Tìm x:
a, x – 2364 + 1256 = 38678
b, (x- 4639) × 2 = 2368
c, (x∶ 2) – 23684 = 1286
d, (x – 1689)∶ 2 = 165089
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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
\(a,\Leftrightarrow x^2-6x+9-x^2+4=6\\ \Leftrightarrow-6x=-7\Leftrightarrow x=\dfrac{7}{6}\\ b,\Leftrightarrow x\left(x-12\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=12\end{matrix}\right.\)
\(a,\Leftrightarrow x^2+10x-25=0\)
( Không biết có nhầm đề không ;-; )
\(b,\Leftrightarrow\left(\left(x+2\right)+2\right)^2=0\)
\(\Leftrightarrow\left(x+4\right)^2=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(a,x^2+10x=25< =>x^2+10x-25=0\)
\(< =>x^2+10x+25-50=0\)
\(< =>\left(x+5\right)^2-\left(\sqrt{50}\right)^2=0\)
\(< =>\left(x+5+\sqrt{50}\right)\left(x+5-\sqrt{50}\right)=0\)
\(=>\left[{}\begin{matrix}x=\sqrt{50}-5\\x=-\sqrt{50}-5\end{matrix}\right.\)
b, \(\left(x+2\right)^2+4\left(x+2\right)+4=0\)
\(< =>x^2+4x+4+4x+8+4=0\)
\(< =>x^2+8x+16=0\)
\(< =>\left(x+4\right)^2=0< =>x=-4\)
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
\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)
\(\Rightarrow2x=-4\Rightarrow x=-2\)
b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)
\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)
a. \(x^2-2x+2\left|x-1\right|-7=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2\left(x-1\right)-7=0\\x^2-2x-2\left(x-1\right)-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\\x^2-4x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\\left(x-5\right)\left(x+1\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm3\\x=5\\x=-1\end{matrix}\right.\)
b: Ta có: \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\cdot\left(x^2+5x\right)=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
a. `4x^2-20x+25=0`
`<=>(2x)^2-2.2x.5 +5^2=0`
`<=>(2x-5)^2=0`
`<=>2x-5=0`
`<=>x=5/2`
b. `(x-5)(x+5)-(x-3)^2=2(x-7)`
`<=>x^2-25-x^2+6x-9=2x-14`
`<=>6x-34=2x-14`
`<=>4x=20`
`<=>x=5`
\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)
\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)
Tìm x: