Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(2x-\dfrac{3}{4}\right)^2=\left(3-x\right)^2\)
\(\Rightarrow2x-\dfrac{3}{4}=3-x\)
\(3x=3\dfrac{3}{4}\)
\(x=\dfrac{5}{4}\)
a) \(\left(2x+3\right).\left(\frac{1}{2}.x-\frac{3}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+3=0\\\frac{1}{2}.x-\frac{3}{2}=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=-3\\\frac{1}{2}.x=\frac{3}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=\frac{3}{2}:\frac{1}{2}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=3\end{cases}}\)
Vậy x = \(-\frac{3}{2}\) hoặc x = 3
b)\(\left(\frac{1}{2}-x\right)^2=\frac{64}{49}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)^2=\left(\frac{8}{7}\right)^2\) hoặc \(\left(\frac{1}{2}-x\right)^2=\left(-\frac{8}{7}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}-x=\frac{8}{7}\\\frac{1}{2}-x=-\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}-\frac{8}{7}\\x=\frac{1}{2}+\frac{8}{7}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{9}{14}\\x=\frac{23}{14}\end{cases}}\)
Vậy x = \(-\frac{9}{14}\) hoặc x = \(\frac{23}{14}\)
c) \(\frac{1}{2}.\left(x-4,5\right)=\frac{3}{4}.x=\frac{5}{12}\) ( câu này mik ko hiểu cho lắm)
k mik nha mn!
2^2x*3^2x=36^24
=>(2*3)^2x=36^24
=>6^2x=36^24
=>36^x=36^24
=>x=24 ok nha
2/3^2x-1=3/243
2/3^2x-1=(2/3)^5
suy ra 2x-1=5
2x =5+1
2x =6
x = 6:2
x = 3
vay x=3
\(a;-\frac{3}{9}+\frac{7}{15}=-x+\left(-\frac{5}{18}\right)\)
\(\Rightarrow\frac{2}{15}=-x-\frac{5}{18}\)
\(\Rightarrow-x=\frac{2}{15}+\frac{5}{18}=\frac{37}{90}\)
\(b;2x-\frac{3}{4}=x-\frac{1}{2}\)
\(\Rightarrow2x-x=-\frac{1}{2}+\frac{3}{4}\)
\(\Rightarrow x=\frac{1}{4}\)
a)-3/9+7/15=-x+(-5/18) b)2x-3/4= x-1/2
=>-x+(-7/15)=2/15 =>2x-x=3/4-1/2
=>-x=2/15-(-7/15) =>x=1/4
=>-x=3/5
=>x=+3/5
Ta có : \(\left(3-x\right)\left(\left|x+5\right|\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
\(\Rightarrow\left(x-1\right)^2-\left(2x-3\right)^2=0\\ \Rightarrow\left(x-1-2x+3\right)\left(x-1+2x-3\right)=0\\ \Rightarrow\left(2-x\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)
Nguyễn Hoàng Minh thank you