a) \(A=x^2-2xy+5y^2+4y+51=x^2-2xy+y^2+4y^2+4y+1+50\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)

Dấu \(=\)khi \(\hept{\begin{cases}x-y=0\\2y+1=0\end{cases}}\Leftrightarrow x=y=-\frac{1}{2}\).

b) \(B=\frac{121}{-4x^2-12x+2}=\frac{121}{-4\left(x^2+3x+\frac{9}{4}\right)+11}=\frac{121}{-4\left(x+\frac{3}{2}\right)^2+11}\)

c) \(C=\frac{9}{-2x^2+4x-7}=\frac{9}{-2\left(x^2-2x+1\right)-5}\)

d) \(D=10x^2+4y^2-4xy+8x-4y+20\)

\(=9x^2+6x+1+x^2+4y^2+1+2x-4y-4xy+18\)

\(=\left(3x+1\right)^2+\left(x-2y+1\right)^2+18\)

e) \(E=9x^2+2y^2+6xy-6x-8y+10\)

\(=y^2-6y+9+9x^2+y^2+1+6xy-6x-2y\)

\(=\left(y-3\right)^2+\left(3x+y-1\right)^2\)

47694 tấn = 47694000kg / 295749 tạ = 29574900kg / 396857295 kg=39685729500dg

47694 tan = ..47694000. kg

295749 ta =..29,574900 ..kg

đaay nhé tham khảo phần c thì mik ko bt

1. \(-x^2-2y^2+2xy-4x+2y-12\)

\(=-y^2-2y-1-x^2-y^2-4-4x+4y+2xy-7\)

\(=-\left(y+1\right)^2-\left(x-y+2\right)^2-7\le-7\)

Dấu \(=\)khi \(\hept{\begin{cases}y+1=0\\x-y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=-1\end{cases}}\).

2. \(-10x^2-y^2+6xy+10x-2y+2\)

\(=-x^2+4x-4-9x^2-y^2-1+6x-2y+6xy+7\)

\(=-\left(x-2\right)^2-\left(3x-y-1\right)^2+7\le7\)

3. \(-x^2-5y^2-4xy+2x-2y-5\)

\(=-y^2-6y-9-x^2-4y^2+1-4xy+2x+4y+5\)

\(=-\left(y+3\right)^2-\left(x+2y-1\right)^2+5\le5\)

4. \(-x^2-26y^2+10xy-14x+76y-59\)

\(=-y^2-6y-9-x^2-25y^2-49-14x+70y+10xy-1\)

\(=-\left(y+3\right)^2-\left(x-5y+7\right)^2-1\le-1\)

5. Bạn thử tự vận dụng như cách làm bên trên nhé. 

a) \(16x^2-1=0\)

\(\Rightarrow16x^2=1\)

\(\Rightarrow x^2=\frac{1}{16}\)

\(\Rightarrow x^2=\left(\pm\frac{1}{4}\right)^2\)

\(\Rightarrow x=\orbr{\begin{cases}\frac{1}{4}\\\frac{-1}{4}\end{cases}}\)

b) \(x^2+\frac{1}{4}=0\)

Ta có: \(x^2\ge0\forall x\Rightarrow x^2+\frac{1}{4}\ge\frac{1}{4}>0\)

=> Vô nghiệm

c) \(x^3+3x^2-\left(x+3\right)=0\)

\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)

\(\Rightarrow\left(x^2-1\right)\left(x+3\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\)

Trường hợp 1: \(x-1=0\Rightarrow x=1\)

Trường hợp 2: \(x+1=0\Rightarrow x=-1\)

Trường hợp 3: \(x+3=0\Rightarrow x=-3\)

\(8x^3-2x=0\)

\(\Rightarrow2x\left(4x^2-1\right)=0\)

\(\Rightarrow2x\left(2x-1\right)\left(2x+1\right)=0\)

Trường hợp 1: \(2x=0\Rightarrow x=0\)

Trường hợp 2: \(2x-1=0\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Trường hợp 3: \(2x+1=0\Rightarrow2x=-1\Rightarrow x=\frac{-1}{2}\)