\(\frac{x+2}{\sqrt{x}+1}+\frac{x-3\sqrt{x}+2}{x-1}\)
rút gọn
bài 1: rút gọn:
C=\(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
bài 2 :rút gọn
E=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
Rút gọn B=\(\left(\frac{3x +3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+2}-2\right):\frac{1}{x-1}\)
Rút gọn A=\(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-3\frac{\sqrt{x}-1}{x-5\sqrt{x}+6}\)
ĐK \(x\ne\left\{2;3\right\}\)
Ta có \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-3.\frac{\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+2\sqrt{x}+3-3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=-\frac{1}{\sqrt{x}-3}\)
Rút gọn: \(A=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
rút gọn P=\(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9-x}{1+\sqrt{x}-6}\right)\)
\(P=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(x-9\right)+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-x+9+2x-4\sqrt{x}-5}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}=\dfrac{3}{\sqrt{x}-2}\)
I .cho C= \(\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)
a, rút gọn C
b, tính C vs x=\(\frac{4}{9}\)
c, tìm x để GTTĐ của C =\(\frac{1}{3}\)
II. cho P = \(\hept{\frac{\sqrt{x}-2}{x-1}}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1})X\frac{\left(1-x\right)^2}{2}\)
a, rút gọn P
b, chứng minh rằng nếu 0<x<1 thì P>0
III. Cho Q= \(\frac{2\sqrt{x-9}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
a, rút gọn Q
b, tìm các gtri x nguyên để Q có gtri nguyên
\(P=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)Rút gọn biểu thức P
RÚT GỌN BIỂU THỨC: \(P=\left(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}\right)\left(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}\right)\)
ĐKXĐ: \(x\ge1\); x khác 2; 3
Ta có:
\(\frac{1}{\sqrt{x}-\sqrt{x-1}}=\frac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}=\sqrt{x}+\sqrt{x-1}\)
\(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-1-2}=\sqrt{x-1}+\sqrt{2}\)
=> \(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x}+\sqrt{x-1}-\left(\sqrt{x-1}+\sqrt{2}\right)=\sqrt{x}-\sqrt{2}\)
\(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}=\frac{2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
=> \(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{2-\sqrt{x}}{\sqrt{x}}\)
rút gọn
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(\Leftrightarrow A=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)
Dòng 3 thêm cho mik dấu tương đương ở đầu dòng hộ :> Xin lỗi vì sơ suất
rút\:gọn\:\frac{2\sqrt{x}}{\sqrt{x}-2}-\frac{5\sqrt{x}-2}{x-2\sqrt{x}}-\frac{\sqrt{x}+1}{\sqrt{x}}
\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-2}{x-2\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}}\left(x>0;x\ne4\right)\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{5\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2x-5\sqrt{x}+2-x+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)