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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}
1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0
\(\Rightarrow\)(x3+64)-x(x2+2x+x+2)+3x2=0
\(\Rightarrow\)x3+64-x3-2x2-x2-2x+3x2=0
\(\Rightarrow\)-2x+64=0
\(\Rightarrow\)-2x=-64
\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50
\(\Rightarrow\)8x-24x2+2-6x+24x2-60x-4x+10=50
\(\Rightarrow\)-62x+12=50
\(\Rightarrow\)-62x=50-12
\(\Rightarrow\)-62x=38
\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
`Answer:`
Bài 1:
a) \(7+2x=22-3x\)
\(\Leftrightarrow2x+3x=22-7\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=3\)
b) \(8x-3=5x+12\)
\(\Leftrightarrow8x-5x=12+3\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
c) \(x-12+4x=25+2x-1\)
\(\Leftrightarrow x-12+4x-25-2x+1=0\)
\(\Leftrightarrow\left(x+4x-2x\right)+\left(1-12-25\right)=0\)
\(\Leftrightarrow3x-36=0\)
\(\Leftrightarrow x=12\)
d) \(x+2x+3x-19=3x+5\)
\(\Leftrightarrow6x-19=3x+5\)
\(\Leftrightarrow6x-3x=5+19\)
\(\Leftrightarrow3x=24\)
\(\Leftrightarrow x=8\)
Bài 2:
a) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}}\)
b) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow2x+7=0\text{ hoặc }x-5=0\text{ hoặc }5x+1=0\)
\(\Leftrightarrow x=-\frac{7}{2}\text{ hoặc }x=5\text{ hoặc }x=-\frac{1}{5}\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x^2=-1\text{(Loại)}\end{cases}}}\)
d) \(\left(x^2-4\right)+\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow x^2-4+\left(3x-2x^2-6+4x\right)=0\)
\(\Leftrightarrow x^2-4=\left(-2x^2+7x-6\right)=0\)
\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)
\(\Leftrightarrow-x^2+7x-10=0\)
\(\Leftrightarrow x^2-5x-2x+10=0\)
\(\Leftrightarrow x.\left(x-5\right)-2.\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right).\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}}\)
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
1, \(27x^2\left(x+3\right)-12x\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(27x^2-12x\right)=0\)
\(\Leftrightarrow3x\left(x+3\right)\left(9x-4\right)=0\Leftrightarrow x=0;x=-3;x=\dfrac{4}{9}\)
2, \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-12\right)=0\Leftrightarrow x=\dfrac{1}{4}\)
3, \(\left(2x-1\right)^2=49\Leftrightarrow\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
4, \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\Leftrightarrow\left(5x-3\right)^2=\left(4x-7\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=4x-7\\5x-3=7-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{9}\end{matrix}\right.\)
1) \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
\(\Leftrightarrow3x\left(x+3\right)\left(9x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\9x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{4}{9}\end{matrix}\right.\)
Vậy: ...
2) \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x\right)^2-2\cdot4x\cdot1+1^2=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-12\right)=0\)
\(\Leftrightarrow-13\left(4x-1\right)=0\)
\(\Leftrightarrow4x-1=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
Vậy: ...