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Nguyễn Huệ Lam
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Nguyễn Huệ Lam
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alibaba nguyễn
10 tháng 8 2017 lúc 15:30

\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)

\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)

\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)

\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)

\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)

alibaba nguyễn
10 tháng 8 2017 lúc 15:35

\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)

\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)

\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)

\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)

alibaba nguyễn
10 tháng 8 2017 lúc 15:57

\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)

Đặt \(\hept{\begin{cases}\sqrt{2+x}=a\\\sqrt{2-x}=b\end{cases}\Rightarrow}a^2+b^2=4\)

\(\Rightarrow C=\frac{\sqrt{2ab}.\left(a^3-b^3\right)}{a^2+b^2+ab}=\frac{\sqrt{2ab}.\left(a-b\right)\left(a^2+b^2+ab\right)}{a^2+b^2+ab}\)

\(=\sqrt{2ab}.\left(a-b\right)=\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{2+x}-\sqrt{2-x}\right)\)

Trang Nguyễn
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An Thy
30 tháng 6 2021 lúc 8:51

a) ĐKXĐ có thêm \(x\ne4\)

 \(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)

\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)

\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)

 \(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)

\(=\dfrac{x+1}{\sqrt{x}+3}:\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+1}{\sqrt{x}+1}\)

Nguyễn Ngọc Lộc
30 tháng 6 2021 lúc 8:55

Hiếu Lê
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Nguyễn Lê Phước Thịnh
23 tháng 9 2021 lúc 20:35

\(A=\dfrac{3x}{x-2}\cdot\sqrt{x^2-4x+4}\)

\(=\dfrac{3x}{x-2}\cdot\left(x-2\right)\)

=3x

\(B=\dfrac{-5y}{x+3}\cdot\sqrt{x^2+6x+9}\)

\(=\dfrac{-5y}{x+3}\cdot\left|x+3\right|\)

\(=\pm5y\)

Nguyễn Quỳnh Anh
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Quynh Existn
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Nguyễn Huy Tú
21 tháng 7 2021 lúc 19:00

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Trần Huyền
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l҉o҉n҉g҉ d҉z҉
19 tháng 9 2020 lúc 14:45

Mình tách thành hai phần nhìn cho dễ hiểu nhé !

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

+) \(\frac{x-3\sqrt{x}}{x-9}-1=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}-1=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}\)

+) \(\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)

\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{9-x+x-9-x+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

=> \(\frac{-3}{\sqrt{x}+3}\div\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{-3}{\sqrt{x}+3}\times\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{4-x}\)

\(=\frac{3\left(\sqrt{x}-2\right)}{x-4}=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3}{\sqrt{x}+2}\)

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Minh Anh Vũ
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Nguyễn Huệ Lam
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katherina
10 tháng 8 2017 lúc 15:29

Hỏi đáp Toán

Khánh Chi Trần
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Nguyễn Lê Phước Thịnh
12 tháng 4 2022 lúc 8:23

\(A=\dfrac{x+3}{\left(x-3\right)^2}:\dfrac{12-x^2+x+x^2-9}{x\left(x-3\right)}\)

\(=\dfrac{x+3}{\left(x-3\right)^2}\cdot\dfrac{x\left(x-3\right)}{x+3}=\dfrac{x}{x-3}\)