a/ Chứng minh:

\(\left(x+a\right)\left(x+b\right)\)

\(=x^2+bx+ax+ab\)

\(=x^2+\left(ax+bx\right)+ab\)

\(=x^2+x\left(a+b\right)+ab=VP\) (đpcm)

b/ Chứng minh:

\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)

\(=x^3+cx^2+ax^2+acx+bx^2+bcx+abx+abc\)

\(=x^3+\left(ax^2+bx^2+cx^2\right)+\left(abx+bcx+acx\right)+abc\)

\(=x^3+x^2\left(a+b+c\right)+x\left(ab+bc+ac\right)+abc=VP\) (đpcm)

a. \(VT=\left(x+a\right)\left(x+b\right)=x^2+ã+bx+ab=x^2+\left(a+b\right)x+ab=VP\)

B. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)

\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)

\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)

từ A=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)

=>A=x²-ax-bx+ab+x²-bx-cx+bc+x²-cx-ax+ac

=>A=3x²-2ax-2bx-2cx+ab+bc+ac

=>A=3x²-2x(a+b+c)+ab+bc+ac

mà a+b+c=2x(gt)

=>A=3x²-2x.2x+ab+bc+ac

=>A=3x²-4x²+ab+bc+ac

=>A=ab+bc+ac-x²=VP

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

Với a = 1, b = 4, c = 2, d = 3 thì  a + b = 5 =c + d.

          Biến đổi:  P(x)  = (x + 1)(x + 4)( x + 2)( x + 3) – 15

                               = (x² + 5x + 4)(x² + 5x + 6) – 15

          Đặt y = x² + 5x + 4 thì P(x) trở thành

          Q(y) = y(y + 2) – 1

           = y² +2y – 15

           = y² – 3y + 5y – 15

           = y(y – 3) + 5( y – 3)

           = (y – 3)(y + 5)

     Do đó: P(x) = (x² +5x + 1)(x² + 5x + 9)

VP là j vậy

a)(x+a)(x+b)

=x(x+b)+a(x+b)

=x²+xb+ax+ab

=x²+(a+b).x+a.b

Vậy (x+a)(x+b)=x²+(a+b).x+a.b

b)(x+a)(x+b)(x+c)

=x(x+b)(x+c)+a(x+b)(x+c)

=(x²+xb)(x+c)+(ax+ab)(x+c)

=x²(x+c)+xb(x+c)+ax(x+c)+ab(x+c)

=x³+x².c+x².b+xbc+ax²+axc+abx+abc

=x³+(a+b+c).x²+(ab+bc+ca).x+abc

Vậy (x+a)(x+b)(x+c)=x³+(a+b+c).x²+(ab+bc+ca).x+abc

c)(a+b+c)(a²+b²+c²-ab-bc-ca)

=a(a²+b²+c²-ab-bc-ca)+b(a²+b²+c²-ab-bc-ca)+c(a²+b²+c²-ab-bc-ca)

=a³+ab²+ac²-a².b-abc-a².c+ba²+b³+bc²-ab²-b².c-bca+ca²+cb²+c³-cab-bc²-c².a

=a³+b³+c³ -abc-bca-cab

=a³+b³+c³ -3abc

Vậy (a+b+c)(a²+b²+c²-ab-bc-ca)=a³+b³+c³ -3abc

ngu như con hà cày

a ) VP = \(\left(x+a\right).\left(x+b\right)=x^2+bx+ax+ab\)

     VT = \(x^2+\left(a+b\right).x+ab=x^2+ax+bx+ab\)

\(\Rightarrow VT=VP\)

b ) VP : \(\left(x+a\right).\left(x+b\right)\left(x+c\right)=\left(x^2+bx+ax+ab\right).\left(x+c\right)\) ( Vế đầu áp dụng luôn ở câu a )

\(=x^2.x+x^2.c+bx.x+bx.c+ax.x+ax.c+ab.x+ab.c\)

\(=x^3+cx^2+bx^2+cbx+ax^2+cax+abx+abc\)

\(=x^3+\left(cx^2+bx^2+ax^2\right)+\left(cbx+cax+abx\right)+abc\)

\(=x^3+\left(a+b+c\right)x^2+\left(ab+ac+bc\right).x+abc\)

Vậy \(\left(x+a\right).\left(x+b\right).\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)

a) VP =\(\left(x+a\right)\left(x+b\right)=x^2+bx+\text{ax+ab}\)

\(VT=x^2+\left(a+b\right).x+ab=x^2+ax+bx+ab\\ =>VT=VP\)

b) VP : \(\left(x+a\right).\left(x+b\right).\left(x+c\right)=\left(x^2+bx+ax+ab\right).\left(x+c\right)\)( Vế đầu áp dụng luôn ở câu a )

\(=x^2.x+x^2.c+bx.x+bx.c+\text{ax}.x+\text{ax}.c+ab.c+ab.c\\ =x^3+cx^2+bx^2-cbx+\text{ax}^2+ca.x+ab.x+abc\\ \)

\(=x^3+\left(cx^2+bx^2+\text{ax}^2\right)-\left(cbx+c\text{ax}+abx\right)+abc\\ =x^3-\left(a+b+c\right)x^2+\left(ab+ac+bc\right).x+abc\)

Vậy \(\left(x+a\right)\left(x-b\right)\left(x+c\right)=x^3+\left(a+b+c\right).x^2+\left(ab+ca+bc\right).x+abc\)

TC:a+b+cd=2p=>b+c=2p-a

=>(b+c)²=(2p-a)²

=>b²+2bc+c²=4p²-4pa+a²

=>b²+2bc+c²-a²=4p²-4pa

=>2bc+b²+c²-a²=4p(p-a) ĐPCM

Ta có : \(a+b+cd=2p\Rightarrow b+c=2p-a\)

\(\Rightarrow\left(b+c\right)^2=\left(2p-a\right)^2\)

\(\Rightarrow b^2+2bc+c^2=4p^2-4pa+a^2\)

\(\Rightarrow b^2+2bc+c^2-a^2=4p^2-4pa\)

\(\Rightarrow2bc+b^2+c^2-a^2=4p\left(p-a\right)\)

\(\RightarrowĐPCM\)