Question

[minh huong]

giải phương trình:

a,\(\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}\)

b,CM:\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)

Answer 1

[Võ Hồng Phúc]

a. ĐKXĐ :\(x\ge2\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-2}=\sqrt{x+3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\text{ }\left(\text{TM}\right)\\\text{vô nghiệm}\end{matrix}\right.\)

b. \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\) \(\left(1\right)\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge a^2+b^2+c^2+d^2+2ac+2bd\)

\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge ac+bd\) \(\left(2\right)\)

\(+\) Nếu \(ac+bd< 0\) thì \(\left(2\right)\) được chứng minh

\(+\) Nếu \(ac+bd\ge0\), ta có :

\(\left(2\right)\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge a^2c^2+b^2d^2+2abcd\)

\(\Leftrightarrow a^2c^2+a^2d^2+b^2c^2+b^2d^2\ge a^2c^2+b^2d^2+2abcd\)

\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) \(\left(3\right)\)

Bất đẳng thức \(\left(3\right)\) đúng \(\forall a,d,b,c\in R\)

Vậy bất đẳng thức một được chứng minh