Đặt A = \(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)=> \(A^3=18+3A\Leftrightarrow A^3-3A-18=0\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\Leftrightarrow A-3=0\Leftrightarrow A=3\)

\(\dfrac{\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}}=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{3}=\dfrac{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{3}=\dfrac{1}{3}\)

\(\sqrt{\dfrac{1}{125}}.\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}=\sqrt{\dfrac{1}{125}.\dfrac{32}{35}:\dfrac{56}{225}}=\sqrt{\dfrac{36}{1225}}=\dfrac{6}{35}\)

ĐKXĐ: \(x,y\ge0\)

Từ hệ pt trên suy ra \(\sqrt{x}+\sqrt{y+9}=\sqrt{y}+\sqrt{x+9}\Leftrightarrow x+y+9+2\sqrt{xy+9x}=x+y+9+2\sqrt{yx+9y}\Leftrightarrow2\sqrt{xy+9x}=9+2\sqrt{xy+9y}\Leftrightarrow x=y\)

Vậy hệ pt thành \(\sqrt{x}+\sqrt{x+9}=9\Leftrightarrow2x+9+2\sqrt{x^2+9x}=81\Leftrightarrow2\sqrt{x^2+9x}=72-2x\)

( đk : x \(\le\) 36, y \(\le36\))

\(\Leftrightarrow4\left(x^2+9x\right)=5184-288x+4x^2\Leftrightarrow324x=5184\Leftrightarrow x=16\) (tmđk)

b. C = \(sin^4a+sin^2a.cos^2a+cos^2a=\left(1-cos^2\right)^2+\left(1-cos^2a\right)cos^2a+cos^2a=1-2cos^2+cos^4a+cos^2a-cos^4a+cos^2a=1\)

a. \(\dfrac{1+2sin\alpha cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{sin^2\alpha+2sin\alpha cos\alpha+cos^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)

ĐKXĐ: x > 1

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

ac sao là phân giác của góc bac được bạn