Bài 6: 

a²+b²=(a+b)²-2ab

<=> 2010  =36-2ab   

<=>ab=-987

M=a³+b³

=(a+b)(a²-ab+b²)

=6(a²+987+b^2)

=6(2010+987)

=17982

a)a+b+c=9

=>(a+b+c)²=81

=>a²+b²+c²+2ab+2bc+2ca=81

Từ a²+b²+c²=141=>2ab+2bc+2ca=81-141=-60

=>2(ab+bc+ca)=-60=>ab+bc+ca=-30

b)x+y=1

=>(x+y)³=1

=>x³+3x²y+3xy²+y³=1

=>x³+y³+3xy(x+y)=1

=>x³+y³+3xy=1(Do x+y=1)

c)a³-3ab+2c=(x+y)³-3(x+y)(x²+y²)+2(x³+y³)

=x³+3x²y+3xy²+y³-3x³-3y³-3x²y-3xy²+2x³+2y³=0

d)đang tìm hướng giải

b) \(x^3-y^3-3xy\)

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

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

\(=\left(x-y\right)\left(1-xy\right)-3xy\)

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

ĐÂY NÀY:

( x +y) ^2 = a^2 => x^2 + 2xy + y^2 = a^2 

=> 2xy = a^2 - ( x^2  + y^2) = a^2 -b

=> xy = a^2-b/2

Ta có E = x^3 + y^3 = ( x+ y)(  x^2 - xy + y^2)

 E = a ( b - a^2-b/2)

a, Từ x+y=1

=>x=1-y

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

\(=3y^2-3y+1=3\left(y^2-y+\frac{1}{3}\right)=3\left(y^2-2.y.\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\right)\)

\(=3\left[\left(y-\frac{1}{2}\right)^2+\frac{1}{12}\right]=3\left(y-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\) với mọi y

=>GTNN của x³+y³ là 1/4

Dấu "=" xảy ra \(< =>\left(y-\frac{1}{2}\right)^2=0< =>y=\frac{1}{2}< =>x=y=\frac{1}{2}\) (vì x=1-y)

Vậy .......................................

b) Ta có: \(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{y+x}\)

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

\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{y+z}-\left(x+y+z\right)\)

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

Đặt \(A=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}\)

\(A=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{y+x}+1\right)-3\)

\(=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{y+x}-3\)

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

\(=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\ge\frac{9}{2}-3=\frac{3}{2}\)

(phần này nhân phá ngoặc rồi dùng biến đổi tương đương)

\(=>P=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}-1\right)\ge2\left(\frac{3}{2}-1\right)=1\)

=>minP=1

Dấu "=" xảy ra <=>x=y=z

Vậy.....................

a, Thay x = -2 ; y = 3 ta được 

\(A=\dfrac{4\left(-2\right)-5.3}{8\left(-2\right)-7.3}=\dfrac{-8-15}{-16-21}=\dfrac{23}{37}\)

b, Ta có \(\dfrac{x}{y}=\dfrac{5}{4}\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow x=5k;y=4k\)

Thay vào ta được \(A=\dfrac{4.5k-5.4k}{8.5k-7.4k}=\dfrac{0}{40k-28k}=0\)