Tìm x
a) 4(x + 1)2 + (2x + 1)2 - 8(x – 1)(x + 1) - 11=0
b)(x + 3)2 – (x – 4)(x + 8) – 1 = 0
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AL
4
TK
1
KV
16 tháng 10 2023
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
DT
1
28 tháng 12 2021
\(a,\Leftrightarrow x^2-2x-x^2+1=0\\ \Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(2x-1-x-4\right)\left(2x-1+x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(3x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
KG
0
NH
2
29 tháng 7 2021
`a)4x(x-2)+x-2=0`
`<=>(x-2)(4x+1)=0`
`<=>[(x-2=0),(4x+1=0):}`
`<=>[(x=2),(x=-1/4):}`
Vậy `S={2;-1/4}.`
`b)(3x-1)^3-9=0`
`<=>(3x-1-3)(3x-1+3)=0`
`<=>(3x-4)(3x+2)=0`
`<=>[(3x-4=0),(3x+2=0):}`
`<=>[(x=4/3),(x=-2/3):}`
Vậy `S={4/3;-2/3}.`
`c)x^3-8+(x-2)(x+1)=0`
`<=>(x-2)(x^2+2x+4)+(x-2)(x+1)=0`
`<=>(x-2)(x^2+3x+5)=0`
Mà `x^2+3x+5=(x+3/2)^2+11/4>=11/4>0`
`<=>x-2=0`
`<=>x=2`
Vậy `S={2}`
29 tháng 7 2021
a) Ta có: \(4x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b)Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
NH
2
6 tháng 8 2021
a, \(4x\left(x-2\right)+x-2=0\Leftrightarrow\left(4x+1\right)\left(x-2\right)=0\Leftrightarrow x=-\dfrac{1}{4};x=2\)
b, \(\left(3x-1\right)^2-9=0\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\Leftrightarrow x=\dfrac{4}{3};x=-\dfrac{2}{3}\)
c, \(x^3-8+\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\ne0\right)=0\Leftrightarrow x=2\)
6 tháng 8 2021
a) Ta có: \(4x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b) Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
16 tháng 11 2021
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
VT
1
14 tháng 6 2021
a) \(\left(x+2\right)\left(x^2-2x+4\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4+x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{23}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(N\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{-2\right\}\)
b) \(9x^2-4-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(3x-2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2-3x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\cdot4=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(S=\left\{\dfrac{2}{3}\right\}\)
a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)
\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)
\(\Leftrightarrow12x=-2\)
hay \(x=-\dfrac{1}{6}\)
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)
\(\Leftrightarrow2x=-40\)
hay x=-20