ĐKXĐ: x>=1
\(\sqrt{\left(x-1\right)^3}=8\)
=>\(\left(x-1\right)^3=8^2=64\)
=>x-1=4
=>x=4+1=5(nhận)
\(\sqrt{\left(x-1\right)^3}=\sqrt{4^3}\)
\(\Rightarrow x-1=4\)
\(x=4+1\)
\(x=5\)
Vậy \(x=5\)
Tìm x:
1) \(\text{(x−1):0,16=−9:(1−x)}\)
2) \(\left(\left|x\right|-\dfrac{3}{2}\right)\left(2x^2-10\right)=0\)
3)\(8\sqrt{x}=x^2\left(x\ge0\right)\)
Tìm x biết:
a)\(\left|\sqrt{2}-x\right|=\sqrt{2}\)
b)\(\left|x-1\right|=\sqrt{3}+2\)
c)\(\left|1-2x\right|=\sqrt{5}-1\)
d)\(\left|1-x\right|=\sqrt{2}-0,\left(1\right)\)
e)\(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)
f)\(\left|x-\sqrt{2}\right|=1,\left(4\right)\)
tìm x biết
a)\(\frac{3.\left(x-1\right)}{2}=\frac{8}{27\left(x-1\right)}\)
b)\(x-3\sqrt{x}=0\) với \(x\ge0\)
tìm x biết : x.(x+2/3)-x.(x-3/4)=7/12
b)\(\sqrt{x^2}+1\)=x+2
c)\(\left(2x+1\right)^5\)=\(\left(2x+1\right)^{2019}\)
tìm x biết x ko thuoc {1;3;8;20} va
\(\frac{2}{\left(x-1\right).\left(x-3\right)}+\frac{3}{\left(x-3\right).\left(x-8\right)}+\frac{12}{\left(x-8\right).\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Tìm x biết \(\left|\sqrt{x+1}-0,5\right|-0,6=\sqrt{\left(-3\right)^2}+0,4\)
Tìm x biết:
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=\frac{3}{4}\)
1.Tìm các số x, y, z thỏa mãn đẳng thức\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)
2.Tìm x,y,z biết : \(x+y=x\div y=3\left(x-y\right)\)
Tìm x biết \(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x+8\right)\left(x+20\right)}-\frac{1}{20}=-\frac{3}{4}\)
Tìm x biết:
\(\sqrt{1+2+3+...+\left(x-1\right)+x+\left(x-1\right)+...+1+2+3}=2010\)
