\(2x^2+x-3=2\sqrt{2x+3}\left(ĐKXĐ:x\ge-\frac{3}{2}\right)\)

Đặt \(\sqrt{2x+3}=2y-1\left(y\ge\frac{1}{2}\right)\)

Ta có hpt \(\hept{\begin{cases}2x^2+x-3=2\left(2y-1\right)\\\left(2y-1\right)^2=2x+3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x^2+x-3-4y+2=0\\2y^2-2y-x-1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x^2+x-4y=1\left(1\right)\\2y^2-2y-x=1\left(2\right)\end{cases}}\)

Lấy (1)-(2)\(\Leftrightarrow2\left(x^2-y^2\right)+x-4y+2y+x=0\)

\(\Leftrightarrow2\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)=0\)

Nhận thấy \(x+y+1\ge-\frac{3}{2}+\frac{1}{2}+1=0\)

Dấu "=" xảy ra <=>x=-3/2;y=1/2(thỏa mãn)

Nếu x=y

\(\Rightarrow\sqrt{2x+3}=2y-1\Leftrightarrow\sqrt{2x+3}=2x-1\)

\(\Leftrightarrow\hept{\begin{cases}2x+3=4x^2-4x+1\left(3\right)\\x\ge\frac{1}{2}\end{cases}}\)

\(\left(3\right)\Leftrightarrow4x^2-6x-2=0\Leftrightarrow2x^2-3x-1=0\)

Dùng đenta giải ra nghiệm rồi kết luận nha

\(VT\ge2\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)-3\)

\(\ge2\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=VP^{\left(đpcm\right)}\)

bỏ số 14 cuối nha mọi ng, mình nhầm

Tớ giải bừa

\(\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}+1\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)

\(=\left(a-b\right)^2\left(\sqrt{\frac{a+b}{a-b}}\right)^2-1^2\)

\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)

\(=2ab-2b^2\)

E=\(|x-2005|+|2006-x|\ge|x-2005+2006-x|=1\)

Dấu = xảy ra khi \(\orbr{\begin{cases}x\ge2006\\x\le2005\end{cases}}\)