Tìm x,biết:
\(x^3+x^2-2x-8=0\)
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29 tháng 8 2015
Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
TQ
1
T
11 tháng 12 2023
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) $x^3-2x^2-4x+8=0$
$\Rightarrow(x^3-2x^2)-(4x-8)=0$
$\Rightarrow x^2(x-2)-4(x-2)=0$
$\Rightarrow (x^2-4)(x-2)=0$
$\Rightarrow (x^2-2^2)(x-2)=0$
$\Rightarrow (x-2)(x+2)(x-2)=0$
$\Rightarrow (x-2)^2(x+2)=0$
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
$\text{#}Toru$
NT
3
23 tháng 8 2021
3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
21 tháng 7 2020
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 = 4
<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)
<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy S = { 5 ; 1 }
b) x2 - 9 = 0
<=> x2 = 9
<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
Vậy S = { 3 ; -3 }
c) x( x - 2x ) - x2 - 8 = 0
<=> x2 - 2x2 - x2 - 8 = 0
<=> -2x2 - 8 = 0
<=> -2x2 = 8
<=> x2 = -4 ( vô lí )
<=> x = \(\varnothing\)
Vậy S = { \(\varnothing\)}
21 tháng 7 2020
d) 2x( x - 1 ) - 2x2 + x - 5 = 0
<=> 2x2 - 2x - 2x2 + x - 5 = 0
<=> -x - 5 = 0
<=> -x = 5
<=> x = -5
Vậy S = { -5 }
e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0
<=> x2 - 3x - ( x2 - x - 2 ) = 0
<=> x2 - 3x - x2 + x + 2 = 0
<=> - 2x + 2 = 0
<=> -2x = -2
<=> x = 1
Vậy S = { 1 }
f) x( 3x - 1 ) - 3x2 - 7x = 0
<=> 3x2 - x - 3x2 - 7x = 0
<=> -8x = 0
<=> x = 0
Vậy S = { 0 }
8 tháng 12 2017
a) \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy \(x=1;-3\)
b) \(x^2-4x+8=2x-1\)
\(\Leftrightarrow x^2-4x+8-2x+1=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x=3
9 tháng 12 2017
a) \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy \(x=1;-3\)
b) \(x^2-4x+8=2x-1\)
\(\Leftrightarrow x^2-4x+8-2x+1=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
13 tháng 12 2023
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
LQ
1
18 tháng 7 2021
( x - 1 )( x + 2 ) - x - 2 = 0
<=> ( x - 1 )( x + 2 ) - ( x + 2 ) = 0
<=> ( x + 2 )( x - 2 ) = 0
<=> x = ±2
( 2x - 7 )3 = 8( 7 - 2x )2
<=> ( 2x - 7 )3 - 8( 2x - 7 )2 = 0
<=> ( 2x - 7 )2( 2x - 15 ) = 0
<=> x = 7/2 hoặc x = 15/2
HT
0
\(x^3+x^2-2x-8=0\)
\(\Leftrightarrow\left(x^3+3x^2+4x\right)-\left(2x^2+6x+8\right)=0\)
\(\Leftrightarrow x\left(x^2+3x+4\right)-2\left(x^2+3x+4\right)=0\)
\(\Leftrightarrow\left(x^2+3x+4\right)\left(x-2\right)=0\)(1)
Ta thấy \(x^2+3x+4\)
\(=x^2+2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+4\)
\(=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0;\forall x\)
\(\Rightarrow\left(1\right)\)xảy ra \(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy x=2
\(\Leftrightarrow\left(x^3-8\right)+\left(x^2-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\\left(x+\frac{3}{2}\right)^2+\frac{7}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x\in\varnothing\end{cases}\Rightarrow x=2.}\)
Vậy ........