Tìm x:a, 172 x2 - 79: 983 = 2-3
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\(a,\Leftrightarrow x^2-x-x^2+4x-4=2\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(6-x\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3-6+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\\ c,\Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
a: Ta có: \(2x^3-18x=0\)
\(\Leftrightarrow2x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b: Ta có: \(\left(3x-2\right)\left(2x+1\right)-6x\left(x+2\right)=11\)
\(\Leftrightarrow6x^2+3x-4x-2-6x^2-12x=11\)
\(\Leftrightarrow-13x=13\)
hay x=-1
c: Ta có: \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8=3-3x^2\)
\(\Leftrightarrow3x=12\)
hay x=4
a) 2x3-18x=0
⇔ 2x(x2-9)=0
⇔ 2x(x-3)(x+3)=0
⇔ \(\left\{{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b)(3x-1)(2x+1)-6x(x+2)=11
⇔ 6x2+x-1-6x2-12x=11
⇔ -11x=12
\(\Leftrightarrow x=-\dfrac{12}{11}\)
c) (x-1)3-(x+2).(x2-2x+4)=3.(1-x2)
⇔ x3-3x2+3x-1-x3-8-3+3x2=0
⇔ 3x=12
⇔ x=4
a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow x^2+7x-x^2-x+6=0\)
hay x=-1
b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow x+2=0\)
hay x=-2
b. (x + 2)2 - x2 + 4 = 0
<=> (x + 2 - x)(x + 2 + x) + 4 = 0
<=> 2(2 + 2x) + 4 = 0
<=> 4(1 + x) + 4 = 0
<=> 4(1 + x) = -4
<=> 1 + x = -1
<=> x = -1 - 1
<=> x = -2
a) (5 + x)(x - 5) - x(x + 5) = 10
x² - 25 - x² - 5x = 10
-5x = 10 + 25
-5x = 35
x = 35 : (-5)
x = -7
b) x.(2x + 3) - 2(x² + x) = 2
2x² + 3x - 2x² - 2x = 2
x = 2
a: \(\left(x+5\right)\left(x-5\right)-x\left(x+5\right)=10\)
=>\(x^2-25-x^2-5x=10\)
=>-5x-25=10
=>-5x=35
=>x=-7
b: \(x\left(2x+3\right)-2\left(x^2+x\right)=2\)
=>\(2x^2+3x-2x^2-2x=2\)
=>x=2
a: \(x^2-4x=3\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
b: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
Bài 2:
a: Ta có: \(2x+79=x+45\)
nên 2x-x=45-79
hay x=-24
b: Ta có: \(15-\left(x+7\right)=\left(x-8\right)+22\)
\(\Leftrightarrow8-x-x-14=0\)
\(\Leftrightarrow2x=22\)
hay x=11
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1+1=2 2 nhân 2 bằng 4