\(\left(\dfrac{\sqrt{x}}{2\sqrt{x}-2}+\dfrac{3-\sqrt{x}}{2x-2}\right):\left(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{x\sqrt{x}-1}\right)\)
1) Rút gọn
2)Tìm x để P=3
3)Tính P tại x = 15+\(6\sqrt{6}\)
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Những câu hỏi liên quan
7 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2\)
\(=\dfrac{x-2\sqrt{x}+1-\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{4x}\)
\(=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{4x}\)
\(=\dfrac{-4\sqrt{x}\cdot\left(x-1\right)}{4x}\)
\(=\dfrac{-x+1}{\sqrt{x}}\)
b) Để P=2 thì \(-x+1=2\sqrt{x}\)
\(\Leftrightarrow-x+1-2\sqrt{x}=0\)
\(\Leftrightarrow x+2\sqrt{x}-1=0\)
\(\Leftrightarrow x+2\sqrt{x}+1-2=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=\sqrt{2}\\\sqrt{x}+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{2}-1\\\sqrt{x}=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\Leftrightarrow x=3-2\sqrt{2}\)
Vậy: Để P=2 thì \(x=3-2\sqrt{2}\)
5 tháng 5 2021
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: \(B=\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\right):\left(\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\sqrt{x}-6+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=1\)
5 tháng 5 2021
Câu 3:
Ta có: \(Q=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)
\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)
\(=\dfrac{a+\sqrt{a}}{\sqrt{a}-2}\cdot\dfrac{\sqrt{a}-2}{\sqrt{a}+1}\cdot\dfrac{\sqrt{a}-2}{1}\)
\(=\sqrt{a}\left(\sqrt{a}-2\right)\)
\(=a-2\sqrt{a}\)
20 tháng 6 2023
a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1
23 tháng 8 2021
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
23 tháng 8 2021
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: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
11 tháng 5 2022
\(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{2\left(2\sqrt{x}-3\right)-\left(\sqrt{x}-1\right)}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right):\left(\dfrac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\right).\left(\dfrac{2\sqrt{x}-3}{2\sqrt{x}+1}\right)\)
\(P=\dfrac{\left(3\sqrt{x}-5\right)\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(2\sqrt{x}+1\right)}\)
\(P=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
7 tháng 2 2021
a) Ta có: \(P=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)
\(=\left(\dfrac{x+1}{x+1}+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{x+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{x-2\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{1}\cdot\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b) Để \(P=5\) thì \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}=5\)
\(\Leftrightarrow x+\sqrt{x}+1=5\left(\sqrt{x}-1\right)\)
\(\Leftrightarrow x+\sqrt{x}+1=5\sqrt{x}-5\)
\(\Leftrightarrow x+\sqrt{x}+1-5\sqrt{x}+5=0\)
\(\Leftrightarrow x-4\sqrt{x}+6=0\)
\(\Leftrightarrow x-2\cdot\sqrt{x}\cdot2+4+2=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+2=0\)(Vô lý)
Vậy: Không có giá trị nào của x để P=5
AT
10 tháng 7 2021
a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{1}{x-1}\)
7 tháng 2 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
a) Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2x-\sqrt{x}+2\sqrt{x}+2\)
\(=2-x\)
b) Để P=3 thì 2-x=3
hay x=-1(Không thỏa mãn ĐKXĐ)
Vậy: Không có giá trị nào của x để P=3
c) Thay \(x=7+2\sqrt{3}\) vào P, ta được:
\(P=2-7-2\sqrt{3}=-5-2\sqrt{3}\)
Vậy: Khi \(x=7+2\sqrt{3}\) thì \(P=-5-2\sqrt{3}\)
1) Ta có: \(P=\left(\dfrac{\sqrt{x}}{2\sqrt{x}-2}+\dfrac{3-\sqrt{x}}{2x-2}\right):\left(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}+\dfrac{\sqrt{x}+2}{x\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}}{2\left(\sqrt{x}-1\right)}+\dfrac{3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x-2\sqrt{x}+1+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\)
\(=\dfrac{x+\sqrt{x}+3-\sqrt{x}}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+3}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x-\sqrt{x}+3}\)
\(=\dfrac{\left(x+3\right)\left(x+\sqrt{x}+1\right)}{2\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+3\right)}\)
B=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{6}{\sqrt{x}-1}-\dfrac{\sqrt{x}+15}{x+2\sqrt{x}-3}\) Bạn ơi giúp mk câu này vs ạ !