Cho biểu thứ
\(\left(\sqrt{1999}+\sqrt{1997}+...+\sqrt{1}\right)-\left(\sqrt{1998}+\sqrt{1996}+..+\sqrt{2}\right).\)
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AT
2 tháng 7 2021
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
2 tháng 7 2021
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=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: \(A=\left(\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}-1}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
17 tháng 8 2021
1, A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
ĐKXĐ: a≥0
A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)+1\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1}{a+1}-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right).\left(\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{a+1-2\sqrt{a}}\right)\)
A=\(\sqrt{a}+1\)
Vậy A=\(\sqrt{a}+1\)
2, a=1996-2\(\sqrt{1995}\)
a=\(1995-2\sqrt{1995}+1\)
a=\(\left(\sqrt{1995}-1\right)^2\) (TMĐKXĐ)
thay a=\(\left(\sqrt{1995}-1\right)^2\) vào A ta có:
A=\(\sqrt{\left(\sqrt{1995}-1\right)^2}+1\)
A=\(\sqrt{1995}\)
Vậy a=1996-2\(\sqrt{1995}\) thì A=\(\sqrt{1995}\)
25 tháng 7 2021
\(A=\left(\sqrt{x}+\dfrac{4\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4}{2\sqrt{x}-x}\right)\)ĐK : x > 0 ; x \(\ne\)4
\(=\left(\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}\right):\left(\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)=\dfrac{x\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(x-4\right)}\)
\(=\dfrac{x}{\sqrt{x}-2}\)
6 tháng 4 2022
\(P=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
