\(P=\left(\dfrac{\sqrt{x}+1}{x-1}+\dfrac{x}{x-\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\) với x > 0; x \(\ne\) 1
a) Rút gọn P
b) Tìm GTNN của biểu thức M = P.\(\sqrt{x}\) khi x > 1
Rút gọn P = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{3\sqrt{x}-1}{x-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{4\sqrt{x}+1}{x+\sqrt{x}}\right)\) với x>0 ; x≠1
\(=\dfrac{x+\sqrt{x}-3\sqrt{x}+1}{x-1}:\dfrac{x+2\sqrt{x}+1-4\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)
Rút gọn biểu thức dạng chữ:
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\left(x+\sqrt{x}\right)\) với x ≥0, x ≠1
A= \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}}{4-x}\right):\dfrac{\sqrt{x}+1}{x-4}\) với x ≥0, x ≠ 4
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\right):\dfrac{1}{x+6\sqrt{x}+9}\) với x ≥ 0, x ≠ 9
Hộ vs ạ
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
3.
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)
\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)
B = \(\left(\dfrac{x}{\sqrt{x}-1}+\dfrac{2x-\sqrt{x}}{\sqrt{x}-x}\right)\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\)với (x>0;x\(\ne\)1)
\(B=\left(\dfrac{x}{\sqrt{x}-1}+\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}\right).\left(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\\ =\left(\dfrac{x}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\right).\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\\ =\left(\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\right).\left(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right)\\ =\left(\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\right)\\ =\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\right).\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\\ =\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
A= \(\left[\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\right)+\dfrac{\sqrt{x}}{\sqrt{x}+3}+3\left(\dfrac{\sqrt{x}}{x-9}\right)\right]:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{1}{1}\right)\)với x>= 0 , x #9
\(A=\left(\dfrac{3\sqrt{x}}{\sqrt{x}+3}+\dfrac{3\sqrt{x}}{x-9}\right):\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{3x-9\sqrt{x}+3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
* Cho biểu thức
A= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\) với x > 0, x ≠ 1
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-1\right)^2\left(\sqrt{x}-2\right)}\)
\(\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)với\) x>0,x khác 1
Ta có: \(\left(\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\sqrt{x}+\dfrac{\sqrt{x}}{x-1}\right)\)
\(=\dfrac{x\sqrt{x}+1-\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x\sqrt{x}-\sqrt{x}+\sqrt{x}}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+\sqrt{x}}{x\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{x\sqrt{x}}\)
\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)
= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)=\(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
Rút gọn \(\left(\dfrac{1}{\sqrt{x}}-\sqrt{x}\right):\left(\dfrac{\sqrt{x}-1}{x+\sqrt{x}}+\dfrac{1-\sqrt{x}}{\sqrt{x}}\right)\) với x>0,x khác 1
\(=\left(\dfrac{1-x}{\sqrt{x}}\right):\dfrac{\sqrt{x}-1+1-x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{1-x}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\)
\(=\dfrac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
A1= \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)với x≠1, x≥ 0
A2= \(\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x\sqrt{x}-x+\sqrt{x}-1}\right]:\left(1-\dfrac{\sqrt{x}}{x+1}\right)\)với x≥0, x≠1 và -1
\(A_1=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A_2=\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{x-\sqrt{x}+1}{x+1}\\ A_2=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x-\sqrt{x}+1}\\ A_2=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)
Rút gọn các biểu thức sau:
a) \(\left(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)\)
b) \(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\) với x>0
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)