Rút gọn biểu thức :
\(A=\left(\frac{a\sqrt{a}-3}{a-2\sqrt{a}-3}-\frac{2\left(\sqrt{a}-3\right)}{\sqrt{a}+1}+\frac{\sqrt{a}+3}{3-\sqrt{a}}\right):\frac{a+8}{a-1}\)
Rút gọn biểu thức \(B=\left(\frac{\sqrt{a-2}+2}{3}\right)\left(\frac{\sqrt{a-2}}{3+\sqrt{a-2}}+\frac{a+7}{11-a}\right):\left(\frac{3\sqrt{a-2}+1}{a-3\sqrt{a-2}-2}-\frac{1}{\sqrt{a-2}}\right)\)
Bài 1:Rút gọn
\(a,\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(b,\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(c,\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)\times\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\left(a\ne1;a\ge0\right)\)
Bài 2: Rút gọn biểu thức
\(P=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
rút gọn biểu thức \(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a+1}}{\sqrt{a}}+\left(\sqrt{a-}\frac{1}{\sqrt{a}}\right).\left(\frac{3\sqrt{a}}{\sqrt{a-1}}-\frac{2+\sqrt{a}}{\sqrt{a+1}}\right)\)
\(\left(\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}\right):\left(1+\sqrt{\frac{a+1}{a-1}}\right)\)
\(\left(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a^3}+3\right)\left(a-b\right)}\right)\)
Rút gọn 2 biểu thức trên?
Ai giúp mình với, tks nhiều
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
Rút gọn biểu thức:
A= \(\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{3\sqrt{ab}}{a\sqrt{a}+b\sqrt{b}}\right).\left[\left(\frac{1}{\sqrt{a}-\sqrt{b}}-\frac{3\sqrt{ab}}{a\sqrt{a}-b\sqrt{b}}\right):\frac{a-b}{a+\sqrt{ab}+b}\right]\)
Rút gọn biểu thức: C = \(\frac{\sqrt{a}-a^3}{1-a^2}:\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
Cho biểu thức P=\(\left[\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right]:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)\)
a/ Rút gọn P
b/ Tìm a để \(\frac{1}{P}-\frac{\sqrt{a}+1}{8}\ge1\)
cho biểu thức P=\(\left[\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right]:\left(\frac{1}{\sqrt{a}+1}+\frac{1}{\sqrt{a}-1}\right)\)
a/ Rút gọn P
b/ Tìm a để \(\frac{1}{P}-\frac{\sqrt{a}+1}{8}\ge1\)
1) cho biểu thức P=\(\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
a) tìm ĐKXĐ của x để P có nghĩa
b) rút gọn P
c) tìm các giá trị của x để P=\(\frac{6}{5}\)
2) cho biểu thức P=\(\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\left(a>0;a\ne1,-1\right)\)
a) rút gọn P
b)tính giá trị biểu thức P khi \(a=19-8\sqrt{3}\)
c) tìm giá trị của a để P<1
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
2)a) \(P=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
\(=\frac{a-2\sqrt{a}+1}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}=\sqrt{a}-1\)
b) \(19-8\sqrt{3}=\left(\sqrt{3}-4\right)^2\Rightarrow P=\sqrt{\left(\sqrt{3}-4\right)^2}-1=4-\sqrt{3}-1=3-\sqrt{3}\)
c) P < 1 <=> \(\sqrt{a}-1< 1\Leftrightarrow a< 4\)
Kết hợp với điều kiện : \(P< 1\Leftrightarrow\hept{\begin{cases}0< a< 4\\a\ne1\end{cases}}\)