\(\frac{4212}{14640}=\frac{4212:2}{14640:2}=\frac{2106}{7320}\)

\(\frac{6318}{21960}=\frac{6318:3}{21960:3}=\frac{2106}{7320}\)

Vậy\(\frac{2106}{7320}=\frac{4212}{14640}=\frac{6318}{21960}\)

a) \(x^3+2x^2y+xy^2-9x\)

\(=x\left(x^2+2xy+y^2-9\right)\)

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

b) \(2x-2y-x^2+2xy-y^2\)

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

c) \(x^2-2x-4y^2-4y\)

\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)

\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

Áp dụng tc của dãy tỉ số bằng nhau ta cól

\(\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{8}{-2}=-4\)

=> \(\begin{cases}x=-12\\y=-20\end{cases}\)

\(\left(x-4\right)+8=2x\)

\(\Leftrightarrow2x-x=8-4\)

\(\Leftrightarrow x=4\)

\(B=x-x^2=-\left(x^2-x+\frac{1}{4}\right)+\frac{1}{4}=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì: \(-\left(x-\frac{1}{2}\right)^2\le0\)

=> \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Vậy GTLN của B là \(\frac{1}{4}\) khi \(x=\frac{1}{2}\)

a) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{1}{5}\end{array}\right.\)

b) \(x+1=\left(x+1\right)^2\)

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

\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-1\\x=0\end{array}\right.\)

c) \(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

\(\Leftrightarrow x=0\)

Có: \(\left(x-y\right)^2=x^2-2xy+y^2=56-2\cdot20=56-40=16\)

B = 2(x^2 +3x +4) = 2(x^2 + 2x.3/2 + 9/4 +7/4) = 2(x+3/2)^2+7/2 \(\ge\frac{7}{2}\)

Vậy MinB = 7/2 khi x= -3/2

\(\left(x+y\right)^2=9\)

\(\Leftrightarrow x^2+2xy+y^2=9\)

\(\Leftrightarrow x^2+y^2=9-2xy=9-2\cdot\left(-5\right)=19\)

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

b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)

c) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)

d) \(y^2\left(x-1\right)-7y^3+7xy^3\)

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

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