\(\dfrac{x+1}{x+2}=\dfrac{-3}{2}\)
=>\(\dfrac{2\left(x+1\right)}{2\left(x-2\right)}=\dfrac{-3\left(x-2\right)}{2\left(x-2\right)}\)
=>2(x+1)=-3(x-2)
=>2x+2=-3x+6
=>5x=4
=>x=\(\dfrac{4}{5}\)
Còn nốt 2 câu cíu với =)
A) 3 . (x - \(\dfrac{1}{2}\)) - 3 . (x - \(\dfrac{1}{3}\)) = x
B) \(\dfrac{1}{2}\) . (x + 2) - 4 . (x - \(\dfrac{1}{4}\)) = \(\dfrac{1}{2}\) . x
\(B = 2x^{3} - \dfrac{1}{3}x^{2} - |x - \dfrac{1}{4}| \) với \(x = \dfrac{1}{2}\)
Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
Tìm \(\left(x\right)\):
a) \(\left(x-\dfrac{1}{2}\right)^2\):\(\dfrac{2}{9}\) =\(\left(-1\dfrac{1}{3}\right)\):\(\left(\dfrac{1}{2}-x\right)\)
b) \(\dfrac{3.x-1}{4}\)=\(\dfrac{2.x-5}{3}\)
c) \(\dfrac{3}{-2}\)=\(\dfrac{x-3}{3.x+1}\)
Cú tui mấy ông bà ơi:)) Làm dc thì làm giúp tui nhen, cảm ơn nè
\(\left(\dfrac{-2}{3}\right)^2.x=\left(\dfrac{-2}{3}\right)^5\)
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\left(\dfrac{2}{3}x-1\right)\left(\dfrac{3}{4}x+\dfrac{1}{2}\right)=0\)
\(\dfrac{4}{9}:x=3\dfrac{1}{3}:2,25\)
\(1\dfrac{1}{3}:0,8=\dfrac{2}{3}:0,1x\)
\(3^{2x-1}+2.9^{x-1}=405\)
\(\left(\dfrac{1}{3}\right)^{x-1}+5.\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\)
\(\dfrac{3}{5}.\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}.\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\)
Tìm x ( Giúp với mình cần gấp )
72x+72x+3=344 (5-x)(9x2-4)=0 \(\left|2-2x\right|\)-3,75=(-0,5)2
\(\sqrt{x-1}\)+\(\dfrac{2}{3}\)=1 (\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)x)2=2\(\dfrac{1}{4}\) giúp mình với!!!! cảm ơn nhìu=))❤
tìm x;y
A) \(\dfrac{2}{5}x-\dfrac{1}{3}=-1\dfrac{1}{2}:\dfrac{5}{4}\)
B) x;y tỉ lệ thuận với 5 và 3 và x+y=32
c) x;y tỉ lệ nghịch với 5 và 3 và x+y = 32
1/ 3 x (x - \(\dfrac{1}{2}\)) - 3 x (x - \(\dfrac{1}{3}\)) - (x - \(\dfrac{1}{2}\)) = 0
2/ \(\dfrac{-2x+1}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{-1}{3}\)
Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)
b) \(\dfrac{1}{3}\)x + \(\dfrac{2}{5}\) . (x + 1) = 0
