\(x+1+\sqrt{x-1}=2\sqrt{x}\)
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Những câu hỏi liên quan
23 tháng 8 2021
1: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=x-\sqrt{x}+1\)
23 tháng 8 2021
2: Ta có: \(A=\left(\dfrac{x+2\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{x+2\sqrt{x}-x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{1}{x-1}\)
3: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
12 tháng 8 2023
1: \(=\left(1+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{\sqrt{x}-1+\sqrt{x}}{\sqrt{x}-1}:\dfrac{x-9+x-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{2x+\sqrt{x}-11}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(2x+\sqrt{x}-11\right)}\)
2: \(=\dfrac{x-1-2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-1\right)}:\dfrac{\sqrt{x}+1-2}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
11 tháng 5 2022
Câu 3:
\(L=\left(\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)^2\cdot\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
\(=\dfrac{a-\sqrt{a}-2-\left(a+\sqrt{a}-2\right)}{a-1}\cdot\dfrac{1}{\sqrt{a}}=\dfrac{-2}{a-1}\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 4 2022
\(y'=\dfrac{\left(x+\sqrt{x^2+1}\right)'}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}.\sqrt{x+\sqrt{x^2+1}}}\)
\(=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 8 2023
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5x-7\sqrt{x}+2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=-\dfrac{5\sqrt{x}-2}{\sqrt{x}+3}\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 8 2023
\(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\) (ĐK: \(x\ge0\))
\(=\dfrac{\left(\sqrt{x}\right)^3+1^3}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=x-\sqrt{x}+1\)
______________
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\) (ĐK: \(x\ge0;x\ne9\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
17 tháng 8 2021
1: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
Lời giải:
ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow (x-2\sqrt{x}+1)+\sqrt{x-1}=0$
$\Leftrightarrow (\sqrt{x}-1)^2+\sqrt{x-1}=0$
Ta thấy: $(\sqrt{x}-1)^2\geq 0; \sqrt{x-1}\geq 0$ với mọi $x\geq 1$
Do đó để tồng trên bằng $0$ thì:
$(\sqrt{x}-1)^2=\sqrt{x-1}=0$
$\Leftrightarrow x=1$ (thỏa mãn).