Đặt \(\hept{\begin{cases}\sqrt{x-1}=a\\\sqrt{x+1}=b\end{cases}\left(a;b>0\right)\Rightarrow}b^2-a^2=0\)

\(b^2+2+ab=3b+a\)

\(\Leftrightarrow a\left(b-1\right)+\left(b^2-2b+1\right)-\left(b-1\right)=0\)

\(\Leftrightarrow a\left(b-1\right)+\left(b-1\right)^2-\left(b-1\right)=0\)

\(\Leftrightarrow\left(b-1\right)\left(a+b-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}b=1\\a+b=2\end{cases}}\)

Tự làm nốt nhé~~~

\(\sqrt{98}-\sqrt{72}-0,5\sqrt{8}.\)

\(=\sqrt{7\cdot7\cdot2}-\sqrt{6\cdot6\cdot2}-0,5\sqrt{2\cdot2\cdot2}\)

\(=7\sqrt{2}-6\sqrt{2}-0,5\cdot2\sqrt{2}\)

\(=7\sqrt{2}-6\sqrt{2}-\sqrt{2}\)

\(=0\)

\(\sqrt{98}-\sqrt{72}-\frac{\sqrt{8}}{2}\)

=\(7\sqrt{2}-6\sqrt{2}-\sqrt{2}\)

\(=0\)

\(D=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)

   \(=\left(2\sqrt{3}-1\right)^2\left(\sqrt{3}+2\right)^2-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)

   \(=\left(4+3\sqrt{3}\right)^2-8\sqrt{28+6\sqrt{3}}\)\(=\left(4+3\sqrt{3}\right)^2-8\left(3\sqrt{3}+1\right)\)

   \(=43+24\sqrt{3}-24\sqrt{3}-8=35\)

\(a^2+b^2=\frac{9a^2}{9}+\frac{16b^2}{16}\ge\frac{\left(3a+4b\right)^2}{9+16}=\frac{5^2}{25}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\frac{3a}{9}=\frac{4b}{16}=\frac{3a+4b}{9+16}=\frac{5}{25}=\frac{1}{5}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=\frac{3}{5}\\b=\frac{4}{5}\end{cases}}\)

\(x^2+\left(m+2\right)x+m-1\)

\(\Delta=b^2-4ac=\left(m+2\right)^2-4.1.\left(m-1\right)\)

                             \(=m^2+4m+4-4m+4\)

                             \(=m^2+8\)

Vì \(m^2\ge0\forall m\Rightarrow m^2+8\ge8>0\forall m\Rightarrow\Delta>0\forall m\)

Vậy phương trình luôn có hai nghiệm phân biệt với mọi m

Áp dụng hệ thức vi-ét ta có:

\(\hept{\begin{cases}x_1+x_2=\frac{-b}{a}=-\left(m+2\right)\\x_1x_2=\frac{c}{a}=m-1\end{cases}}\)

Theo bài ra ta có:

\(A=x_1^2+x_2^2-3x_1x_2\)

\(=\left(x_1+x_2\right)^2-2x_1x_2-3x_1x_2\)

\(=\left(x_1+x_2\right)^2-5x_1x_2\)

Đến đây dễ r:)

\(\frac{3x+2\sqrt{x}+2}{\sqrt{x}}=3\sqrt{x}+2+\frac{2}{\sqrt{x}}\ge2+2\sqrt{6}\)

\("="\Leftrightarrow x=\frac{2}{3}\)

Bui Huyen đề yêu cầu tìm max mà nhỉ ?

Đặt \(A=\frac{3x+2\sqrt{x}+2}{\sqrt{x}}\)

\(\Leftrightarrow A\sqrt{x}=3x+2\sqrt{x}+2\)

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

Phương trình có nghiệm \(\Leftrightarrow\Delta\ge0\)

\(\Leftrightarrow\left(A-2\right)^2-4\cdot\left(-3\right)\cdot\left(-2\right)\ge0\)

\(\Leftrightarrow\left(A-2\right)^2-24\ge0\)

\(\Leftrightarrow\left(A-2\right)^2\ge24\)

\(\Leftrightarrow\orbr{\begin{cases}A-2\ge\sqrt{24}\\A-2\le-\sqrt{24}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}A\ge\sqrt{24}+2\\A\le-\sqrt{24}+2\end{cases}}\)

Vậy \(maxA=-\sqrt{24}+2\)

Mặt khác \(A\ge0\)do đó A không có GTLN ???