Tìm x
a)x^3(x-3)<0
b)120-4(1-x)=106-3x
tìm X
a) 2/3 + 1/3 :3 x X= 20%
b)3-X/5-X = 6/11
Lời giải:
a.
$\frac{2}{3}+\frac{1}{3}:3\times x=20\text{%}$
$\frac{2}{3}+\frac{1}{9}\times x=\frac{1}{5}$
$\frac{1}{9}\times x=\frac{1}{5}-\frac{2}{3}=\frac{-7}{15}$
$x=\frac{-7}{15}: \frac{1}{9}=\frac{-21}{5}$
b.
$\frac{3-x}{5-x}=\frac{6}{11}$
$\Rightarrow 6(5-x)=11(3-x)$
$\Rightarrow 30-6x=33-11x$
$\Rightarrow 5x=3$
$\Rightarrow x=\frac{3}{5}$
TÌM X
a) /x-17/=2,3
b) /x+3/4/=0
c)/x+3/4/+1/3=0
a) \(\left|x-17\right|=2,3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-17=2,3\\x-17=-2,3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=19,3\\x=14,7\end{matrix}\right.\)
b) \(\left|x+\dfrac{3}{4}\right|=0\)
\(\Leftrightarrow x+\dfrac{3}{4}=0\Leftrightarrow x=-\dfrac{3}{4}\)
c) \(\left|x+\dfrac{3}{4}\right|+\dfrac{1}{3}=0\)
\(\Leftrightarrow\left|x+\dfrac{3}{4}\right|=-\dfrac{1}{3}\)( vô lý do \(\left|x+\dfrac{3}{4}\right|\ge0\forall x\))
Vậy \(S=\varnothing\)
Tìm x
a) (x-3)2-2.(x-3)=0
b) x:0,25+x:0,2+x:0,1+x=34
a) Ta có: \(\left(x-3\right)^2-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
b) Ta có: \(x:0.25+x:0.2+x:0.1+x=34\)
\(\Leftrightarrow4x+5x+x+x=34\)
\(\Leftrightarrow11x=34\)
hay \(x=\dfrac{34}{11}\)
bài 3:tìm x
a)x(x+4)+(5+x)(5-x)=3
b)(2x-3)^2=(4x+3)^2
a) \(\Rightarrow x^2+4x+25-x^2=3\Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\)
b) \(\Rightarrow\left(4x+3-2x+3\right)\left(4x+3+2x-3\right)=0\)
\(\Rightarrow2\left(x+3\right).6x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
tìm x
a) x mũ 2 -9 +(x+3 )2 =0
b) x (x -3 ) -x mũ 2 +x =8
Tìm x
a.(x-2)(x2+2x+4)-x(x-3)(x+3)=26
b.(x-3)(x2+3x+9)-x(x+4)(x-4)=21
`a)(x-2)(x^2+2x+4)-x(x-3)(x+3)=26`
`<=>x^3-2^3-x(x^2-9)=26`
`<=>x^3-8-x^3+9x=26`
`<=>9x-8=26`
`<=>9x=34`
`<=>x=34/9`
`b)(x-3)(x^2+3x+9)-x(x+4)(x-4)=21`
`<=>x^3-3^3-x(x^2-16)=21`
`<=>x^3-27-x^3+16x=21`
`<=>16x-27=21`
`<=>16x=48`
`<=>x=3`
bài 3: tìm x
a)10+3(x-1)=10+6x
b) (x+1) (x-2)=0
a: =>10+3x-3=6x+10
=>3x-3=6x
=>-3x=3
=>x=-1
b: =>x+1=0 hoặc x-2=0
=>x=-1 hoặc x=2
a) \(10+3\left(x-1\right)=10+6x\)
\(\Rightarrow10+3x-3=10+6x\)
\(\Rightarrow3x-6x=10-10+3\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-\dfrac{3}{3}\)
\(\Rightarrow x=-1\)
b) \(\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Tìm x
a, (x-2)^2-3x^2+6x=0
b, x^3+27+(x+3)(x-9)=0
\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
Tìm x
a) (x-1)^10 + (x-2)^10 =1
b) (x+3)^100 + (x+4)^100 = 1/3
Tìm x
a) (x-1)^10 + (x-2)^10 =1
b) (x+3)^100 + (x+4)^100 = 1/3