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11 tháng 10 2015
\(\left(\sqrt{x+5}-\sqrt{x}\right)\left(\sqrt{x+5}+\sqrt{x}\right)=\sqrt{x+5}+\sqrt{x}\)
=> \(x+5-x=M\Rightarrow M=5\)
b ) tương tự
11 tháng 10 2015
b) N.N' = \(\left(\sqrt{25-x^2}-\sqrt{15-x^2}\right).\left(\sqrt{25-x^2}+\sqrt{15-x^2}\right)=\left(25-x^2\right)-\left(15-x^2\right)=10\)
=> 2.N = 10 => N = 10:2 =5
1 tháng 7 2023
Sửa đề: căn x-5/căn x-3
a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\)
b: x-5căn x+6=0
=>căn x=2 hoặc căn x=3
=>x=9(loại) hoặc x=4(nhận)
Khi x=4 thì A=5/(2+3)=5/5=1
HT
1
7 tháng 11 2016
Ta có
\(\left(\sqrt{25-x^2}+\sqrt{15-x^2}\right)\left(\sqrt{25-x^2}-\sqrt{15-x^2}\right)=25-x^2-15+x^2=10\)
=> Số cần tìm bằng 5
15 tháng 8 2021
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
15 tháng 8 2021
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
15 tháng 8 2021
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
24 tháng 8 2021
Ta có: \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5}{\sqrt{x}+3}\)
7 tháng 7 2021
đK: \(x\ge0;x\ne25;x\ne9\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right]:\left[\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right]\)
\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right]:\dfrac{25-x-\left(x-9\right)+\left(x-25\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\sqrt{x}-3}{\sqrt{x}+5}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{\sqrt{x}+5}{-\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\)
17 tháng 2 2021
a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)
Ta có: \(A\cdot\left(\sqrt{25-x^2}-\sqrt{15-x^2}\right)=\left(25-x^2-15+x^2\right)=10\)
Do đó A = 10/2 = 5