\(\Delta'=\left(m+1\right)^2-\left(m^2+2\right)=2m-1\)

Để phương trình có hai nghiệm \(x_1,x_2\)thì \(\Delta'\ge0\Rightarrow2m-1\ge0\Leftrightarrow m\ge\frac{1}{2}\).

Theo Viete ta có: \(\hept{\begin{cases}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{cases}}\)

\(A=\sqrt{2\left(x_1^2+x_2^2\right)+16}-3x_1x_2\)

\(A=\sqrt{2\left(x_1+x_2\right)^2-4x_1x_2+16}-3x_1x_2\)

\(A=\sqrt{2\left(2\left(m+1\right)\right)^2-4\left(m^2+2\right)+16}-3\left(m^2+2\right)\)

\(A=\sqrt{4m^2+16m+16}-3\left(m^2+2\right)\)

\(A=2\sqrt{\left(m+2\right)^2}-3\left(m^2+2\right)\)

\(A=2\left|m+2\right|-3\left(m^2+2\right)\)

\(A=2m+4-3m^2-6\)(vì \(m\ge\frac{1}{2}\)nên \(m+2>0\))

\(A=-3m^2+2m-2\)

Ta lập bảng biến thiên với hàm \(f\left(x\right)=-3x^2+2x-2\)với \(x\ge\frac{1}{2}\).

Kết quả: \(maxA=-\frac{7}{4}\)

suy ra \(a=-7,b=4\Rightarrow2a-3b=-26\)

\(\left(x-y\right)^2=49\Leftrightarrow\orbr{\begin{cases}x-y=7\\x-y=-7\end{cases}}\)

- \(\hept{\begin{cases}x-y=7\\3x+4y=84\end{cases}\Leftrightarrow\hept{\begin{cases}x=16\\y=9\end{cases}}}\)

- \(\hept{\begin{cases}x-y=-7\\3x+4y=84\end{cases}\Leftrightarrow\hept{\begin{cases}x=8\\y=15\end{cases}}}\)

PT <=> \(x^2-1=2x\sqrt{x\left(x-2\right)}\)

bình phương 2 vế ta được : \(x^4-2x^2+1=4x^3\left(x-2\right)\)

\(x^4-2x^2+1=4x^4-8x^3\)

\(-3x^4-2x^2+8x^3+1=0\)

a, ĐK : \(x\ne-1;-2\)

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

Khử mẫu : \(2x+4-3x-3=x^2+x+2x+2\)

\(\Leftrightarrow-x+1=x^2+3x+2\Leftrightarrow-x^2-4x-1=0\)

giải delta nốt nhé ! 

b;c tương tự