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a: \(=\dfrac{2^6\cdot3^3}{3^{-4}\cdot2^6}=\dfrac{3^3}{3^{-4}}=3^7\)

c: \(=5^4\cdot5^3\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{4}{100}\)

\(=5^7\cdot\left(\dfrac{2}{5}\right)^5\cdot\left(\dfrac{1}{5}\right)^2\)

\(=5^2\cdot\left(\dfrac{1}{5}\right)^2\cdot5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)

27 tháng 7 2018

a, x:(1/2)3=-1/2

x:1/8= -1/2

x= -1/2.1/8

x=-1/16

b,(3/4)5.x=(3/4)7

x=(3/4)7:(3/4)5

x= (3/4)2

c,(2/5)^8:x=(2/5)^6

x=.......

như cái trên nha lm giống thế

27 tháng 7 2018

thanks nha

1: Tìm x

a) Ta có: \(\left(2x-1\right)^3=-27\)

\(\Leftrightarrow2x-1=-3\)

\(\Leftrightarrow2x=-3+1=-2\)

hay x=-1

Vậy: x=-1

b) Ta có: \(\left(2x-3\right)^4=625\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=-5\\2x-3=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5+3=-2\\2x=5+3=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

Vậy: \(x\in\left\{-1;4\right\}\)

c) Ta có: \(\left(x-2\right)^5=\left(x-2\right)^7\)

\(\Leftrightarrow\left(x-2\right)^5-\left(x-2\right)^7=0\)

\(\Leftrightarrow\left(x-2\right)^5\left[1-\left(x-2\right)^2\right]=0\)

\(\Leftrightarrow\left(x-2\right)^5\cdot\left[1-\left(x-2\right)\right]\cdot\left[1+\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)^5\cdot\left(1-x+2\right)\cdot\left(1+x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)^5\cdot\left(-x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^5=0\\-x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-x=-3\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)

Vậy: \(x\in\left\{1;2;3\right\}\)

d) Ta có: \(5^{x+2}+5^{x+3}=750\)

\(\Leftrightarrow5^{x+2}\cdot1+5^{x+2}\cdot5=750\)

\(\Leftrightarrow5^{x+2}\left(1+5\right)=750\)

\(\Leftrightarrow5^{x+2}\cdot6=750\)

\(\Leftrightarrow5^{x+2}=125\)

\(\Leftrightarrow x+2=3\)

hay x=1

Vậy: x=1

3 tháng 4 2018

Căng, sự thật là nó rất căng

Nhg dù sao thì.....

1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)

Xét \(A\left(x\right)=0\)

\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)

\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)

\(\Rightarrow-3x^2-12x+15=0\)

\(\Rightarrow-3x^2+3x-15x+15=0\)

\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)

Xét \(B\left(x\right)=0\)

\(\Rightarrow x^3+x^2-4x-4=0\)

\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)

Đó là những j mình biết khocroikhocroi