rút gọn bt : \(C=\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\)
\(C=\left(1\cdot \frac{x-3\sqrt{x}}{x-9}\right)chia\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{ }x}-6\right)\)
cái bên trên là \(\frac{9-x}{x+\sqrt{x}-6}\) nha chứ không phải là \(\frac{9-x}{x+\sqrt{x}}-6\)
Rút gọn bt C
RÚT GỌN BT:
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\) với x lớn hơn 0; x khác 9;x khác 25
`Answer:`
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{9-x}\right):\left(\frac{\sqrt{x}-1}{x-3\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\left(ĐK:x>0;x\ne9;x\ne25\right)\)
\(=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=-\frac{3\sqrt{x}-x+2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)
\(=-\frac{\sqrt{x}\left(3+\sqrt{x}\right)}{3+\sqrt{x}}.\frac{\sqrt{x}}{5-\sqrt{x}}\)
\(=-\sqrt{x}.\frac{\sqrt{x}}{5-\sqrt{x}}\)
\(=\frac{x}{\sqrt{x}-5}\)
rút gọn bt Q=\(\frac{\sqrt{x}-3}{2+\sqrt{x}}\)+\(\frac{\sqrt{x}-2}{3+\sqrt{x}}\)\(-\)\(\frac{9-x}{x+\sqrt{x}-6}\)
Tìm đk và rút gọn:
A=\(\frac{x+3+2\sqrt{x^2-9}}{-2x+6+\sqrt{x^2-9}}\)
Rút gọn biểu thức:
\(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\left(x>3\right)\)
Rút gọn biểu thức
\(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\)
\(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}=\frac{\left(\sqrt{x+3}\right)^2+2\sqrt{x+3}\sqrt{x-3}}{2.\left(\sqrt{x-3}\right)^2+\sqrt{x+3}\sqrt{x-3}}\)
\(=\frac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2\sqrt{x-3}+\sqrt{x+3}\right)}=\frac{\sqrt{x+3}}{\sqrt{x-3}}\)
\(=\frac{\sqrt{x^2-9}}{x-3}\)
Rút gọn biểu thức
A = \(\frac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\)
B = \(\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
Giúp mk vs , mk đang cần gấp
Rút gọn biểu thức : A= \(\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\)
\(A=\dfrac{x+3+2\sqrt{x^2-9}}{2x-6+\sqrt{x^2-9}}\left(x>3\right)\\ A=\dfrac{\left(x+3\right)+2\sqrt{\left(x-3\right)\left(x+3\right)}}{2\left(x-3\right)+\sqrt{\left(x-3\right)\left(x+3\right)}}\\ A=\dfrac{\sqrt{x+3}\left(\sqrt{x+3}+2\sqrt{x-3}\right)}{\sqrt{x-3}\left(2+\sqrt{x+3}\right)}\)
Tới đây chịu rùi, hình như đề sai đk?
cho biểu thức M=\(\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\) : \(2\sqrt{1+\frac{2x}{3-x}}\)
rút gọn M