K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

5 tháng 8 2018

\(a,16^x:4^x=16\)

\(\left(4^x\right)^2:4^x=4^2\)

\(\Rightarrow4^x=4^2\Leftrightarrow x=2\)

\(b,2^{-1}.2^x+4.2^x=72\)

\(\Rightarrow2^{x-1}+2^{x+2}=72\)

\(\Rightarrow2^{x-1}\left(1+2^3\right)=72\)

\(\Rightarrow2^{x-1}=72:9=8=2^3\)

\(\Rightarrow x=4\)

\(c,\left(2^x+1\right)^3=-64\)

\(\Rightarrow2^x+1=-4\)

\(\Rightarrow2^x=-5\)

5 tháng 8 2018

a) 16^x : 4^x = 16 

<=> ( 16 : 4 )^x = 16 

<=> 4^x = 16

<=> 4^x = 4^2 

=> x = 2

Vậy x =2

4 tháng 10 2021

a. x=5/6

 

4 tháng 10 2021

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

25 tháng 10 2018

a) \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)

\(\dfrac{13}{3}:\dfrac{x}{4}=20\)

\(\dfrac{x}{4}=\dfrac{13}{3}:20\)

\(\dfrac{x}{4}=\dfrac{13}{3}\cdot\dfrac{1}{20}\)

\(\dfrac{x}{4}=\dfrac{13}{60}\)

\(x=\dfrac{13}{60}\cdot4\)

Vậy \(x=\dfrac{13}{15}\)

b)\(2^3:4.2^{\left(x+1\right)}=64\)

\(8:4.2^{\left(x+1\right)}=64\)

\(2.2^{\left(x+1\right)}=64\)

\(2\cdot2^x.2=64\)

\(4.2^x=64\)

\(2^x=64:4\)

\(2^x=16\)

\(2^x=2^4\)

Vậy \(x=4\)

15 tháng 3 2019

Tìm x

a) x= \(\frac{13}{15}\)

b) x= 4

Chúc bạn học tốthaha

12 tháng 7 2019

a)\(\left(\frac{1}{3}\right)^{-1}-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^4.2^3=3-1+\frac{1}{16}.8=3-1+\frac{1}{2}=\frac{5}{2}\\ \)

b)\(2^2.2^3.\left(\frac{2}{3}\right)^{-2}=2^5.\frac{9}{4}=72\)

c)\(\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3:\left(\frac{-2}{3}\right)^{-3}=\left(\frac{3}{4}\right)^5:\left(\frac{3}{2}\right)^3=\frac{9}{128}\)

2)

\(3^{x+1}=9^x\Leftrightarrow3^x.3=9^x\Rightarrow3=9^x:3^x\Rightarrow3=3^x\Rightarrow x=1\)

\(\left(x-0,1\right)^2=6,25\Leftrightarrow\left(x-0,1\right)^2=2,5^2\Rightarrow\left(x-0,1\right)=2,5\Rightarrow x=2,5+0,1=2,6\)

\(3^{2x-1}=243\Leftrightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow2x=6\Rightarrow x=3\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\Rightarrow x=1\)

3 tháng 12 2018

easy

\(\left(x-3\right)^2=16\)

\(\Rightarrow\left(x-3\right)^2=4^2\)

\(\Rightarrow x-3=4\)

\(\Rightarrow x=4+3\)

\(\Rightarrow x=7\)

3 tháng 7 2019

a) \(\left|2-\frac{3}{2}x\right|-4=x+2\)

=> \(\left|2-\frac{3}{2}x\right|=x+2+4\)

=> \(\left|2-\frac{3}{2}x\right|=x+6\)

ĐKXĐ : \(x+6\ge0\) => \(x\ge-6\)

Ta có: \(\left|2-\frac{3}{2}x\right|=x+6\)

=> \(\orbr{\begin{cases}2-\frac{3}{2}x=x+6\\2-\frac{3}{2}x=-x-6\end{cases}}\)

=> \(\orbr{\begin{cases}2-6=x+\frac{3}{2}x\\2+6=-x+\frac{3}{2}x\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{2}x=-4\\\frac{1}{2}x=8\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{8}{5}\\x=16\end{cases}}\) (tm)

b) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

=> \(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

=> \(\left(4x-1\right)^{20}.\left[\left(4x-1\right)^{10}-1\right]=0\)

=> \(\orbr{\begin{cases}\left(4x-1\right)^{20}=0\\\left(4x-1\right)^{10}-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}4x-1=0\\\left(4x-1\right)^{10}=1\end{cases}}\)

=> \(\orbr{\begin{cases}4x=1\\4x-1=\pm1\end{cases}}\)

=> x = 1/4

hoặc x = 0 hoặc x = 1/2

26 tháng 6 2018

b) (x-1)x+2 =(x-1)2

=> x+2=2

=> x=0

vậy

c) (1-3x)3=-64

=> 1-3x=-4

=> -3x=-4-1

=> -3x=-5

=> x=5/3

vậy ....

26 tháng 6 2018

Làm nốt ::v

\(a.\left(x+2\right)^2=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\Leftrightarrow\left(x+2\right)^2=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{\dfrac{1}{6}}\\x+2=-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{1}{6}}-2\\x=-\sqrt{\dfrac{1}{6}}-2\end{matrix}\right.\)

KL...........