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)

Câu 1. D

Câu 4. A, C

Câu 5. Xem lại đề!

Câu 5. D

Câu 7. B

Câu 8. A

Bạn không hiểu chỗ nào cứ hỏi lại mình nhé!

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

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

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

B1

a, \(=>A=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x.2y=4xy\)

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

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

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

d, \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)

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

\(=\left(a+b-c+b-c\right)\left(a+b-c-b+c\right)\)

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

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

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

B2:

\(\)\(x+y=3=>\left(x+y\right)^2=9=>x^2+2xy+y^2=9\)

\(=>xy=\dfrac{9-\left(x^2+y^2\right)}{2}=\dfrac{9-\left(17\right)}{2}=-4\)

\(=>x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(17+4\right)=63\)

Bài 1: 

a) Ta có: \(\left(x+y\right)^2-\left(x-y\right)^2\)

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

=4xy

b) Ta có: \(\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)

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

\(=\left(2y\right)^2=4y^2\)

c) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)

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

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

\(=x^6-1\)

d) Ta có: \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(b-c\right)^2\)

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

\(=\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)+\left(a+b+c-b+c\right)\left(a+b+c+b-c\right)\)

\(=a\cdot\left(a+2b-2c\right)+\left(a+2c\right)\left(a-2b\right)\)

\(=a^2+2ab-2ac+a^2-2ab+2ac-4bc\)

\(=2a^2-4bc\)

Bài 2: 

Ta có: x+y=3

nên \(\left(x+y\right)^2=9\)

\(\Leftrightarrow2xy+17=9\)

\(\Leftrightarrow2xy=-8\)

hay xy=-4

Ta có: \(x^3+y^3\)

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

\(=3^3-3\cdot\left(-4\right)\cdot3\)

\(=27+36=63\)

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

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

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

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

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

\(\Rightarrowđpcm\)

Ta có: (x+a)(x+b)(x+c) = x³ + (a+b+c)x² +(ab+bc+ca)x + abc

VT = (x²+ax+bx+ab)(x+c)

= x³ + ax² + bx² + abx + cx² + cax + bcx + abc ⁽¹⁾

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

= x³ + ax² + bx² + abx + cx² + cax + bcx + abc ⁽²⁾

Từ (1) và (2), suy ra:

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

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

Vậy (x + a) . (x + b) = x² + (a + b) . x + ab.

b)\(\left(x+a\right).\left(x+b\right).\left(x+c\right)=\left(x^2+bx+ax+ab\right).\left(x+c\right)\)(Vế đầu mình 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 (x + a) . (x + b) . (x + c) = x³ + (a + b + c) . x² + (ab + bc + ca) . x + abc.