\(\left\{{}\begin{matrix}x^3+xy^2+3\left(x-2y\right)=0\\x^2+xy=3\end{matrix}\right.\)\(\Rightarrow x^3+xy^2+\left(x^2+xy\right)\left(x-2y\right)=0\)\(\Leftrightarrow x^3+xy^2+x^3-x^2y-2xy^2=0\Leftrightarrow2x^3-x^2y-xy^2=0\)\(\Leftrightarrow x\left(2x+y\right)\left(x-y\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=-2x\\x=y\end{matrix}\right.\)

+) \(x=0\Rightarrow0y=3\)(vô nghiệm)

+) y=-2x \(\Rightarrow x^2-2x^2=3\Leftrightarrow-x^2=3\)(vô nghiệm)

+) x=y\(\Rightarrow2x^2=3\Leftrightarrow x^2=\dfrac{3}{2}\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt{\dfrac{3}{2}}\\x=y=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)

\(x+\sqrt{5-x^2}+x\sqrt{5-x^2}=5\)

ĐKXĐ: \(-\sqrt{5}\le x\le\sqrt{5}\)

Đặt \(\sqrt{5-x^2}=t\)(\(t\)\(\ge0\))

Ta có: \(\left\{{}\begin{matrix}x+t+xt=5\\x^2+t^2=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(x+t\right)+2xt=10\\\left(x+t\right)^2-2xt=5\end{matrix}\right.\)

\(\Rightarrow\left(x+t\right)^2+2\left(x+t\right)-15=0\Leftrightarrow\left(x+t+5\right)\left(x+t-3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=-5-x\\t=3-x\end{matrix}\right.\)(giải hai trường hợp rồi kết luận nghiệm)

Ta có: \(\left|A-B\right|=\left|A\right|-\left|B\right|\Leftrightarrow\left(A-B\right)^2=\left(\left|A\right|-\left|B\right|\right)^2\)\(\Leftrightarrow A^2-2AB+B^2=A^2-2\left|A\right|\left|B\right|+B^2\)\(\Leftrightarrow\left|AB\right|=AB\)\(\Leftrightarrow AB\ge0\)

\(\sqrt{x+3}+\sqrt{2x-1}=4-x\)(1)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow\sqrt{x+3}-2+\sqrt{2x-1}-1+x-1=0\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x+3}+2}+\dfrac{2\left(x-1\right)}{\sqrt{2x-1}+1}+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x+3}+2}+\dfrac{2}{\sqrt{2x-1}+1}+1\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\dfrac{1}{\sqrt{x+3}+2}+\dfrac{2}{\sqrt{2x-1}+1}+1\)>0)

\(\Leftrightarrow x=1\)(thỏa mãn)

Vậy phương trình có nghiệm là x=1

Ta có: \(\left\{{}\begin{matrix}x^4+2x^3y+x^2y^2=2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+xy\right)^2=2x+9\\x^2+2xy=6x+6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+xy\right)^2=2x+9\\xy=3x+3-\dfrac{x^2}{2}\end{matrix}\right.\) \(\Rightarrow\left(\dfrac{x^2}{2}+3x+3\right)^2=2x+9\)( đến đây là phương trình 1 ẩn rồi, tự giải tiếp)

Ta có: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Ta có: \(\left\{{}\begin{matrix}x^3=3x+8y\\y^3=8x+3y\end{matrix}\right.\)

\(\Rightarrow x^3-y^3=5y-5x\)\(\Leftrightarrow x^3-y^3+5x-5y=0\)\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+5\right)=0\)

\(\Leftrightarrow x=y\)(vì \(x^2+xy+y^2+5>0\))

Thay \(x=y\) vào phương trình \(x^3=3x+8y\) ta được

\(x^3=11x\)\(\Leftrightarrow x\left(x^2-11\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\x=y=\sqrt{11}\\x=y=-\sqrt{11}\end{matrix}\right.\)

Ta có: \(y=\sqrt{3+x}+\sqrt{5-x}\)

ĐKXĐ: \(-3\le x\le5\)

\(y^2=3+x+5-x+2\sqrt{\left(3+x\right)\left(5-x\right)}=8+2\sqrt{\left(3+x\right)\left(5-x\right)}\)\(\ge8\)

\(\Rightarrow y\ge2\sqrt{2}\)

Dấu "=" xảy ra khi và chỉ khi \(\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)(thỏa mãn)

Vậy min y = \(2\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

mặt khác \(y^2\) = \(8+2\sqrt{\left(3+x\right)\left(5-x\right)}\le8+3+x+5-x=16\)

\(\Rightarrow y\le4\)

Dấu"=" xảy ra khi và chỉ khi \(3+x=5-x\Leftrightarrow x=1\)(thỏa mãn)

Vậy max y = 4 \(\Leftrightarrow x=1\)

Ta có: \(tan\alpha=2\Leftrightarrow\dfrac{sin\alpha}{cos\alpha}=2\Leftrightarrow sin\alpha=2cos\alpha\)

A = \(\dfrac{16cos^2\alpha+6cos^2\alpha}{20cos^2\alpha-2cos^2\alpha}=\dfrac{22cos^2\alpha}{18cos^2\alpha}=\dfrac{11}{9}\)