a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
Tìm x :
1) \(\left(-0,75x+\dfrac{5}{2}\right).\dfrac{4}{7}-\left(-\dfrac{1}{3}\right)=-\dfrac{5}{6}\)
2) \(\left(4x-9\right)\left(2,5+\dfrac{-7}{3}x\right)=0\)
3) \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
4)\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
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
Tìm x,y biết :
a) \(\left|3.x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}.y+\dfrac{3}{5}\right|\)= 0
b)\(\left|\dfrac{3}{2}.x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}.y-\dfrac{1}{2}\right|\le0\)
\(3^{2x-1}+2.9^{x-1}=405\)
\(\left(\dfrac{1}{3}\right)^{x-1}+5.\left(\dfrac{1}{3}\right)^{x+1}=\dfrac{14}{9^3}\)
\(\dfrac{3}{5}.\left(3x^3-\dfrac{8}{9}\right)-\dfrac{1}{2}.\left(\dfrac{3}{2}-1\right)=-\dfrac{1}{4}\)
Tìm x ( Giúp với mình cần gấp )
\(x^2-19=5.9;\left(2x+1\right)^3=-0,001;\left(\dfrac{5}{6}\right)^{2x-1}=\left(\dfrac{5}{6}\right)^5;\left(\dfrac{1}{3}x-\dfrac{2}{3}\right)^3=27;\left(\dfrac{1}{32}\right)^x=\left(\dfrac{1}{2}\right)^{15}\)
\(\dfrac{3}{4}+\dfrac{1}{4}:x=-2\dfrac{1}{2}\)
\(\left(\dfrac{2}{3}\right)^{100}:x=\left(\dfrac{-2}{3}\right)^{98}\)
\(\dfrac{3}{2}:\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{4}\)
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
Bài 1: Thu gọn đơn thức ( a là hằng só )
a) \(1\dfrac{1}{4}x^2y\left(\dfrac{-5}{6}xy\right)^0.\left(-2\dfrac{1}{3}xy\right)\)
b) \(\dfrac{1}{2}x.\dfrac{1}{4}x^2.\dfrac{x^3}{8}.2y.4y^2.x^3\)
c) \(\left(\dfrac{-9}{2}\right)^3.3xy\left(4a^2x^3\right)\left(4\dfrac{1}{3}ay^2\right)\)
d) \(\left(\dfrac{1}{3}xy^2\right)^3\left(\dfrac{-3}{7}x^4y\right)^2\)
