Tìm x : \(\frac{2x-3}{3}\)= \(\frac{27}{2x-3}\)
Tìm x:
\(\frac{2x-1}{-27}=\frac{3}{1-2x}\)
Câu 1: Tìm x
1/\(\frac{1+3x}{2}=\frac{7}{3}-\frac{x+1}{6}\)
2,\(\left|\frac{2x+1}{3}\right|=\left|\frac{x-1}{5}\right|\)
3,\(\left(\frac{4}{3}+2x\right)^3=\frac{8}{-27}\)
Tìm x, biết: \(\frac{2x-1}{-27}=\frac{3}{1-2x}\)
\(\frac{2x-1}{-27}=\frac{3}{1-2x}\)
\(\Rightarrow\left(2x-1\right).\left(1-2x\right)=3.\left(-27\right)\)
\(-4x^2=-81\) ( chỗ này bn tự phân tích ra nha!)
\(x^2=\frac{-81}{-4}=\frac{81}{4}=\left(\frac{9}{2}\right)^2=\left(-\frac{9}{2}\right)^2\)
=> x = 9/2 hoặc x = -9/2
Tìm x trong các tỉ lệ thức sau:
\(\frac{2x-1}{3}=\frac{7}{27.\left(2x+1\right)}\)
Bài 8: Tìm số nguyên x biết
a) \(\left(\frac{-12}{27}+\frac{2}{3}\right)+\frac{-2}{9}\le x\le\left(\frac{11}{7}+\frac{2}{5}\right)+\frac{7}{5}+\frac{3}{7}\) b\(\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\)
c)\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
Tìm x:
a)\(2.\left(3x-\frac{1}{2}\right)-2x=\frac{1}{2}\left(2x-3\right)\)
b)\(\left(2x-\frac{3}{5}\right)^2=\frac{4}{25}\)
c)\(\left(3x-1\right)^3=27\)
d)\(5-\left|x\right|=2\)
e)|2x+1|-3=3
f)|3-2x|=5
\(\left(5-x\right)\left(3x-\frac{1}{4}\right)=0\)
cho biểu thức P =\(\frac{x^2+2}{x^2+2x-3}:\left(\frac{x+3}{3x-3}-\frac{x+1}{2x+6}+\frac{x^2-27}{6x^2+12x-18.}\right)\)
tìm điều kiện để P đc xác định
rút gọn biểu thức
tìm các giá trị x để P có giá trị âm
Tìm x:
a)\(\frac{1}{\left(2x-3\right)^2}=9\)
b)\(\frac{1}{\left(x-1\right)^3}=-\frac{1}{27}\)
c) (x-1)2=(2x-5)2
a, \(\frac{1}{\left(2x-3\right)^2}=9\Leftrightarrow\left(\frac{1}{2x-3}\right)^2=3^2\Leftrightarrow\frac{1}{2x-3}=3\Leftrightarrow1=6x-9\Leftrightarrow x=\frac{5}{3}\)
b, \(\frac{1}{\left(x-1\right)^3}=-\frac{1}{27}\Leftrightarrow\left(\frac{1}{x-1}\right)^3=\left(\frac{-1}{3}\right)^3\Leftrightarrow\frac{1}{x-1}=\frac{1}{-3}\Leftrightarrow x-1=-3\Leftrightarrow x=-2\)
c, \(\left(x-1\right)^2=\left(2x-5\right)^2\Leftrightarrow\orbr{\begin{cases}x-1=2x-5\\x-1=-2x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}}\)