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Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
a.1+3+5+7+...+2x+1=225
[(2x+1-1):2+1]x(2x+1+1):2=225
(x+1)x(2x+2):2=225
(x+1)x(x+1)=225
(x+1)2=225
(x+1)2=152
x+1=15
x=14
_______________
b. 130-[5.(9-x)+43]=47
5.(9-x)+43=83
5.(9-x)=40
9-x=8
x=1
_______________
c.16x<324
24x<220
=>x∈{0;1;2;3;4}
a: =>(x+1)^2=225
=>x+1=15
=>x=14
b: =>[5*(9-x)+43]=130-47=83
=>5(9-x)=40
=>9-x=8
=>x=1
c: =>2^4x<2^20
=>4x<20
=>0<x<5
11: Ta có: \(\left(x+3\right)^3=125\)
\(\Leftrightarrow x+3=5\)
hay x=2
12: Ta có: \(\left(2x\right)^4=16\)
\(\Leftrightarrow x^4=1\)
hay \(x\in\left\{1;-1\right\}\)
Bài 1
a) \(x=x^5\)
\(x^5-x=0\)
\(x\left(x^4-1\right)=0\)
\(x=0\) hoặc \(x^4-1=0\)
* \(x^4-1=0\)
\(x^4=1\)
\(x=1\)
Vậy x = 0; x = 1
b) \(x^4=x^2\)
\(x^4-x^2=0\)
\(x^2\left(x^2-1\right)=0\)
\(x^2=0\) hoặc \(x^2-1=0\)
*) \(x^2=0\)
\(x=0\)
*) \(x^2-1=0\)
\(x^2=1\)
\(x=1\)
Vậy \(x=0\); \(x=1\)
c) \(\left(x-1\right)^3=x-1\)
\(\left(x-1\right)^3-\left(x-1\right)=0\)
\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)
\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)
*) \(x-1=0\)
\(x=1\)
*) \(\left(x-1\right)^2-1=0\)
\(\left(x-1\right)^2=1\)
\(x-1=1\) hoặc \(x-1=-1\)
**) \(x-1=1\)
\(x=2\)
**) \(x-1=-1\)
\(x=0\)
Vậy \(x=0\); \(x=1\); \(x=2\)
1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
a) \(x=\dfrac{25}{72}\)
b)\(x=-\dfrac{1}{4}\)
\(x=\dfrac{3}{2}\)
c)\(x=\dfrac{5}{4}\) hoặc
x \(=\dfrac{8}{5}\)
d và e chịu vì mk kg giỏi lắm về mũ
f)\(x=-2\)
G)\(x=-\dfrac{5}{12}\)
a)5x+1=125
=>5x+1=53
=>x+1=3
=>x=2
vậy x=2
b)42x+1=64
=>42x+1=43
=>2x+1=3
=>x=1
vậy x =1
e)=>43x+2017=42020-3
=>3x+2017=2017
=>x=0
vậy x=0
f)=>2x+2x x 23=144
=>2x x (1+23)=144
=>2x x 9=144
=>2x=16
=>2x=24
=>x=4
vậy x=4