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12 tháng 9 2016
Ta có \(\frac{2\cos^2-1}{\sin+\cos}=\frac{\cos^2-\sin^2}{\sin+\cos}=\frac{\left(\cos+\sin\right)\left(\cos-\sin\right)}{\sin+\cos}\)
= cos - sin
12 tháng 9 2018
\(A=2\sqrt{27}-\sqrt{75}-\sqrt{\frac{4}{3}}\)\(=2\sqrt{9.3}-\sqrt{25.3}-\sqrt{\frac{4.3}{9}}\)\(=2.3\sqrt{3}-5\sqrt{3}-\frac{2}{3}\sqrt{3}\)\(=6\sqrt{3}-5\sqrt{3}-\frac{2}{3}\sqrt{3}\)\(=\frac{1}{3}\sqrt{3}\)\(=\frac{\sqrt{3}}{3}\)
10 tháng 9 2020
ĐKXĐ: x \(\ge\)0; x khác 9 (1)
a) B = \(\frac{1}{3-\sqrt{x}}+\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{x-9}\)
B = \(\frac{-\left(\sqrt{x}+3\right)+\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
B = \(\frac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
B = \(\frac{-4\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
B = \(\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\)
B = \(\frac{4}{3-\sqrt{x}}\)
b) B > A <=> \(\frac{4}{3-\sqrt{x}}>1\) <=> \(\frac{4}{3-\sqrt{x}}-1>0\)
<=> \(\frac{4-3+\sqrt{x}}{3-\sqrt{x}}>0\)
<=> \(\frac{\sqrt{x}+1}{3-\sqrt{x}}>0\)
Do \(\sqrt{x}+1>0\) => \(3-\sqrt{x}>0\) <=> \(\sqrt{x}< 3\)
<=> \(x< 9\)
Kết hợp với đk (1)
=> \(0\le x< 9\)
AP
11 tháng 7 2019
\(\left(\sqrt{5}+2\right).\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\left(\sqrt{5}+2\right).\sqrt{17-4\sqrt{\left(\sqrt{5}\right)^2+2.2.\sqrt{5}+2^2}}=\left(\sqrt{5}+2\right).\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}=\left(\sqrt{5}+2\right).\sqrt{17-4.\sqrt{5}-8}=\left(\sqrt{5}+2\right).\sqrt{9-4\sqrt{5}}=\left(\sqrt{5}+2\right).\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}=\left(\sqrt{5}+2\right).\sqrt{\left(\sqrt{5}-2\right)^2}=\left(\sqrt{5}+2\right).\left(\sqrt{5}-2\right)=\left(\sqrt{5}\right)^2-4=5-4=1\)
Y
3 tháng 7 2019
\(\sqrt{15-10\sqrt{2}}\\ =\sqrt{15-2\cdot\sqrt{5}\cdot\sqrt{5}\cdot\sqrt{2}}\\ =\sqrt{\left(\sqrt{10}\right)^2-2\cdot\sqrt{10}\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}\\ =\sqrt{10}-\sqrt{5}\\ =\sqrt{5}\left(\sqrt{2}-1\right)\)
27 tháng 5 2018
9 T I C H sai buồn
\(A=\frac{\sqrt{x^3}}{\sqrt{xy}-2y}-\frac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}.\frac{1-x}{1-\sqrt{x}}..\)
nhờ vào năng lực rinegan tối hậu của ta , ta có thể dễ dàng nhìn thấy mẫu chung
\(x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=\sqrt{x}\left(\sqrt{x}-2\sqrt{xy}\right)+\left(\sqrt{x}-2\sqrt{y}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+1\right)\)
\(A=\frac{\sqrt{x^3}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}-\frac{2x\left(x-1\right)}{\left(\sqrt{x}-2\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}.\)
\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
\(A=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\sqrt{x}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\left(\sqrt{x}-2\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x}{\sqrt{y}}\)
b) thay y=625 vào ta được
\(\frac{x}{\sqrt{625}}=\frac{x}{25}< 0.2\Leftrightarrow x< 5\)
vậy \(0< x< 5\)
TT
1
12 tháng 12 2023
a: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}\)
\(=\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}\)
\(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}=\sqrt{20}=2\sqrt{5}\)
c: \(\dfrac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}\)
\(=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{10}+\sqrt{6}}{5-3}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
e: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right)\cdot\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
=6-121
=-115
g: \(\dfrac{1}{\sqrt{5}+1}+\dfrac{1}{\sqrt{5}-2}-\dfrac{1}{3-\sqrt{5}}-\sqrt{5}\)
\(=\dfrac{\sqrt{5}-1}{4}+\dfrac{\sqrt{5}+2}{1}-\dfrac{1\left(3+\sqrt{5}\right)}{4}-\sqrt{5}\)
\(=\dfrac{1}{4}\sqrt{5}-\dfrac{1}{4}+\sqrt{5}+2-\dfrac{3}{4}-\dfrac{\sqrt{5}}{4}-\sqrt{5}\)
\(=-1+2=1\)
HK
3
HP
6 tháng 2 2022
\(X=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{2-\sqrt{x}}\left(đk:x\ge0;x\ne4\right)\)
\(X=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-2}\)
\(X=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(X=\dfrac{3+2\sqrt{x}-4-\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(X=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(X=\dfrac{1}{\sqrt{x}+1}\)
HP
6 tháng 2 2022
\(S=\left(\dfrac{1}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\right)\left(đk:x\ge0;x\ne1\right)\)
\(S=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{1-\sqrt{x}}{x+4\sqrt{x}+4}\right)\)
\(S=\dfrac{\sqrt{x}-2+x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{x+4\sqrt{x}+4}{1-\sqrt{x}}\)
\(S=\dfrac{x+3\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{1-\sqrt{x}}\)
\(S=\dfrac{\left(x+3\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(1-\sqrt{x}\right)}\)
(đến đoạn này thì trong ngoặc ko tách ra đc nữa nên mik nghĩ là đến đây là xong, nếu sai thì bn nói mik)
a) Ta có: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=2\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2a-4}{a+2}\)