Tìm x.
(8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0
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PB
1
8 tháng 7 2018
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
Vậy S = { 0, 6}
5 tháng 11 2023
2: \(3x\left(x-4\right)+2x-8=0\)
=>\(3x\left(x-4\right)+2\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
3: 4x(x-3)+x2-9=0
=>\(4x\left(x-3\right)+\left(x+3\right)\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(4x+x+3\right)=0\)
=>\(\left(x-3\right)\left(5x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{5}\end{matrix}\right.\)
4: \(x\left(x-1\right)-x^2+3x=0\)
=>\(x^2-x-x^2+3x=0\)
=>2x=0
=>x=0
5: \(x\left(2x-1\right)-2x^2+5x=16\)
=>\(2x^2-x-2x^2+5x=16\)
=>4x=16
=>x=4
SX
1
19 tháng 8 2015
(8 - 5x) (x + 2) + 4(x - 2) (x + 1) + 2(x - 2) (x + 2) = 0
=> (x + 2) [ (8 - 5x) + 4(x + 1) + 2(x - 2)] = 0
=> (x + 2) (8 - 5x + 4x + 4 + 2x - 4) = 0
=> (x + 2) (x + 8) = 0
=> x + 2 = 0 hoặc x + 8 = 0
=> x = -2 hoặc x = -8
DQ
0
VT
3
3 tháng 7 2019
a) (x+2)(x+3)-(x-2)(x+5)=0
\(x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(2x+16=0\)
\(2x=-16\)
\(x=-8\)
Vậy......
b) (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0
\(8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2+4x-4x-8=0\)
\(-6x+x^2=0\)
\(x\left(-6+x\right)=0\)
=> x=0 hoặc -6+x=0 <=>x=6
Vậy \(x\in\left\{0;6\right\}\)
3 tháng 7 2019
a) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)x+\left(x+2\right).3-\left(x+5\right)x+\left(x+5\right).2=0\)
\(\Leftrightarrow x^2+2x+3x+6-x^2+5x+2x+10=0\)
\(\Leftrightarrow12x+16=0\)
\(\Leftrightarrow12x=-16\)
\(\Leftrightarrow x=\frac{-4}{3}\)
Vậy...
VT
2
3 tháng 7 2019
\(a,\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(x^2+5x+6-x^2-3x+10=0\)
\(2x+16=0\)
\(2x=-16\)
\(x=-8\)
\(b,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(8x+16-5x^2-10x+4x^2-4x-8+2x^2-8=0\)
\(x^2-6x=0\)
\(x\left(x-6\right)=0\)
\(\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)
3 tháng 7 2019
\(a,\)\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow x^2+5x+6-x^2-3x+10=0\)
\(\Rightarrow2x=-16\Leftrightarrow x=-8\)
\(b,\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow8x+16-5x^2-10x+4\left(x^2-x+2\right)+2\left(x^2-4\right)=0\)
\(\Rightarrow8x+16x-5x^2-10x+4x^2-4x+8+2x^2-8=0\)
\(\Rightarrow x^2+10x=0\Rightarrow x\left(x+10\right)=0\Rightarrow x\in\left\{0;-10\right\}\)
QT
0
SX
0
(8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
=> 8(x + 2) - 5x(x + 2) + 4[x(x + 1) - 2(x + 1)] + 2(x2 - 4) = 0
=> 8x + 16 - 5x2 - 10x + 4(x2 + x - 2x - 2) + 2x2 - 8 = 0
=> 8x + 16 - 5x2 - 10x + 4x2 + 4x - 8x - 8 + 2x2 - 8 = 0
=> (8x - 10x + 4x - 8x) + (16 - 8 - 8) + (-5x2 + 4x2 + 2x2) = 0
=> 0 + x2 = 0
=> x2 = 0 => x = 0
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(-5x^2-2x+16+4\left(x^2-x-2\right)+2\left(x^2-4\right)=0\)
\(-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(x^2-6x=0\)
\(x\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)