Rút gọn các biểu thức:
a) \(\sqrt{\dfrac{x-2\sqrt{x+1}}{x+2\sqrt{x+1}}}\)(x lớn hơn hoặc bằng 0)
b) \(\dfrac{x-1}{\sqrt{y-1}}\)\(\sqrt{\dfrac{y-2\sqrt{y+1}}{\left(x-1\right)^2}}\) (x khác 1, y khác 1 và y lớn hơn hoặc bằng 0)
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Những câu hỏi liên quan
17 tháng 1
a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(x+\sqrt{x}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\left(x+\sqrt{x}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\cdot\left(\sqrt{x}+1\right)\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}}{x-1}=\dfrac{2x}{x-1}\)
b: Để Q là số nguyên thì \(2x⋮x-1\)
=>\(2x-2+2⋮x-1\)
=>\(2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;2;3\right\}\)
29 tháng 12 2022
\(P=\left(\dfrac{1}{2\sqrt{x}}-\dfrac{x}{2\sqrt{x}}\right)^2.\left(\dfrac{\left(\sqrt{x}-1\right)^2}{x-1}-\dfrac{\left(\sqrt{x}+1\right)^2}{x-1}\right)\)
\(=\left(\dfrac{1-x}{2\sqrt{x}}\right)^2.\left(\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\dfrac{\left(1-x\right)^2}{2\sqrt{x}}.\dfrac{-4\sqrt{x}}{-\left(1-x\right)}\)
\(=\left(1-x\right).2\sqrt{x}\)
\(=2\sqrt{x}-2x\sqrt{x}\)
24 tháng 10 2023
a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)
\(=1-x\sqrt{x}-x\sqrt{x}\)
\(=1-2x\sqrt{x}\)
b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\left(\dfrac{\left(1-\sqrt{a}\right)\cdot\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\)
\(=\left(\dfrac{1}{\sqrt{a}+1}\right)^2\cdot\left(a+\sqrt{a}+1+\sqrt{a}\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)
Y
17 tháng 8 2023
\(a,A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\\ =2.2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\sqrt{3^2}-1}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =-\dfrac{2\left(\sqrt{3}-1\right)}{2}+\left|\sqrt{3}+1\right|\\ =-\sqrt{3}+1+\sqrt{3}+1\\ =2\)
\(B=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\left(dk:x\ge0,x\ne1\right)\\ =\left(1+\dfrac{\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\right)\left(1-\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{1-\sqrt{x}}\right)\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\\ =1-x\)
\(b,A=4\sqrt{B}\Leftrightarrow4\sqrt{1-x}=2\\ \Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\\ \Leftrightarrow\left|1-x\right|=\dfrac{1}{4}\)
\(\Leftrightarrow1-x=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
Vậy \(x=\dfrac{3}{4}\) thì \(A=4\sqrt{B}\).
H9
HT.Phong (9A5)
CTVHS
17 tháng 8 2023
a) \(A=2\sqrt{20}-\dfrac{2}{\sqrt{3}+1}-\sqrt{80}+\sqrt{4+2\sqrt{3}}\)
\(A=2\cdot2\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}-4\sqrt{5}+\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2}\)
\(A=4\sqrt{5}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-4\sqrt{5}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(A=-\left(\sqrt{3}-1\right)+\sqrt{3}+1\)
\(A=-\sqrt{3}+1+\sqrt{3}+1\)
\(A=2\)
\(B=\left(1+\dfrac{x+\sqrt{x}}{1+\sqrt{x}}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
\(B=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)
\(B=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)\)
\(B=1^2-\left(\sqrt{x}\right)^2\)
\(B=1-x\)
b) Ta có: \(A=4\sqrt{B}\)
\(\Rightarrow2=4\sqrt{1-x}\)
\(\Leftrightarrow\sqrt{1-x}=\dfrac{1}{2}\)
\(\Leftrightarrow1-x=\dfrac{1}{4}\)
\(\Leftrightarrow x=1-\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)
22 tháng 8 2021
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(x-y\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}+x\sqrt{y}+y\sqrt{x}-y\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)( do \(x\ge1\))
22 tháng 8 2021
a: Ta có: \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
\(=\sqrt{xy}\)
b: Ta có: \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\dfrac{ \left|\sqrt{x}-1\right|}{\left|\sqrt{x}+1\right|}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
8 tháng 7 2021
a) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\sqrt{\dfrac{\left(\sqrt{x+1}\right)^2}{\left(\sqrt{x}+1\right)^2}}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1};x\ge0\)
8 tháng 7 2021
b) Ta có: \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}\)
\(=\dfrac{1}{x-1}\)
H
20 tháng 8 2023
\(A=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\\ =\dfrac{x+2-\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}+1-x-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{x+2-2x+4\sqrt{x}-\sqrt{x}-1+x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)^2}\)
H9
HT.Phong (9A5)
CTVHS
20 tháng 8 2023
\(A=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(A=\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(A=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)
\(A=\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(A=\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+1\right)}\)
a) \(\sqrt{\dfrac{x-2\sqrt{x+1}}{x+2\sqrt{x+1}}}\) = \(\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}\) = \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}\)
b) \(\dfrac{x-1}{\sqrt{y}-1}\)\(\sqrt{\dfrac{y-2\sqrt{y+1}}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\) \(\sqrt{\dfrac{\left(y-1\right)^4}{\left(x-1\right)^4}}\)
= \(\dfrac{x-1}{\sqrt{y}-1}\)\(\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^2}\)
= \(\dfrac{\sqrt{y-1}}{x-1}\)
Chúc bạn học tốt :3
Thanks anyway <3