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24 tháng 9 2015

1. \(2^x=4^{y-1}\Rightarrow2^x=\left(2^2\right)^{y-1}=2^{2y-2}\Rightarrow x=2y-2\)

\(27^y=3^{x+8}\Rightarrow\left(3^3\right)^y=3^{x+8}\Rightarrow3^{3y}=3^{x+8}\Rightarrow3y=x+8\)

ta có: x=2y-2

mà 3y=x+8

=> 3y=2y-2+8

=> 3y-2y+2-8=0

=> y-6=0

=> y=6

x=2y-2

=> x=2.6-2=12-2=10

Vậy x=10; y=6.

2.a.\(\left(-\frac{1}{3}\right)^{n-5}=\frac{1}{81}\)

\(\Rightarrow \left(-\frac{1}{3}\right)^{n-5}=\left(-\frac{1}{3}\right)^4\)

=> n-5=4

=> n=4+5

=> n=9

b.\(2^{-1}.2^n+4.2^n=9.2^5\)

\(\Rightarrow2^n.\left(2^{-1}+4\right)=9.32\)

=> 2n.(2-1+4)=288

=> 2n.(1/2+4)=288

=> 2n.9/2=288

=> 2n=288:9/2

=> 2n=64

=> 2n=26

Vậy n=6.

22 tháng 10 2019

1.

a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)

b) x=0

d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)

e) \(x=\frac{2}{3}\)

4 tháng 10 2021

a. x=5/6

 

4 tháng 10 2021

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

4 tháng 9 2020

Bài 1:

Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)

\(\Leftrightarrow2x=\frac{1440}{144}=10\)

\(\Rightarrow x=5\)

Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)

=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)

23 tháng 6 2015

1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)

   \(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)

 ( . là nhân nha) 

    \(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)

   \(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)

\(x=\frac{113}{8}\)

23 tháng 6 2015

\(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)

   \(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{49}{81}=\frac{56}{81}\)

\(4y=\frac{7}{81}\)

y      =  7/81:4

y       = 7/324

a)x ∈ ∅

b) x=3

13 tháng 9 2021

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

Bài 3: 

a: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\\dfrac{3}{4}x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}3x+2>0\\\dfrac{2}{3}x-5< 0\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< \dfrac{15}{2}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x+2=0\\\dfrac{2}{5}x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{3}{4}=-2\\\dfrac{2}{5}x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=6:\dfrac{2}{5}=15\end{matrix}\right.\)