a.

\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)

TH1:

\(x+\frac{1}{2}=0\)

\(x=-\frac{1}{2}\)

TH2:

\(x-\frac{3}{4}=0\)

\(x=\frac{3}{4}\)

Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)

b.

\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)

TH1:

\(\frac{1}{2}x-3=0\)

\(\frac{1}{2}x=3\)

\(x=3\div\frac{1}{2}\)

\(x=3\times2\)

\(x=6\)

TH2:

\(\frac{2}{3}x+\frac{1}{2}=0\)

\(\frac{2}{3}x=-\frac{1}{2}\)

\(x=-\frac{1}{2}\div\frac{2}{3}\)

\(x=-\frac{1}{2}\times\frac{3}{2}\)

\(x=-\frac{3}{4}\)

Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)

c.

\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)

\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)

\(-\frac{4}{3}x=\frac{13}{3}\)

\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)

\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)

\(x=-\frac{13}{4}\)

d.

\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)

\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)

\(x=5\)

a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)

=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

=>  x + 1 = 0

=> x = -1

b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

=> x - 2021 = 0

=> x = 2021

c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)

=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)

=> \(-\frac{1}{12}x+6=7\)

=> \(-\frac{1}{12}x=1\)

=> x = -12

Thiếu điều kiện xy = 1; x+y khác 0 nhá bn

Bài này tương tự câu 1 ở đây

 nhiều quá trời, lm chắc mỏi tay luôn

\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\) 

              \(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)

             \(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .

\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\) 

 \(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)            

              \(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)

              \(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)

\(2^x=2\Rightarrow x=1\)

\(3^x=3^4\Rightarrow x=4\)

\(7^x=7^7\Rightarrow x=7\)

\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)

\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)

\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)

\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)

\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)

\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)

\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)

\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)

\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)

\(\left(-2\right)^{4x+2}=64\)

\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)

\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)

\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)

\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)

                                      \(2x-5x=-4+1\) 

                                           \(-3x=-3\Rightarrow x=1\)

\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)

 \(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)

\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)

\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)

\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)

 \(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)

\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)

\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).

hehe. đánh tới què tay, hoa mắt lun r nekkk!!

người kiểu gì mà bài này ko làm đc

a) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)\(5\)

=> \(\frac{2}{3}-\left(\frac{1}{3}x-\frac{1}{2}\right)-\left(x+\frac{1}{2}\right)=5\)

=>\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

=>\(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5\)

=>\(\frac{2}{3}-\frac{4}{3}x=5\)

=>\(\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)

=>\(x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{4}\)

b)\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)

=>\(4x-x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=> \(3x-\frac{1}{2}=2x-\left(-\frac{9}{2}\right)\)

=>\(x=-\left(-\frac{9}{2}\right)+\frac{1}{2}=5\)

a) \(x=\frac{9}{10}\)

b) \(x=\frac{-4}{3}\)

c) \(x=\frac{1}{42}\)

d) \(x=\frac{-47}{10}\)

ko có thời gian nên mình chỉ cho đáp án thôi nhé

thông cảm cho mình ngen

đúng thì k đấy

chúc bạn học giỏi

làm chi tiết cho mk nhé

ai làm chi tiết mk k cho nhìu

\(-4\frac{1}{2}\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

<=>  \(-1,5\le x\le\frac{11}{18}\)

đến đây tự làm

mk k biết điều kiện của x  nên giúp đến đó