- a)\(\sqrt{x+1}\)=x
b)\(\sqrt{x}\)=x
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
13 tháng 11 2021
Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
13 tháng 11 2021
a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
HS
1 tháng 8 2018
Câu (A) đề có sao không nhỉ?
\(B=\dfrac{1}{a^2-\sqrt{x}}:\dfrac{\sqrt{a}+1}{a\sqrt{a}+a+\sqrt{a}}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}.\left(a\sqrt{a}-1\right)}.\dfrac{a\sqrt{a}+1+\sqrt{a}}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}.\left(\sqrt{a}-1\right).\left(a+\sqrt{a}+1\right)}.\dfrac{\sqrt{a}.\left(a+\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{a}-1}.\dfrac{1}{\sqrt{a}+1}\)
\(\Leftrightarrow\dfrac{1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)
\(\Leftrightarrow\dfrac{1}{a-1}\)
HS
1 tháng 8 2018
\(E=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x+1}{\sqrt{x}}\)
1 tháng 2 2021
Ta có: \(P=\dfrac{A}{B}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{\sqrt{x}}:\left(\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{\sqrt{x}}:\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
9 tháng 7 2022
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
H
11 tháng 7 2023
\(A+B=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\left(\text{đ}pcm\right)\)
11 tháng 7 2023
A+B
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
14 tháng 4 2023
a: Khi x=25 thì \(A=\dfrac{7\cdot5-2}{5-2}=\dfrac{33}{3}=11\)
b: P=A*B
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x}-1}-\dfrac{4\sqrt{x}}{x-1}\right)\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+2\sqrt{x}+2-4\sqrt{x}}{x-1}\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{x-3\sqrt{x}+2}{x-1}\cdot\dfrac{7\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\cdot\left(7\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{7\sqrt{x}-2}{\sqrt{x}+1}\)
31 tháng 8 2021
a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=1-\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=1-\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)
b: Ta có: P=A:B
\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)