Xét khai triển:

\(\left(1+x\right)^{90}=C_{90}^0+C_{90}^1x+C_{90}^2x^2+...+C_{90}^{90}x^{90}\)

Thay \(x=2\) ta được:

\(3^{90}=C_{90}^0+2C_{90}^1+2^2C_{90}^2+...+2^{90}C_{90}^{90}\)

Vậy \(B=3^{90}\)

\(X=\left(a+b\right)^n=\sum\limits^n_{k=0}C^k_n.a^k.b^{n-k}\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

\(\Rightarrow A=\sum\limits^{90}_{k=2}C^k_{90}.2^k=...\)

Hoặc có thể làm như vầy: \(A=X-C^0_{90}.2^0-C^1_{90}.2=3^{90}-1-90.2=...\)

a, (x+2)(x−2)−(x−3)(x+1)(x+2)(x-2)-(x-3)(x+1)

=x2−4−(x2−3x+x− 3)=x2-4-(x2-3x+x- 3)

=x2− 4−x2+2x+3=x2- 4-x2+2x+3

=2x−1=2x-1

2.

a, x2−4+(x−2)2x2-4+(x-2)2

=(x−2)(x+2) +(x−2)2=(x-2)(x+2) +(x-2)2

=(x−2)(x+2+x−2)=(x-2)(x+2+x-2)

=2x(x−2)=2x(x-2)

b, x3−2x2+x−xy2x3-2x2+x-xy2

=x(x2− 2x+1−y2)=x(x2- 2x+1-y2)

=x[(x−1)2 −y2]=x[(x-1)2 -y2]

=x(x−1−y)(x−1+y)=x(x-1-y)(x-1+y)

c, x3−4x2−12x+27x3-4x2-12x+27

=(x3+27)−(4x2+12x)=(x3+27)-(4x2+12x)

=(x+3)(x2−3x+9)−4x(x+3)=(x+3)(x2-3x+9)-4x(x+3)

=(x+3)(x2−3x+9−4x)=(x+3)(x2-3x+9-4x)

=(x+3)(x2−7x+9)

ta có , theo định lí viet nên : \(\hept{\begin{cases}x_1+x_2=-m\\x_1x_2=\frac{m^2-2}{2}\end{cases}\Rightarrow}x_1x_2=\frac{\left(x_1+x_2\right)^2-2}{2}\Leftrightarrow x_1^2+x_2^2=2\)

.ta có 

\(A=2x_1x_2+\frac{3}{x_1^2+x_2^2+2x_1x_2+2}=2x_1x_2+\frac{3}{2x_1x_2+4}\)

Mà \(2=x_1^2+x_2^2\ge2\left|x_1x_2\right|\Rightarrow-1\le x_1x_2\le1\)

trên đọna [-1,1] hàm trên đồng biến nên : \(min=-2+\frac{3}{-2+4}=-\frac{1}{2}\)

\(m=2+\frac{3}{2+4}=\frac{5}{2}\)

=\(\frac{5}{2}\)nha

ab(x² + y²) + xy(a² + b²)

= abx² + aby² + xya² + xyb² 

= ax(ay + bx) + by(ay + bx) 

= (ax + by)(ay + bx) 

ab(x²+y²)−xy(a²+b²)ab(x²+y²)−xy(a²+b²)

=abx²+aby²−a²xy−b²xy

=(abx²−b²xy)−(a²xy−aby²)

=bx(ax−by)−ay(ax−by)

=(ax−by)(bx−ay)

= abx² aby²+a²xy+b²xy

=(abx²+a²xy)+(aby²+b²xy)

=ax(bx+ay)+by(ay+bx)

=(bx+ay)(ax+by)

Đây ạ!!

a) Ta có : \(\frac{-2}{x}=\frac{y}{3}\Leftrightarrow xy=-6\) 

Vì x < 0 < y nên

b) Ta có : \(\frac{x-3}{y-2}=\frac{3}{2}\)

\(\Leftrightarrow2\left(x-3\right)=3\left(y-2\right)\)

\(\Leftrightarrow2x-6=3y-6\)

\(\Leftrightarrow2x=3y\)

Mà x - y = 4 => x = 4 + y,do đó \(2\left(4+y\right)=3y\)

=> 8 + 2y = 3y

=> 3y - 2y = 8

=> y = 8

Thay y = 8 vào x - y = 4 ta có :

x - 8 = 4 => x = 4 + 8 = 12

Vậy x = 12,y = 8

Trả lời:

1) \(A=x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x\) (đpcm)

2, \(B=-4x^2+4x-2=-\left(4x^2-4x+2\right)=-\left[\left(4x^2-4x+1\right)+1\right]\)

\(=-\left[\left(2x-1\right)^2+1\right]=-\left(2x-1\right)^2-1\le-1< 0\forall x\) (đpcm)