Sửa: \(\left(\dfrac{3}{4}-x\right)^3=\dfrac{1}{64}=\left(\dfrac{1}{4}\right)^3\)
\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2}\)
3/4 - x =1/64
x =3/4-1/64
x =48/64 - 1/64
x =47/64
Vậy x= 47/64
Sửa: \(\left(\dfrac{3}{4}-x\right)^3=\dfrac{1}{64}=\left(\dfrac{1}{4}\right)^3\)
\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2}\)
3/4 - x =1/64
x =3/4-1/64
x =48/64 - 1/64
x =47/64
Vậy x= 47/64
\(\dfrac{0,8:\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{17}}+\left(1,2.0,5\right):\dfrac{4}{5}\)
(mn giải giúp mik với ạ! iu mn nhiều![]()
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Cho \(\dfrac{b+c-5}{a}=\dfrac{a+c+2}{b}=\dfrac{a+b+3}{c}=\dfrac{1}{a+b+c}\left(a,b,c\ne0,a+b+c\ne0\right)\)
Tính \(\left(a-3b\right)\left(b-c\right)\left(3c-a\right)\)
Ai giúp mik đi, mik cho 5 coin
Tìm x, y, z biết
a) \(\dfrac{x}{y+z+1}\) =\(\dfrac{y}{x+y+2}=\dfrac{z}{x+y-3}\)
b)\(6\left(x-\dfrac{1}{y}\right)=3\left(y-\dfrac{1}{2}\right)=2\left(z-\dfrac{1}{x}\right)=xyz-\dfrac{1}{xyz}\)
Giúp mik nha!
Tìm x, y, z biết \(\dfrac{-2.\left(x-3\right)}{5}=\dfrac{y+4}{-4}=\dfrac{3.\left(z-5\right)}{2}\) và x - y + z = -1
Cho 3 số x,y,z thỏa mãn điều kiện
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{5x+7y-7}{4x}\)
Hãy tính giá trị biểu thức : B=\(\left(1+\dfrac{x}{y}\right)\cdot\left(1+\dfrac{y}{z}\right)\cdot\left(1+\dfrac{z}{x}\right)\)
Tìm \(x\) trong các tỉ lệ thức sau :
a) \(\left(\dfrac{1}{3}.x\right):\dfrac{2}{3}=1\dfrac{3}{4}:\dfrac{2}{5}\)
b) \(4,5:0,3=2,25:\left(0,1.x\right)\)
c) \(8:\left(\dfrac{1}{4}.x\right)=2:0,02\)
d) \(3:2\dfrac{1}{4}=\dfrac{3}{4}:\left(6.x\right)\)
( 2 . \(2^x\) -8 ) .\(\left(\dfrac{x}{3}-\dfrac{3}{4}\right)=0\)
giúp mk với ạ
Tìm x, y, z
\(\left\{{}\begin{matrix}\dfrac{-x+2}{-5}=\dfrac{y-1}{2}=\dfrac{z+5}{3}\\-2y+4x-3z=35\end{matrix}\right.\)
Giúp mk đi, mọi ngừi ^^ (23h58p là e phải nộp ròi ah) ![]()
Cho \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}\)
CMR \(\dfrac{a^3+c^3+m^3}{b^3-d^3-n^3}\) = \(\left(\dfrac{a+c-m}{b+d-m}\right)^3\)
mọi người ơi giup mik với ai làm đc mik tick cho