\(M=\left[x+\left(y-z\right)-2x\right]+y+z-\left(2-x-y\right)\)

\(=-x+y-z+y+z-2+x+y\)

\(=3y-2\)

\(N=x-\left[x-\left(y-z\right)-x\right]\)

\(=x-\left(-y+z\right)\)

\(=x+y-z\)

\(M+N=3y-2+x+y-z=x+4y-z-2\)

\(M-N=\left(3y-2\right)-\left(x+y-z\right)\)

\(=3y-2-x-y+z\)

\(=-x+2y+z-2\)

\(M=\left[x+\left(y-z\right)-2x\right]+y+z-\left(2-x-y\right)\\ M=x+y-z-2x+y+z-2+x+y\\ M=3y-2\)

\(N=x-\left[x-\left(y-z\right)-x\right]\\ N=x-\left(x-y+z-x\right)\\ N=x-x+y-z+x\\ N=x+y-z\)

\(M+N=3y-2+x+y-z\\ M+N=x+4y-z-2\)

\(M-N=3y-2-\left(x+y-z\right)\\ M-N=3y-2-x-y+z\\ M-N=-x+2y+z-2\)

mik đag cần gấp các bn giải nhanh dùm mik nha

Ủa pt hàm là \(f\left(f\left(x\right)+y\right)=2x+f\left(f\left(x\right)-y\right)\) hay \(f\left(f\left(x\right)+y\right)=2x+f\left(f\left(y\right)-x\right)\) vậy bạn?

Vì nếu pt hàm là \(f\left(f\left(x\right)+y\right)=2x+f\left(f\left(x\right)-y\right)\)

Nếu ta thế \(y=0\) thì:

\(f\left(f\left(x\right)\right)=2x+f\left(f\left(x\right)\right)\Leftrightarrow2x\equiv0\) điều này vô lý nên ko thể tồn tại 1 hàm như vậy

1a.

\(2P=1-\frac{bc}{2a^2+bc}+1-\frac{ca}{2b^2+ca}+1-\frac{ab}{2c^2+ab}\)

\(\Rightarrow2P=3-\left(\frac{bc}{2a^2+bc}+\frac{ca}{2b^2+ca}+\frac{ab}{2c^2+ab}\right)\)

\(\Rightarrow2P=3-\left(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{c^2a^2}{2b^2ca+c^2a^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\right)\)

\(\Rightarrow2P\le3-\frac{\left(ab+bc+ca\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=3-1=2\)

\(\Rightarrow P\le1\)

\(P_{max}=1\) khi \(a=b=c\)

1b.

\(Q=\frac{a^2}{5a^2+b^2+c^2+2bc}+\frac{b^2}{5b^2+a^2+c^2+2ca}+\frac{c^2}{5c^2+a^2+b^2+2ab}\)

\(Q=\frac{a^2}{a^2+b^2+c^2+\left(2a^2+bc\right)+\left(2a^2+bc\right)}+\frac{b^2}{a^2+b^2+c^2+\left(2b^2+ca\right)+\left(2b^2+ca\right)}+\frac{c^2}{a^2+b^2+c^2+\left(2c^2+ab\right)+\left(2c^2+ab\right)}\)

\(\Rightarrow Q\le\frac{1}{9}\left(\frac{a^2}{a^2+b^2+c^2}+\frac{b^2}{a^2+b^2+c^2}+\frac{c^2}{a^2+b^2+c^2}+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)

\(\Rightarrow Q\le\frac{1}{9}\left(1+2\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\right)\)

Theo kết quả câu a ta có:

\(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\le1\)

\(\Rightarrow Q\le\frac{1}{9}\left(1+2\right)=\frac{1}{3}\)

\(Q_{max}=\frac{1}{3}\) khi \(a=b=c\)

2.

Do \(x+y+z=\frac{3}{2}\Rightarrow x< \frac{3}{2}< 2\)

Ta có:

\(VT=x+x.4y\left(\frac{1}{2}+z\right)\le x+x\left(\frac{1}{2}+z+y\right)^2=x+x\left(\frac{1}{2}+\frac{3}{2}-x\right)^2\)

\(\Leftrightarrow VT\le x+x\left(2-x\right)^2=x^3-4x^2+5x\)

\(\Leftrightarrow VT\le x^3-4x^2+5x-2+2=\left(x-1\right)^2\left(x-2\right)+2\le2\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;\frac{1}{2};0\right)\)