Question

[nguyenhoangtung]

\(\sqrt{8x^2-16x+10}+\sqrt{2x^2-4x+4}=\sqrt{7-x^2+2x}\)

Answer 1

[Nguyễn Đức Trí]

\(\sqrt[]{8x^2-16x+10}+\sqrt[]{2x^2-4x+10}=\sqrt[]{7-x^2+2x}\)

\(\Leftrightarrow\sqrt[]{8x^2-16x+10}=\dfrac{1}{4}\sqrt[]{2\left(7-x^2+2x\right)}-\sqrt[]{2x^2-4x+10}\)

\(\Leftrightarrow\sqrt[]{8x^2-16x+10}=\dfrac{1}{4}\sqrt[]{14-2x^2+4x}-\sqrt[]{2x^2-4x+10}\left(1\right)\)

Áp dụng BĐT Bunhiacopxki ta được:

\(\left[\dfrac{1}{4}\sqrt[]{14-2x^2+4x}+\left(-1\right).\sqrt[]{2x^2-4x+10}\right]^2\le\left(\dfrac{1}{16}+1\right)\left(14-2x^2+4x+2x^2-4x+10\right)=\dfrac{17}{16}.24=\dfrac{51}{2}\)

Dấu "=" xảy ra khi và chỉ khi

\(\sqrt[]{14-2x^2+4x}+4\sqrt[]{2x^2-4x+10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}14-2x^2+4x=0\\2x^2-4x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}14+2-2\left(x^2-2x+1\right)=0\\2\left(x^2-2x+1\right)+10-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\left(x-1\right)^2+16=0\\2\left(x-1\right)^2+8=0\end{matrix}\right.\) \(\Leftrightarrow x\in\varnothing\)

\(pt\left(1\right)\Leftrightarrow8x^2-16x+10=\dfrac{51}{2}\)

\(\Leftrightarrow16x^2-32x+20-51=0\)

\(\Leftrightarrow16x^2-32x-31=0\left(2\right)\)

\(\Delta'=256+496=752>0\)

\(\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{47}\)

\(pt\left(2\right)\) có 2 nghiệm phân biệt

\(x=\dfrac{16\pm4\sqrt[]{47}}{16}=\dfrac{4\pm\sqrt[]{47}}{4}\)

Answer 2

[Nguyễn Đức Trí]

Cách giải trên đã sai, mình giải lại

\(\left(1\right)\Leftrightarrow\sqrt[]{8\left(x^2-2x+1\right)+2}+\sqrt[]{2\left(x^2-2x+1\right)+2}=\sqrt[]{8-\left(x^2-2x+1\right)}\)

\(\Leftrightarrow\sqrt[]{8\left(x-1\right)^2+2}+\sqrt[]{2\left(x-1\right)^2+2}=\sqrt[]{8-\left(x-1\right)^2}\left(2\right)\)

Vì \(\left(x-1\right)^2\ge0,\forall x\in R\)

\(\Rightarrow\left\{{}\begin{matrix}8\left(x-1\right)^2+2\ge2,\forall x\in R\\2\left(x-1\right)^2+2\ge2,\forall x\in R\\8-\left(x-1\right)^2\le8,\forall x\in R\end{matrix}\right.\)

Nên khi \(\left(x-1\right)^2=0\Leftrightarrow x=1\)

Thay \(x=1\) vào \(\left(2\right)\) ta được

\(\sqrt[]{8.0+2}+\sqrt[]{2.0+2}=\sqrt[]{8-0}\)

\(\Leftrightarrow\sqrt[]{2}+\sqrt[]{2}=\sqrt[]{8}=2\sqrt[]{2}\left(đúng\right)\)

Vậy nghiệm của phương trình đã cho là \(x=1\)