Bài 1: Tìm x, biết:
6) x\(^8\) - x\(^6\) = 0
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Bài 1
a) (x + 3)(x + 2) = 0
x + 3 = 0 hoặc x + 2 = 0
*) x + 3 = 0
x = 0 - 3
x = -3 (nhận)
*) x + 2 = 0
x = 0 - 2
x = -2 (nhận)
Vậy x = -3; x = -2
b) (7 - x)³ = -8
(7 - x)³ = (-2)³
7 - x = -2
x = 7 + 2
x = 9 (nhận)
Vậy x = 9
1/ `|x|=10<=> x=\pm 10`
2/ `|x-8|=0<=>x-8=0<=>x=8`
3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`
4/ `|x+1|=3`
$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$
5/ `15-x=16-(14-42)`
`<=>15-x=16+28`
`<=>15-x=44`
`<=>x=-29`
6/ `210-(x-12)=168`
`<=>210-x+12=168`
`<=>222-x=168`
`<=>x=54`
\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)
a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)
c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)
1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2
Bài 1:
a) x + 1919 - 3535 = 3636 b) 3434 - x + 611611 = 5656
x + 1919 = 3636 + 3535 3434 - x = 5656 - 611611
x + 1919 = 11101110 3434 - x =
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
x8 - x6 = 0
x( 8 - 6 ) = 0
x2 = 0
x2 = 02
\(\Rightarrow\)x = 0
Vậy x = 0