Tìm x biết:
(x-3)^x+2-(x-3)^x+8=0
\(a,\left(x+2\right)^{10}+\left(x+2\right)^8=0\\ \Leftrightarrow\left(x+2\right)^8\left[\left(x+2\right)^2+1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^8=0\\\left(x+2\right)^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\\left(x+2\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\end{matrix}\right.\\ b,\left(x+3\right)^{10}-\left(x+3\right)^8=0\\ \Leftrightarrow\left(x+3\right)^8\left[\left(x+3\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)^8=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
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) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
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: 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, biết:
a) 7x(x + 1) - 3(x + 1) =0
b) 3 ( x + 8) - x^2 - 8x = 0
c) x^2 - 10x = -25
d) x^2 - 10x = -25
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
Tìm x biết
a) (x-1).(x+2).(x+3).(x+6)-28=0
b) (x^2+4x+3).(x^2+6x+8)=0
1, tìm x biết :
[ x-3 ] + [ x-2 ] + [ x-1 ] + ... + [ x+5 ] = 0
2, tìm x sao cho :
[ x-7 ] . [ x+3 ] < 0
3, cho biểu thức : A = [ 5.x^2 - 8.x^2 - 9.x^2 ] . [ 3y^3] . tìm x, y để A > hoặc = 0
cậu chia từng câu ra cho mình nhé
Tìm x, biết:
a) 2-x = 2 ( x - 2 ) 3 ; b) 8 x 3 - 72x = 0;
c) ( x - 1 , 5 ) 6 + 2 ( 1 , 5 - x ) 2 = 0; d) 2 x 3 +3 x 2 +3 + 2x = 0;
e) x 3 - 4x- 14x(x - 2) = 0; g) x 2 (x + 1)- x(x + 1) + x(x - 1) = 0.
1.Tìm số nguyên x biết
|x-3|=|x+7|
|3x+2|+|x-8|
|x+3|+|y-8|=0
|x-5|+|x-3|=0
(x-2)(x+6)=0
x(3x+6)=0
(2x-4)(x3-1)=0
(x-2)(x+7)>0
(x+3)(x+5)>0
(x+2)(x-5)<0
Tìm x biết 3x + 3^x+1= 3^8 . 2 + 2 . 3^8. 2017^0
\(3^x+3^{x+1}=3^8.2+2.3^8.2017^0\)
\(3^x+3^x.3=3^8.2^2\)
\(3^x=3^8.2^2:3\)
\(3^x=3^7.2^2\)
3x+3x+1=38.2+2.38.201703x+3x+1=38.2+2.38.20170
3x+3x.3=38.223x+3x.3=38.22
3x=38.22:33x=38.22:3
3x=37.223x=37.22