Cho P =( \(\frac{2\text{x}\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\)) . \(\frac{x-1}{2\text{x}+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
Rút gọn P
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Cho biểu thức
A=\(\text{[}1-\frac{\sqrt{x}}{1+\sqrt{x}}\text{]}:\text{[}\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)
a, Rút gọn A
b, Tìm x để A= \(\frac{1}{2}\)
a) A= (\(\left(\frac{1+\sqrt{x}}{1+\sqrt{x}}-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right)\)
A=\(\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}\right)\)
A= \(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
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b) Để A = \(\frac{1}{2}\)
thì \(\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{1}{2}\)
=> 2\(\sqrt{x}-4\)=\(\sqrt{x}+1\)
=> \(\sqrt{x}=5\)
=> x = 25
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Cho biểu thức
A= \(\text{[}1-\frac{\sqrt{x}}{1+\sqrt{x}}\text{]}:\text{[}\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)
a, Rút gọn A
b, Tìm x để A<0
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
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Cho biểu thức: \(B=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}\text{+}2}\text{+}\frac{\sqrt{x}}{1-\sqrt{x}}\)
tìm Đkxđ rồi rút gọn B.
\(ĐKXĐ:x\ge0;x\ne1\)
\(B=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)
\(B=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}+\frac{4\sqrt{x}}{2\left(\sqrt{x}+1\right).2\left(\sqrt{x}-1\right)}\)
\(B=\frac{2\sqrt{x}+2-2\sqrt{x}+2+4\sqrt{x}}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{4\sqrt{x}+4}{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{4\left(\sqrt{x}+1\right)}{4\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)
là \(\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}nha toi bi nham\)
\(\left(\frac{2x\text{√}x+x-\text{√}x}{x\text{√}x-1}-\frac{x+\text{√}x}{x-1}\right)\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
a) Rút gọn
b) min
\(chobi\text{ểu}th\text{ức}y=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)
rút gọn y
Rút gọng:\(\left(\frac{1}{\text{x}-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{\text{x}}+1}{\text{x}-2\sqrt{x}+1}\)
Rút gọn biểu thức:\(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)-\(\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}\)+x+1
Mình cần người giải giúp bài toán giúp mk ha
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rút gon biểu thức \(\text{A=\frac{x-2\sqrt{x}+3}{x\sqrt{x}+1}}+\frac{\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\)
1) B=\(\left(\frac{\sqrt{x}}{2}\text{+}\frac{1}{2\sqrt{x}}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}\text{+}1}-\frac{\sqrt{x}\text{+}1}{\sqrt{x}-1}\right)\)
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