c) ĐK: \(x^2\ge m\)
\(\frac{x^2-4}{\sqrt{x^2-m}+1}=0\)
Do mẫu số luôn lớn hơn 0 nên \(x^2-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy...
d) ĐK: \(7\ge x\ge4\)
\(\sqrt{x+3}-\sqrt{7-x}=\sqrt{2x-8}\)
\(\Leftrightarrow x+3+7-x-2\sqrt{\left(x+3\right)\left(7-x\right)}=2x-8\)
\(\Leftrightarrow10-2\sqrt{-x^2+4x+21}=2x-8\)
\(\Leftrightarrow9-x=\sqrt{-x^2+4x+21}\)
\(\Leftrightarrow x^2-18x+81=-x^2+4x+21\)
\(\Leftrightarrow2x^2-22x+60=0\)
\(\Leftrightarrow x^2-11x+30=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy...
e) ĐK: \(-1\le x\le1\)
\(\sqrt{x+1}+\sqrt{1-x}=2-\frac{x^2}{4}\)
Áp dụng bđt Cô - si :
\(\sqrt{x+1}\le\frac{x+1+1}{2}=\frac{x+2}{2}\)
\(\sqrt{1-x}\le\frac{1+1-x}{2}=\frac{2-x}{2}\)
\(\Rightarrow\sqrt{x+1}+\sqrt{1-x}\le\frac{x+2+2-x}{2}=2\)
Mà \(2-\frac{x^2}{4}\le2\forall x\)
Dấu "=" khi \(x=0\)
Đề kì kì, ta không đánh giá được ?