(2x-1)tất cả mũ 4=16
(2x+1)tất cả mũ 3=125
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1. \(2^x-26=6\)
\(\Rightarrow2^x=6+26\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
2. \(64\cdot4^x=16^8\)
\(\Rightarrow4^3\cdot4^x=4^{16}\)
\(\Rightarrow4^x=4^{16}:4^3\)
\(\Rightarrow4^x=4^{13}\)
\(\Rightarrow x=13\)
3. \(\left(2x-1\right)^4=16\)
\(\Rightarrow\left(2x-1\right)^4=2^4\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
4. \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
a)\(x^{15}=x\Rightarrow x\in\left\{0;1\right\}\) d)Là ý b
b)\(\left(2x+1\right)^3=125\\ \left(2x+1\right)^3=5^3\Rightarrow2x+1=5\\ 2x=4\\ x=2\) e)\(\left(x-3\right)^2=25\\ \Rightarrow\left(x-3\right)^2=5^2\\ \Rightarrow\hept{\begin{cases}x-3=5\Rightarrow x=8\\x-3=-5\Rightarrow x=-2\end{cases}}\)
c)\(\left(x-5\right)^4=\left(x-5\right)^6\\ \Rightarrow x-5\in\left\{0;1\right\}\Rightarrow x\in\left\{5;6\right\}\)
a, \(\left(x+\frac{4}{3}y^2\right)^2\)
\(=x^2+\frac{8}{3}xy^2+\frac{16}{9}y^4\)
b, \(\left(2x-3y\right)^2\)
\(=4x^2-12xy+9y^2\)
c, \(\left(x^2+2x\right)\left(2x-x^2\right)\)
\(=\left(2x+x^2\right)\left(2x-x^2\right)\)
\(=4x^2-x^4\)
d, \(\left(x+\frac{1}{2}\right)^3\)
\(=x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
e, \(\left(2x-\frac{6}{5}y\right)^3\)
\(=8x^3-\frac{72}{5}x^2y+\frac{216}{25}xy^2-\frac{216}{125}y^3\)
Rút gọn hả bạn ?
( 3x - 1 )2 - 9( x - 1 )( x + 1 )
= 9x2 - 6x + 1 - 9( x2 - 1 )
= 9x2 - 6x + 1 - 9x2 + 9
= 10 - 6x
( 2x + 3 )( 2x - 3 ) - ( 2x - 1 )2 - ( x - 1 )
= 4x2 - 9 - ( 4x2 - 4x + 1 ) - x + 1
= 4x2 - x - 8 - 4x2 + 4x - 1
= 3x - 9
2( x - 2y )( x + 2y ) + ( x - 2y )2 + ( x + 2y )2
= [ ( x + 2y ) + ( x - 2y ) ]2
= [ x + 2y + x - 2y ]2
= ( 2x )2 = 4x2
(5 mũ 2)mũ 3 : 5 mũ 2
5 mũ 2 x3 :5 mũ 2
5 mũ 6 : 5 mũ 2 =5 mũ 4=625
a.(3^2+4^2).x=10^2
(9+16).x =100
25.x =100
x =100:25
x =4
b.(x-5)^2 =81
x-5 =9
x =9+5
x =14
c.(2x+1)^3 = 343
2x+1 = 7
2x =7-1
2x =6
x =6:2
x = 3
\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)
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