tìm x , biết:
(x-2)2 - 4x + 8 =0
Bài 1: Tìm x biết a) x^3 - 4x^2 - x + 4= 0 b) x^3 - 3x^2 + 3x + 1=0 c) x^3 + 3x^2 - 4x - 12=0 d) (x-2)^2 - 4x +8 =0
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
Tìm x biết: (x^3-2x^2-4x+8)/(x-2)=0
Tìm x biết: a) (x-3)^{2}-(x+2)(x-2)=-5 b) x^{3}-2x^{2}-4x+8=0
a) $(x-3)^2-(x+2)(x-2)=-5$
$\Rightarrow x^2-2\cdot x\cdot3+3^2-(x^2-2^2)=-5$
$\Rightarrow x^2-6x+9-(x^2-4)=-5$
$\Rightarrow x^2-6x+9-x^2+4=-5$
$\Rightarrow-6x+13=-5$
$\Rightarrow-6x=-18$
$\Rightarrow x=3$
b) $x^3-2x^2-4x+8=0$
$\Rightarrow(x^3-2x^2)-(4x-8)=0$
$\Rightarrow x^2(x-2)-4(x-2)=0$
$\Rightarrow (x^2-4)(x-2)=0$
$\Rightarrow (x^2-2^2)(x-2)=0$
$\Rightarrow (x-2)(x+2)(x-2)=0$
$\Rightarrow (x-2)^2(x+2)=0$
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
$\text{#}Toru$
tìm x<0 biết |x-2|+3|2-x|+|4x-8|=32 vậy x=...
tìm x biết
1) 8x2+30x+7=0
2) (x2-4x)2-8(x2-4x)+15=0
a)8x2+30x+7=0
=>8x2+28x+2x+7=0
=>(8x2+2x)+(28x+7)=0
=>2x(4x+1)+7(4x+1)=0
=>(2x+7)(4x+1)=0
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)
b)(x2-4x)2-8(x2-4x)+15=0
=>x4-8x3+8x2+32x+15=0
=>(x-5)(x+1)(x2-4x-3)=0
\(\Rightarrow\hept{\begin{cases}x=5\\x=-1\\x=2-\sqrt{7};x=\sqrt{7}+2\end{cases}}\)
Tìm x < 0 biết :
/ x - 2 / + 3 / 2 - x / + / 4x - 8 / = 32
Tìm x biết a) 3√x - 7 = 0 b) √x-2 + √4x-8=3
\(a,3\sqrt{x}-7=0\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{x}=7\\ \Leftrightarrow\sqrt{x}=\dfrac{7}{3}\\ \Leftrightarrow x=\dfrac{49}{9}\left(tmdk\right)\)
Vậy \(S=\left\{\dfrac{49}{9}\right\}\)
\(b,\sqrt{x-2}+\sqrt{4x-8}=3\left(dk:x\ge2\right)\\ \Leftrightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}=3\\ \Leftrightarrow\sqrt{x-2}+2\sqrt{x-2}=3\\ \Leftrightarrow3\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\\ \Leftrightarrow x=3\left(tmdk\right)\)
Vậy \(S=\left\{3\right\}\)
a: =>3*căn x=7
=>căn x=7/3
=>x=49/9
b: =>3*căn x-2=3
=>căn x-2=1
=>x-2=1
=>x=3
`a, 3 sqrt x - 7 = 0`
`<=> 3 sqrt x = 7`
`<=> sqrt x = 7/3`
`<=> x = 49/9`.
Vậy `x = 49/9`
`b, sqrt(x-2) + sqrt(4x-8) = 3`
`<=> sqrt(x-2) + 2 sqrt(x-2) = 3`
`<=> 3 sqrt(x-2) = 3`
`<=> sqrt(x-2)=1`
`<=> x-2=1`
`<=> x = 3`
Vậy `x = 3`
Tìm X < 0 biết [x-2] + 3[2 – x] + [4x -8] = 32 . trả lời X =
tìm x biết :
4x(x+1) = 8(x+1)
x(2x+1) +\(\dfrac{1}{3}-\dfrac{2}{3}x=0\)
x(x-4) +(x-4)2 =0
3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
4x.(x+1)-8(x+1)=0
(4x-8)(x+1)=0
suy ra x=2 hoặc x=-1
1) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)