Tim x \(3\left|x-2\right|-\left|x+2\right|=0\)
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B
25 tháng 7 2019
\(1,\left(2x+1\right)\left(x-1\right)-x\left(2x-3\right)+3=0\)
\(\Rightarrow2x^2-2x+x-1-\left(2x^2-3x\right)+3=0\)
\(\Rightarrow2x^2-2x+x-1-2x^2+3x+3=0\)
\(\Rightarrow2x=-2\Rightarrow x=-1\)
\(2,\left(x^2+x-2\right)\left(x^2-x-2\right)-x^2\left(x^2-2\right)+8=0\)
\(\Rightarrow[\left(x^2\right)^2-\left(x-2\right)^2]-x^2\left(x^2-2\right)+8=0\)
\(\Rightarrow x^4-\left(x^2-4x+4\right)-x^4+2x^2+8=0\)
\(\Rightarrow x^4-x^2+4x-4-x^4+2x^2+8=0\)
\(\Rightarrow x^2+4x+4=0\)
\(\Rightarrow\left(x+2\right)^2=0\Rightarrow x=-2\)
31 tháng 7 2016
a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)
\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)
\(\Leftrightarrow40x=46+9+25=80\)
\(\Leftrightarrow x=2\)
b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)
\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)
\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)
\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)
\(\Leftrightarrow2x=-2+6-5=-1\)
\(\Leftrightarrow x=\frac{1}{2}\)
31 tháng 7 2016
x2.(x+3)+y2.(y+5)−(x+y).(x2−xy+y2)=0
<=>\(x^3+3x^2+y^3+5y^2-x^3-y^3=0\)(áp dụng hằng đẳng thức)
<=> \(3x^2+5y^2=0\)
ta thấy \(3x^2\ge0\)với mọi x
\(5y^2\ge0\) với mọi y
=> \(3x^2+5y^2\ge0\)
=> x=0 và y=0
vậy cặp số (x;y)=(0;0)
2T
20 tháng 8 2019
a) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x-3\right)\left(x+3\right)=36\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)
\(\Leftrightarrow8x+76=36\)
\(\Leftrightarrow8x=-40\)
\(\Leftrightarrow x=-5\)
Y
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
TT
0
LT
2
DH
25 tháng 7 2017
a,\(-3x+x-3=0\)
\(\Rightarrow-2x=3\Rightarrow x=-\dfrac{3}{2}\)
b,\(5x\left(x-2\right)-x+2=0\)
\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Chúc bạn học tốt!!!
Ta có \(\left|x-2\right|\ge0\) \(\Rightarrow3\left|x-2\right|\ge0\)
\(\left|x+2\right|\ge0\)
\(\Rightarrow3\left|x-2\right|-\left|x+2\right|\ge0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
Vậy x=2 hoặc x=-2