\(\sqrt{a+2\sqrt{a-1}}\)\(+\)\(\sqrt{a-2\sqrt{a-1}}\)\(\left(1\le a\le2\right)\)Rút gọn biểu thức
K
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Những câu hỏi liên quan
5 tháng 5 2021
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
5 tháng 5 2021
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
28 tháng 1 2020
Đặt: \(a=\sqrt{2+x};b=\sqrt{2-x}\left(a,b\ge0\right)\)
\(\Rightarrow\hept{\begin{cases}a^2+b^2=4\\a^2-b^2=2x\end{cases}}\)
\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a^3-b^3\right)}{4+ab}=\frac{\sqrt{2+ab}\left(a-b\right)\left(a^2+b^2+ab\right)}{4+ab}\)
\(\Rightarrow A=\frac{\sqrt{2+ab}\left(a-b\right)\left(4+ab\right)}{4+ab}=\sqrt{2+ab}\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=\sqrt{4+2ab}\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=\sqrt{\left(a^2+b^2+2ab\right)}\left(a-b\right)=\left(a+b\right)\left(a-b\right)\)
\(\Rightarrow A\sqrt{2}=a^2-b^2=2x\)
\(\Rightarrow A=x\sqrt{2}\)
LD
26 tháng 7 2021
A=\(\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(a-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\left(a+\sqrt{a}\right)}{\left(a-1\right)}\right]\)::::::::\(\left(\dfrac{\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
=\(\left[\dfrac{1}{\sqrt{a}-1}\right]:\left(\dfrac{2\sqrt{a}}{a-1}\right)\)=\(\dfrac{\sqrt{a}-1}{2\sqrt{a}}\)
=\(\dfrac{a^2+a\sqrt{a}+11a+6}{2\sqrt{a}\left(\sqrt{a}+2\right)}\)
26 tháng 7 2021
Ta có: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
26 tháng 12 2021
a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
27 tháng 12 2021
\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)
H
23 tháng 8 2023
ĐK: \(a\ge0;a\ne1\)
Biểu thức trở thành:
\(\left(\dfrac{1-\sqrt{a}^3}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{\left(1-\sqrt{a}\right)\left(1-\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)\\ =\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1-\sqrt{a}}{1+\sqrt{a}}\\ =\left(1+\sqrt{a}+a+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1+\sqrt{a}}\right)\\ =\left(1+2\sqrt{a}+a\right).\left(\dfrac{1-\sqrt{a}}{1+\sqrt{a}}\right)\\ =\dfrac{\left(1+\sqrt{a}\right)^2\left(1-\sqrt{a}\right)}{1+\sqrt{a}}\\ =\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\\ =1-a\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1
\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)
\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)
11 tháng 1
\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)
\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)
\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)
b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)
\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)
\(=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{2}}=a^2\)
15 tháng 8 2023
\(=\dfrac{a^{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}{a^{\left(\sqrt{5}-1\right)+\left(3-\sqrt{5}\right)}}=\dfrac{a}{a^{\sqrt{5}-1+3-\sqrt{5}}}=\dfrac{a}{a^2}=\dfrac{1}{a}\)
