Rút gọn P= \(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{2\sqrt{x}-2+\sqrt{x}-1}\)
Câu 1: Rút gọn biểu thức: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{2}}+\dfrac{6}{x+3\sqrt{x}}\right)\) với x > 0
Câu 2: Rút gọn biểu thức:
\(P=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\) với x > 0; x \(\ne\) 2
Câu 3: Rút gọn biểu thức:
\(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\) với a > 0; a \(\ne\) 4
Câu 1:
Sửa đề: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
Ta có: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
Rút gọn:
1) \(P=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
2) \(P=\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
Giúp mk nhé :3
Rút gọn:
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right).\dfrac{a-4}{\sqrt{4a}}\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
Làm chi tiết giúp mình với vì mình yếu phần này lắm
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}-2\sqrt{3}=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\\x\ne4\end{matrix}\right.\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(P=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{3}\)
\(P=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}a>0\\a\ne4\end{matrix}\right.\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right)\cdot\dfrac{a-4}{\sqrt{4a}}\)
\(M=\dfrac{a+2\sqrt{a}+a-2\sqrt{a}}{a-4}\cdot\dfrac{a-4}{2\sqrt{a}}\)
\(M=\dfrac{2a}{2\sqrt{a}}=\sqrt{a}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(N=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(N=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4-x+9+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(N=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(N=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
+) \(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne9\\x\ne4\end{matrix}\right.\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
\(Q=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\right)\)
\(Q=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4+\sqrt{x}-8-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(Q=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)
\(Q=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
p/s: sorry tại n' câu wa nên mình ko làm chi tiết đc =(( lần sau nhớ chia các câu ra cho dễ nhìn hơn nha, đánh hơi mỏi tay :'( có j ko hỉu cmt dưới nha
Rút gọn:
1) \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
2)\(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
3) \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
4) \(A=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
Mng giúp e vs ạ, cần gấp :<
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
(\(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{x+\sqrt{x}-2}\))
rút gọn có ĐKXĐ !!!
\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{x+\sqrt{x}-2}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}+2+x-\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}:\dfrac{2x-\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{2x-\sqrt{x}-3}=\dfrac{x+8}{2x-\sqrt{x}-3}\)
Ta có: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{x+\sqrt{x}-2}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{x-\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}+2+x-\sqrt{x}+6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}:\dfrac{x-1+x-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+8}{2x-\sqrt{x}-3}\)
rút gọn A
A = \(\left(\dfrac{2}{x-\sqrt{x}}+\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{2\sqrt{x}-x}\)
\(ĐK:x>0;x\ne1\\ A=\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\\ A=\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}=\dfrac{\left(1-\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
\(A=\left(\dfrac{2+x-1-x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}+1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Bài: Rút gọn biểu thức sau
1)(1-\(\dfrac{\sqrt{x}}{1-\sqrt{x}}\)):(\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)-\(\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\)+\(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\))
2)(\(\dfrac{1}{\sqrt{x}+1}\)-\(\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\)):(\(\dfrac{1}{\sqrt{x}-1}\)-\(\dfrac{2}{x-1}\))
1: \(=\left(1+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}-1}:\dfrac{x-9+x-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(2x+\sqrt{x}-11\right)}\)
2: \(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\dfrac{\sqrt{x}+1-2}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
2 a. rút gọn biểu C = \(\dfrac{2x^{\text{2}}-x}{\text{x }-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
b. Rút gọn biểu thức D = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{\text{a}}-1}\right):\dfrac{\sqrt{\text{a}}+1}{a-2\sqrt{a}+1}\)
Vậy khi rút gọn một biểu thức hửu tỉ và một biểu thức chứa căn có tìm điều kiện xác định không?
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
Rút gọn các biểu thức sau:
1.\(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
2.\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
Rút gọn các biểu thức sau:
1.\(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
2.\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
1: Ta có: \(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-x-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\sqrt{x}}{x+\sqrt{x}+1}\)
2) Ta có: \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=-\left(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)