Câu hỏi của Mỹ Hạnh

Question

Bài tập Toán Giải giúp em với ạ!

Nguyen Quynh Huong · Accepted Answer

a, $\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2$

$=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}$

$=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}$

$=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}-y\sqrt{y}+y\sqrt{x}+x\sqrt{y}}{\sqrt{x}+\sqrt{y}}$

$=\dfrac{y\sqrt{x}+x\sqrt{y}}{\sqrt{x}+\sqrt{y}}$

b, $\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}$

$=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$

Câu trả lời:

a, $\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2$

$=\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}$

$=\dfrac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)\left(x-y\right)}{\sqrt{x}+\sqrt{y}}$

$=\dfrac{x\sqrt{x}+y\sqrt{y}-x\sqrt{x}-y\sqrt{y}+y\sqrt{x}+x\sqrt{y}}{\sqrt{x}+\sqrt{y}}$

$=\dfrac{y\sqrt{x}+x\sqrt{y}}{\sqrt{x}+\sqrt{y}}$

b, $\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}$

$=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$