Câu hỏi của Lee Yeong Ji

Question

mn giúp em với ạ, em cảm ơn nhiều <3

An Thy · Accepted Answer

7. Ta có: $\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=x^2+3-x=3$$\Rightarrow\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\Rightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\left(1\right)$Lại có $\left(y+\sqrt{y^2+3}\right)\left(\sqrt{y^2+3}-y\right)=y^2+3-y=3$$\Rightarrow\sqrt{x^2+3}+x=\sqrt{y^2+3}-y\Rightarrow x+y=\sqrt{y^2+3}-\sqrt{x^2+3}\left(2\right)$Lấy $\left(1\right)+\left(2\right)\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0$ Câu trả lời: 7. Ta có: $\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=x^2+3-x=3$$\Rightarrow\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\Rightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\left(1\right)$Lại có $\left(y+\sqrt{y^2+3}\right)\left(\sqrt{y^2+3}-y\right)=y^2+3-y=3$$\Rightarrow\sqrt{x^2+3}+x=\sqrt{y^2+3}-y\Rightarrow x+y=\sqrt{y^2+3}-\sqrt{x^2+3}\left(2\right)$Lấy $\left(1\right)+\left(2\right)\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0$

An Thy · Answer

9. Ta có: $\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}$$=\sqrt{\dfrac{110+2\sqrt{109}}{2}}-\sqrt{\dfrac{110-2\sqrt{109}}{2}}$$=\sqrt{\dfrac{\left(\sqrt{109}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{109}-1\right)^2}{2}}=\dfrac{\sqrt{109}+1}{\sqrt{2}}-\dfrac{\sqrt{109}-1}{\sqrt{2}}$$=\dfrac{2}{\sqrt{2}}=\sqrt{2}$Lại có: $\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}$$=\dfrac{\sqrt{4-2\sqrt{y\left(4-y\right)}}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}\right)^2+2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}$$\dfrac{\sqrt{\left(\sqrt{y}\right)^2-2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}+\sqrt{4-y}\right)^2}$$=\dfrac{\sqrt{\left(\sqrt{y}-\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|$Vì $y>2\Rightarrow\left\{{}\begin{matrix}\sqrt{y}>\sqrt{2}\\sqrt{4-y}< \sqrt{2}\end{matrix}\right.\Rightarrow\sqrt{y}-\sqrt{4-y}>0$$\Rightarrow\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left(\sqrt{y}-\sqrt{4-y}\right)\left(\sqrt{y}+\sqrt{4+y}\right)}{\sqrt{2}\left(y-2\right)}$$=\dfrac{y-\left(4-y\right)}{\sqrt{2}\left(y-2\right)}=\dfrac{2y-4}{\sqrt{2}\left(y-2\right)}=\dfrac{2\left(y-2\right)}{\sqrt{2}\left(y-2\right)}=\sqrt{2}$$\Rightarrow\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}=\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}$Câu trả lời: 9. Ta có: $\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}$$=\sqrt{\dfrac{110+2\sqrt{109}}{2}}-\sqrt{\dfrac{110-2\sqrt{109}}{2}}$$=\sqrt{\dfrac{\left(\sqrt{109}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{109}-1\right)^2}{2}}=\dfrac{\sqrt{109}+1}{\sqrt{2}}-\dfrac{\sqrt{109}-1}{\sqrt{2}}$$=\dfrac{2}{\sqrt{2}}=\sqrt{2}$Lại có: $\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}$$=\dfrac{\sqrt{4-2\sqrt{y\left(4-y\right)}}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}\right)^2+2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}$$\dfrac{\sqrt{\left(\sqrt{y}\right)^2-2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}+\sqrt{4-y}\right)^2}$$=\dfrac{\sqrt{\left(\sqrt{y}-\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|$Vì $y>2\Rightarrow\left\{{}\begin{matrix}\sqrt{y}>\sqrt{2}\\sqrt{4-y}< \sqrt{2}\end{matrix}\right.\Rightarrow\sqrt{y}-\sqrt{4-y}>0$$\Rightarrow\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left(\sqrt{y}-\sqrt{4-y}\right)\left(\sqrt{y}+\sqrt{4+y}\right)}{\sqrt{2}\left(y-2\right)}$$=\dfrac{y-\left(4-y\right)}{\sqrt{2}\left(y-2\right)}=\dfrac{2y-4}{\sqrt{2}\left(y-2\right)}=\dfrac{2\left(y-2\right)}{\sqrt{2}\left(y-2\right)}=\sqrt{2}$$\Rightarrow\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}=\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}$

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