Toán 9

I

iu278

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T

thinhyamai

không chép lại đầu bài nhé !

[TEX]\frac{15\sqrt[]{x}-11}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}[/TEX]+[TEX]\frac{-3\sqrt[]{x}+3}{\sqrt[]{x}-1}-\frac{\sqrt[]{2}x+3}{\sqrt[]{x}+3}[/TEX]
[TEX]=\frac{15\sqrt[]{x}-11-3x-9\sqrt[]{x}+2\sqrt[]{x}+6-\sqrt[]{2}x\sqrt[]{x}+\sqrt[]{2}x-3\sqrt[]{x}+3}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}[/TEX]
[TEX]=\frac{5\sqrt[]{x}-2-3x+\sqrt[]{2}x-\sqrt[]{2}x\sqrt[]{x}}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}[/TEX]
[TEX]=\frac{-3x+5\sqrt[]{x}-2-\sqrt[]{2}x(\sqrt[]{x}-1)}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}[/TEX]
[TEX]=\frac{-3\sqrt[]{x}(\sqrt[]{x}-1)+2(\sqrt[]{x}-1)-\sqrt[]{2}x(\sqrt[]{x}-1)}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}[/TEX]
[TEX]=\frac{(\sqrt[]{x}-1)(2-3\sqrt[]{x}-\sqrt[]{2}x)}{(\sqrt[]{x}-1)(\sqrt[]{x}+3)}[/TEX]
[TEX]=\frac{2-3\sqrt[]{x}-\sqrt[]{2}x}{\sqrt[]{x}+3}[/TEX]
 
M

minhmai2002

a, ĐKXĐ: $x\ne 1;x$\geq$0$

b,$B=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}$

$B=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}+\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}$

$<=>B=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-2x-\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-1)}$

$<=>B=\frac{x+21\sqrt{x}-14}{(\sqrt{x}+3)(\sqrt{x}-1)}$

 
I

iu278

a, ĐKXĐ: $x\ne 1;x$\geq$0$

b,$B=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}$

$B=\frac{15\sqrt{x}-11}{(\sqrt{x}-1)(\sqrt{x}+3)}+\frac{(3\sqrt{x}-2)(\sqrt{x}+3)}{(\sqrt{x}-1)(\sqrt{x}+3)}-\frac{(2\sqrt{x}+3)(\sqrt{x}-1)}{(\sqrt{x}+3)(\sqrt{x}-1)}$

$<=>B=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-2x-\sqrt{x}+3}{(\sqrt{x}+3)(\sqrt{x}-1)}$


$<=>B=\frac{x+21\sqrt{x}-14}{(\sqrt{x}+3)(\sqrt{x}-1)}$

Các bn hình như chép sai đề r , bài này có nhìn bài đứa bên cạnh thấy kq khác cơ mà bài nó thì đc cô chữa r
 
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