Rút gọn (Giải chi tiết từng bước với ạ)
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\) (ĐK: x≥0;x≠9)
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NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 9 2021
ĐKXĐ: \(x\ge0;x\ne3\)
\(B=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3\sqrt{x}-3}{x-9}\)
27 tháng 5 2023
a: \(B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{x-9}=\dfrac{x+9}{x-9}\)
b: \P=A:B
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{x-9}{x+9}=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{x+9}>=\dfrac{-1\cdot3}{9}=\dfrac{-1}{3}\)
Dấu = xảy ra khi x=0
14 tháng 8 2023
Sửa đề: \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}\)
\(=\dfrac{x+3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\dfrac{9\sqrt{x}-9}{x-9}\)
26 tháng 7 2021
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)
=2
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
Y
23 tháng 3 2023
\(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x}{3\sqrt{x}-x}\right).\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\left(dkxd:x\ne0;\pm\sqrt{3}\right)\)
\(=\left(\dfrac{2}{\sqrt{x}-3}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-3\right)\)
\(=\left(\dfrac{2\sqrt{x}-x}{\sqrt{x}\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-3\right)\)
\(=\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}}\)
\(=2-\sqrt{x}\)
Vậy \(A=2-\sqrt{x}\)
9 tháng 9 2021
\(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3\sqrt{x}+1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\\ D=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ D=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
23 tháng 9 2021
\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+7}{4-x}\left(x>0;x\ne4\right)\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ P=\dfrac{\sqrt{x}+6-x-x-3\sqrt{x}-2+2\sqrt{x}+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x+11}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\sqrt{x}}\\ P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}\left(x-4\right)}\)
\(P=\dfrac{-2x\sqrt{x}+11\sqrt{x}+x\sqrt{x}-4\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\\ P=\dfrac{-x\sqrt{x}+8\sqrt{x}+2x-8}{\sqrt{x}\left(x-4\right)}\)
2
3 tháng 7 2023
`\sqrt{[27(x-1)^2]/12} +3/2 - (x - 2)\sqrt{[50x^2]/[8(x-2)^2]}` `(1 < x < 2)`
`=\sqrt{[3(x-1)]^2 .3}/\sqrt{2^2 .3} + 3/2 - (x - 2) \sqrt{(5x)^2 . 2}/\sqrt{[2(x - 2)]^2 . 2}`
`=[3\sqrt{3}|x-1|]/[2\sqrt{3}]+3/2-(x-2)[5\sqrt{2}|x|]/[2\sqrt{2}|x-2|]`
`=[3(x-1)]/2+3/2-[5x(x-2)]/[2(2-x)]` (Vì `1 < x < 2`)
`=3/2x - 3/2 + 3/2 + 5/2x`
`=4x`
Y
12 tháng 3 2023
\(\dfrac{2\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)^2}\)
\(=\dfrac{2x+4\sqrt{x}}{x-4\sqrt{x}+4}\)
Với x ≥ 0; x ≠ 9 ta có:
\(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x-3}\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
Vậy \(A=\dfrac{3}{\sqrt{x}+3}\).
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