6x(x+3y-1)-6x²-8xy

=6x²+18xy-6x-6x²-8xy

=10xy-6x

=2x(5y-3)

Ta có thể xét 2 tam giác bằng nhau để chứng minh .

Không chắc cho lắm .

- Nhân cả tử và mẫu phân thức thứ nhất với a

- Nhân cả tử và mẫu phân thức thứ 2 với ac

- Thay abc =2016 ta có mẫu số chung là :

3ac - 4032 +2016a

- Rút gọn => đáp án : -1

\(P=\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}-\frac{4032-3ac}{3ac-4032+2016a}\)

Ta rút gọn từng biểu thức

\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\)

\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\)

\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}\)

\(=-1+\frac{2016}{3c-2bc+2016}\)

\(\Rightarrow P=-1\)

\(Q=a^2-10a+25-25+4b^2\)

\(Q=\left(a^2-2.5.a+5^2\right)+4b^2-25=\left(a-5\right)^2+4b^2-25\)

\(Q\ge-25\) đẳng thức khi \(\hept{\begin{cases}a=5\\b=0\end{cases}}\)

Q=a²+4b²-10a

=a²-10a+25-25+4b²

=(a-5)²+4b²-25

\(\Rightarrow\left(a-5\right)^2+4b^2\ge0\) voi moi a

\(\Leftrightarrow\left(a-5\right)^2+4b^2\ge-25\)

Vay GTNN la -25

Dau "=" xay ra khi : a-5=0 \(\Rightarrow\)a=5

                              4b=0 \(\Rightarrow\)b=0

Ta co :a+b+c=0

a³+b³+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac²+6abc=0

a³+b³+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac²+3abc+3abc+3abc-3abc=0

a³+b³+c³+(3a²b+3ab²+3abc)+(3b²c+3bc²+3abc)+(3a²c+3ac²+3abc)-3abc=0

a³+b³+c³+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc

Vi : a+b+c=0

\(\Rightarrow\)a³+b³+c³+0+0+0=3abc

\(\Rightarrow\)a³+b³+c³=3abc

\(\Rightarrow\)dpcm

nho k nha

a+b+c = 0

=> a+b = -c

=> (a+b)^3 = (-c) ^3

=> a^3 + b^3 + 3ab(a+b) = -c^3

=> a^3 +b^3 +c^3 = - 3ab(a+b)

=> a^3 + b^3 + c^3 = 3abc ( vì a+b = -c) ( đpcm)

\(D=\frac{x^3+x^2-2x}{x\left|x+2\right|-x^2+4}\)

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

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

\(=\frac{x\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left[x-1\left(x-2\right)\right]}\)

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

nho k nha

a, x²-2x+3

=x²-2x+1+2

=(x-1)²+2

\(\Rightarrow\left(x-1\right)^2\ge0\)voi moi x

Dpcm

b, x²+7x+13 

=x²+7x+\(\frac{49}{4}\)+\(\frac{3}{4}\)

=\(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}\)

\(\Rightarrow\left(x+\frac{7}{2}\right)^2\ge0\)voi moi x

Dpcm

c, x-x²-1

=-x²+x-1

=\(-x^2+2.\frac{1}{2}x-\frac{1}{4}+\frac{5}{4}\)

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

\(=\frac{5}{4}-\left(x-\frac{1}{2}\right)^2\)

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

Dpcm

nho k nha 

\(C=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)

\(=\left(\frac{1}{1-x}+\frac{2}{x+1}+\frac{-5x}{\left(1-x\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\left(\frac{x+1+2-2x-5x}{\left(1-x\right)\left(x+1\right)}\right):\frac{1-2x}{\left(x-1\right)\left(x+1\right)}\)

\(=\left(\frac{3-6x}{\left(1-x\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

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

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

nho k nha

So tu nhien n can tim la :

n=0