(x^2+x+\frac{1}{4}\right)-x^2=0\\ \Leftrightarrow4\left(x+\frac{1}{2}\right)^2-x^2=0\\ \Leftrightarrow\left[2\left(x+\frac{1}{2}\right)-x\right]\left[2\left(x-\frac{1}{2}\right)+x\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2\left(x+\frac{1}{2}\right)-x=0\\2\left(x+\frac{1}{2}\right)+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1-x=0\\2x+1+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-1;-\frac{1}{3}\right\}\)
1.
\(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
2.
\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
3.
\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)
Vì (x-1)2≥0
23>0
\(\Rightarrow\left(x-1\right)^2+23>0\)
Vậy x vô nghiệm
4.
\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
5.
\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
tham khảo
https://hoidapvietjack.com/q/57243/giai-cac-phuong-trinh-sau-a-2x12-2x-12-b-x2-3x-2-5x2-3x60
b) (2x+1)2-2x-1=2
\(< =>4x^2+4x+1-2x-1=2\)
\(< =>4x^2+2x-2=0\)
\(< =>4x^2+4x-2x-2=0\)
\(< =>\left(4x^2+4x\right)-\left(2x+2\right)=0\)
\(< =>4x\left(x+1\right)-2\left(x+1\right)=0\)
\(< =>\left(x+1\right)\left(4x-2\right)=0\)
\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\4x-2=0=>x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy....
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
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}