Ta có: \(\sqrt{\left(2x-1\right)^2}=x+1\)
\(\Leftrightarrow\left|2x-1\right|=x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\left(x\ge\dfrac{1}{2}\right)\\2x-1=-x-1\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)
Tìm x
a)\(\sqrt{2x-1}=3\)
b)\(\sqrt{1-3x}=\dfrac{1}{2}\)
c)\(\sqrt{\left(x-1\right)^2}=\dfrac{1}{2}\)
d)\(\sqrt{\left(1+2x\right)^2}=\dfrac{\sqrt{3}}{2}\)
e)\(\sqrt{\left(1-2x\right)^2=|x-1|}\)
tìm x thoả mãn
\(\left(x+2\right)\left(\sqrt{2x+3}+\sqrt{x+1}\right)+\sqrt{2x^2+5x+3}=1\left(với:x\ge-1\right)\)
Cho M = 1 - \(\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)\(\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
a,Rút gọn M
b,Tìm x thuộc Z sao cho M thuộc Z
P=\(\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}-\sqrt{x}}{\sqrt{2x}-1}\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}+\frac{\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{MTC}-\frac{2x-1}{MTC}\)
=\(\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{MTC}\)
=\(\frac{2x\sqrt{2}+2\sqrt{2x}}{MTC}\)
\(\frac{2x-1}{MTC}+\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{MTC}-\frac{\left(\sqrt{2x}+\sqrt{x}\left(\sqrt{2x}+1\right)\right)}{MTC}\)
=\(\frac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{MTC}\)
=\(\frac{-2\sqrt{x}-2}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x+1}\right)}\)
Cho x là số thực. Tìm GTNN:
\(P=\frac{\sqrt{3\left(2x^2+2x+1\right)}}{3}+\frac{1}{\sqrt{2x^2+\left(3-\sqrt{3}\right)x+3}}+\frac{1}{\sqrt{2x^2+\left(3+\sqrt{3}\right)x+3}}\)
\(\frac{\left(2x\sqrt{2}+2\sqrt{2x}\right)\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right).\left(-2\sqrt{x}-2\right)}\)
=\(\frac{2\sqrt{2x}\left(\sqrt{x}+1\right)}{-2\left(\sqrt{x}+1\right)}\)
=\(-\sqrt{2x}\)
Tìm ĐKXĐ:
a) \(\sqrt{72x}\)
b) \(\dfrac{2x+3}{\sqrt{x^2-4}}\)
c) \(\sqrt{\left(2x+1\right)\left(x+2\right)}\)
d) \(3-\sqrt{16x^2-1}\)
e) \(\sqrt{\dfrac{3+x}{4-x}}\)
Cho M = 1 - \(\left(\frac{2x-1+\sqrt{x}}{1-x}+\frac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right)\)\(\left(\frac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right)\)
a,Rút gọn M
b,Tìm x thuộc Z sao cho M thuộc Z
Tìm x biết :
\(\sqrt{\left(2x-1\right)}^2=5\)
gợi ý:
\(\sqrt{\left(2x-1\right)}^2=\left|2x-1\right|\)
cần giải hai phương trình
2x-1=5
2x-1=-5
