Tìm x biết 8x(x-3)-8(x-1)(x+1)=20
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a, \(x\) + 99: 3 = 55
\(x\) + 33 = 55
\(x\) = 55 - 33
\(x\) = 22
b, (\(x\) - 25):15 = 20
\(x\) - 25 = 20 x 15
\(x\) - 25 = 300
\(x\) = 300 + 25
\(x\) = 325
c, (3\(x\) - 15).7 = 42
3\(x\) - 15 = 42:7
3\(x\) - 15 = 6
3\(x\) = 6 + 15
3\(x\) = 21
\(x\) = 21: 3
\(x\) = 7
8x(x - 3) - 8(x - 1)(x + 1) = 20
=> 8x2 - 24x - 8(x2 - 1) - 20 = 0
=> 8x2 - 24x - 8x2 + 8 - 20 = 0
=> -24x = -12
=> x = 1/2
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8-20=0\Leftrightarrow-24x-12=0\Leftrightarrow-24x=12\Leftrightarrow x=\frac{12}{-24}=\frac{-1}{2}\)
Ta có: 8x.(x - 3) - 8.(x - 1)(x + 1) = 20
=> 8x2 - 24x - 8.(x2 - 1) = 20
=> 8x2 - 24x - 8x2 + 8 - 20 = 0
=> -24x - 12 = 0
=> -24x = 12
=> x = -1/2
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
Áp dụng hằng đẳng thức : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(pt\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow-24x+8=20\Leftrightarrow-24x=12\Leftrightarrow x=\frac{12}{-24}=-\frac{1}{2}\)
Vậy x=-1/2
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
a)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
Lời giải:
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+5\sqrt{x}=-20$
$\Leftrightarrow 5\sqrt{x}-8\sqrt{2x}=-20$
$\Leftrightarrow \sqrt{x}(5-8\sqrt{2})=-20$
$\Leftrightarrow \sqrt{x}=\frac{20}{8\sqrt{2}-5}$
$\Rightarrow x=(\frac{20}{8\sqrt{2}-5})^2$
b. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{5x}-5\sqrt{3x}+4\sqrt{x}=10$
$\Leftrightarrow \sqrt{x}(3\sqrt{5}-5\sqrt{3}+4)=10$
$\Leftrightarrow \sqrt{x}=\frac{10}{3\sqrt{5}-5\sqrt{3}+4}$
$\Rightarrow x=(\frac{10}{3\sqrt{5}-5\sqrt{3}+4})^2$
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow\left(8x^2-8x^2\right)+\left(-24x+8\right)=20\)
\(\Leftrightarrow-24x=20-8\)
\(\Leftrightarrow-24x=12\)
\(\Leftrightarrow x=12:\left(-24\right)\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy: \(x=-\frac{1}{2}\)
\(8x\left(x-2\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-\left(8x^2-8\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow-24x+8=20\)
\(\Leftrightarrow-24x=20-8\)
\(\Leftrightarrow-24x=12\)
\(\Leftrightarrow x=-\frac{1}{2}\)