Rút gọn biểu thức
N=\(\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\)(với x>0, x\(\ne\)2)
Giúp mình với
Rút gọn biểu thức:
A=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\) với x\(\ge\)0,x\(\ne\)4,x\(\ne\)9
`A=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
`A=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4, x ne 9)`
`=(2\sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9-x+9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(x-sqrtx-2)/(x-5sqrtx+6)`
`=((\sqrtx+1)(sqrtx-2))/((sqrtx-2)(sqrtx-3))`
`=(sqrtx+1)/(sqrtx-3)`
Rút gọn biểu thức:
B=\(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{x-2\sqrt{x}}\) với x>0;x\(\ne\)4,x\(\ne\)16
\(=>B=\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-6\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)
Ta có: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{x-2\sqrt{x}}\)
\(=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-6\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
Rút gọn biểu thức:
A=\(\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x-\sqrt{x}}\right).\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\)với x>0;x\(\ne\)1
Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x-\sqrt{x}}\right)\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\left(\sqrt{x}-2\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}-2\right)\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
1.Giải hệ phương trình:
\(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)
2.Rút gọn biểu thức:
B=\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{1}{\sqrt{x}+2}\)với x>0;x\(\ne\)9
1) Ta có: \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\2x+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x=5-y=5-\left(-1\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
2) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+3\sqrt{x}+2+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x-2\sqrt{x}+2x-4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)
\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}-2}\)
\(=3\sqrt{x}\)
Cho biểu thức A= \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\) với x>0 và x\(\ne\)1. Rút gọn biểu thức A
Sửa đề: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2}{x-1}\)
Ai giúp em với em đang cần gấp :
Đây là bài rút gọn biểu thức:
A=\(\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}\)\(\left(x\ne\pm\sqrt{2}\right)\)
B=\(\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}\left(x\ne-\sqrt{5}\right)\)
Khó quá em không nghĩ ra mọi người giúp em nhé
Em cảm ơn
\(A=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)
\(B=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
rút gọn các biểu thức sau
a.A=\(\dfrac{4}{3+\sqrt{7}}+\sqrt{28}\)
b.B=\(\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}+1}-\dfrac{2}{\sqrt{x}}\right):\dfrac{2-\sqrt{x}}{x-1}\) (với x>0; x\(\ne\)1; x\(\ne4\))
a, \(A=\dfrac{4\left(3-\sqrt{7}\right)}{2}+2\sqrt{7}=\dfrac{12}{2}=6\)
b, \(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\sqrt{x}}\right):\dfrac{2-\sqrt{x}}{x-1}\)
\(=\left(\dfrac{\sqrt{x}-2\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\dfrac{2-\sqrt{x}}{x-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\left(\dfrac{2-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\dfrac{x-1}{2-\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2-\sqrt{x}\right)\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
đây bạn nhé
rút gọn các biểu thức
a)P=\(\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
b)Q=\(\left(1+\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}\) với x>0, x\(\ne\)4
a) \(P=\dfrac{1}{2-\sqrt{3}}+\dfrac{1}{2+\sqrt{3}}\)
\(=\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{4}{4-3}\)
\(=4\)
b) \(Q=\left(1+\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}vớix>0,x\ne4\)
\(=\left(\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}}\)
\(=\)\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2}{\sqrt{x}-2}\)
Rút gọn biểu thức:
M=(\(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\)).\(\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)với x\(\ge\)0;x\(\ne\)1
Ta có: \(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\cdot\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\left(\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)