Từ pt 1 ta có thể biến đổi : \(ax+y+z=a^2\)

\(< =>a=\frac{ax+y+z}{a}\)

\(< =>x+y+z=a\)

\(< =>3x+3y+3z=x+ay+z\)

\(< =>2x+y\left(3-a\right)+2z=0\)

\(< =>2a+y-ay=0\)

\(< =>2a+y-ay-2=-2\)

\(< =>a\left(2-y\right)-\left(2-y\right)=-2\)

\(< =>\left(a-1\right)\left(2-y\right)=2.\left(-1\right)=-1.2=-2.1=1.\left(-2\right)\)

\(< =>\left(a;y\right)=\left(3;3\right)=\left(0;0\right)=\left(-1;1\right)=\left(2;4\right)\)

Bạn thay vào là đc :)) giải sai hay đúng cg ko bt nx :(

ĐKXĐ:.............

1.\(\sqrt{x^2-6x+9}=2x-1\)

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

\(\Leftrightarrow\left|x-3\right|=2x-1\)

................

\(2)\sqrt{x+4\sqrt{x}+4}=5x+2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

\(\Leftrightarrow\left|\sqrt{x}+2\right|=5x+2\)

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

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}=4\)

\(\Leftrightarrow\left|x-1\right|+\left|x+2\right|=4\)

\(P=\frac{2005x+2006\sqrt{1-x^2}+2007}{\sqrt{1-x^2}}\)

\(=\frac{2006\left(1+x\right)+\left(1-x\right)}{\sqrt{1-x^2}}+2006\)

\(\ge\frac{2\sqrt{2006\left(1+x\right)\left(1-x\right)}}{\sqrt{1-x^2}}+2006=2\sqrt{2006}+2006\)

Dấu = xảy ra khi:

\(2006\left(1+x\right)=1-x\)

\(\Leftrightarrow x=-\frac{2005}{2007}\)

\(x^{2019}+y^{2019}=2x^{1009}.y^{1009}< =>x^{2020}+x.y^{2019}=2x^{1010}y^{1009}< =\)\(>\left(x^{1010}-y^{1009}\right)^2=y^{2018}\left(1-xy\right)=>\sqrt{1-xy}=\frac{x^{1010}-y^{1009}}{y^{1009}}\)

x;y là số hữu tỉ nên có dạng \(x=\frac{m}{n};y=\frac{p}{q}\left(m;n;p;q\in Z\right)\)=> \(\sqrt{1-xy}=\frac{m^{1010}.q^{1009}-n^{1010}.p^{1009}}{n^{1010}.p^{1009}}=\frac{A}{B}\left(A;B\in Z\right)\)=> \(\sqrt{1-xy}\in Q\)