Áp dụng hằng đẳng thức : \(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy\left(x+y\right)\)

Ta có: \(\left(x+y+z\right)^3-x^3-y^3-z^3=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+z^3+3\left(x+y\right).z.\left(x+y+z\right)-x^3-y^3-z^3\)

\(=x^3+y^3+3xy\left(x+y\right)+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
 

\(MAX\)B=\(\frac{17}{2}\)

B = -2(x² -3x -2)= -2( x² - 2.3x/2 + 9/4 -9/4 -2)

= -2(x-3/2)² + 8,5

GTLN: B = 8,5

1)\(x^4+2x^3+x^2\)

=\(\left(x^4+x^3\right)+\left(x^3+x^2\right)\)đật nhân tử chung ra

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

2) pt => \(\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)

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

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

3)chia tất cả cho 5 pt => \(x^2-2xy+y^2-4x^2\)

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

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

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

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

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

k chi nha

\(\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+...+\frac{x}{10}=\frac{21}{10}\)

\(\frac{1}{10}\left(1+2+3+...+x\right)=\frac{21}{10}\)

\(1+2+3+...+x=21\)

\(\left(1+x\right).x:2=21\)

\(\left(1+x\right).x=42\)

\(x\left(1+x\right)=6.7\)

\(\Rightarrow x=6\)

Phân tích thành nhân tử

18x(x+1)(x-1)

=(18x² +18x)(x-1)

=18x³-18x²+18x²-18x

=18x³-18x

18x(x+1)(x-1)=18x³  -18x

12 đúng k?

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