Ta có:

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

\(=a^2x^2-2abxy+b^2y^2+a^2y^2+2abxy+b^2x^2\) \(=\left(ax-by\right)^2+\left(ay+bx\right)\)

\(=vp\)

\(\Rightarrowđpcm\)

\(\left(2x+3y\right)^2+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)

\(T=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x+4=a\) \(\Rightarrow T=\left(a-1\right)\left(a+1\right)\)

\(=a^2-1=\left(x^2+5x+5\right)^2-1\ge-1\)

Vậy \(Min_T=-1\) khi

\(x^2+5x+5=0\Rightarrow\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{5}{4}=0\)\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2=\dfrac{5}{4}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x+\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}-\dfrac{5}{2}\\x=-\sqrt{\dfrac{5}{4}}-\dfrac{5}{2}\end{matrix}\right.\)

\(1,\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)

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

\(=\left(x-3\right)\left(x-4\right)\)

\(2,6x+3-\left(2x-5\right)\left(2x+1\right)\)

\(=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

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

\(3,\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)

\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

\(4,\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)

\(=\left(3x-2\right)\left(3x-6\right)\)

\(5,\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\)\(=\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\)\(=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)

\(=\left(x-5\right)\left(4x+1\right)\)

6, Tương tự

Theo bài toán :

\(\widehat{A}-\widehat{D}=40^o\Rightarrow\widehat{D}=\widehat{A}-40^o\)

Vì ABCD là hình thang \(\Rightarrow\widehat{A}+\widehat{D}=180^o\Leftrightarrow\widehat{A}+\widehat{A}-40^o=180^o\Rightarrow2\widehat{A}=220^o\Rightarrow\widehat{A}=110^o\)

\(\Rightarrow\widehat{D}=\widehat{A}-40^o=70^o\)

Mà \(\widehat{A}=2\widehat{C}\Rightarrow\widehat{C}=\dfrac{\widehat{A}}{2}=55^o\)

\(\Rightarrow\widehat{D}=360^o-\widehat{A}-\widehat{B}-\widehat{C}=125^o\)

\(B=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}....\dfrac{99^2}{99.100}\)

\(=\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}....\dfrac{99.99}{99.100}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}=\dfrac{1.2.3...99}{2.3.4...100}=\dfrac{1}{100}\)

Bài 1 :

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2017}\right)\) \(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2015}{2016}.\dfrac{2016}{2017}=\dfrac{1.2.3.4....2015.2016}{2.3.4.5...2016.2017}=\dfrac{1}{2017}\)

\(x^3-x^2-7x+a\)

\(=x^3-3x^2+4x^2-12x+5x-15+a\)

\(=x^2\left(x-3\right)+4x\left(x-3\right)+5\left(x-3\right)-15+a\) \(=\left(x-3\right)\left(x^2+4x+5\right)-15+a\)

Nhận thấy :

\(\left(x-3\right)\left(x^2+4x+5\right)⋮x-3\)

Vậy để đa thức \(x^3-x^2-7x+a⋮x-3\) thì -15 + a phải chia hết cho x- 3

\(\Rightarrow a=15\)

\(a,x^2y-xz+z-y=\left(x^2y-y\right)-\left(xz-z\right)\)

\(=y\left(x^2-1\right)-z\left(x-1\right)\)

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

\(=\left(x-1\right)\left(yx+y-z\right)\)

\(b,x^4-x^3+x^2-1\)

\(=x^3\left(x-1\right)+\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)\left(x^3+x+1\right)\)

\(a,3x^3-x^2-21x+7\)

\(=x^2\left(3x-1\right)-7\left(3x-1\right)\)

\(=\left(3x-1\right)\left(x^2-7\right)\)

\(b,x^3-4x^2+8x-8\)

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

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

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

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

\(c,x^3-5x^2-5x+1\)

\(=\left(x^2+1\right)-5x\left(x+1\right)\)

=\(\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)

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