\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)

\(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2-5\right)\left(a^2+5\right)\)

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

\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b-m+n\right)\left(a+b+m-n\right)\)

\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+3^2\right)\)

\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2+\frac{4}{3}x+\frac{1}{9}\right)\)

Tham khảo~

\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)

\(4x^2-9=\left(2x\right)^2-3^2=\left(2x+3\right)\left(2x-3\right)\)

\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2+5\right)\left(a^2-5\right)\)

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

\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b+m-n\right)\left(a+b-m+n\right)\)

\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)

\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2-\frac{4}{3}x+\frac{1}{9}\right)\)

\(3x^2+48+24x-12y^2\)

\(=-3\left(-x^2-16-8x+4y^2\right)\)

\(=-3\left[\left(2y\right)^2-\left(x^2+8x+16\right)\right]\)

\(=-3\left[\left(2y\right)^2-\left(x+4\right)^2\right]\)

\(=-3\left(2y-x-4\right)\left(2y+x+4\right)\)

Max A = -7

sory lộn 

TT

\(A=-2x^2+8x-15\)

\(-A=2x^2-8x+15\)

\(-A=2\left(x^2-4x+4\right)+7\)

\(-A=2\left(x-2\right)^2+7\)

Mà  \(\left(x-2\right)^2\ge0\forall x\Rightarrow2\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge7\)

\(\Leftrightarrow A\le-7\)

Dấu "=" xảy ra khi : 

\(x-2=0\Leftrightarrow x=2\)

Vậy  \(A_{Max}=7\Leftrightarrow x=2\)

\(2bc+b^2+c^2-a^2=\left(b+c\right)^2-a^2=\left(b+c-a\right)\left(b+c+a\right)=2p\left(2p-2a\right)=4p\left(p-a\right)\)

\(mx^2-4mx+4m-nx^2+4nx-4n\)

\(=\left(mx^2-4mx+4m\right)-\left(nx^2-4nx+4n\right)\)

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

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

\(=\left(m-n\right)\left(x-2\right)^2\)

\(mx^2-4mx+4m-nx^2+4nx-4n\)

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

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

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

\(=\left(x-2\right)^2\left(m-n\right)\)

P/s : Phần b ) : \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

a )   \(\left(x+a\right)\left(x+b\right)=x^2+ax+bx+ab=x^2+\left(a+b\right)x+ab\)

b )   \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\) 

\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)

\(=x^2\left(x+c\right)+\left(a+b\right)x\left(x+c\right)+ab\left(x+c\right)\)

\(=x^3+x^2c+\left(ax+bx\right)\left(x+c\right)+abx+abc\)

\(=x^3+x^2c+ax^2+bx^2+axc+bxc+abx+abc\)

\(=x^3+\left(x^2a+x^2b+x^2c\right)+\left(abx+bcx+axc\right)+abc\)

\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)

Sửa đề \(D=25x\left(x.7\right)-7\)

\(\Rightarrow D=25x^2.7-7\)

\(\Rightarrow D=7\left(25x^2-1\right)\)

Do \(25x^2\ge0;1>0\Rightarrow25x^2-1\le-1\)

\(\Rightarrow D\le-7\)

Dấu "=" xảy ra khi \(x=0\)

Vậy Max D = -7 <=> x = 0

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