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

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

                              \(=a^2+b^2+c^2-2ab-2bc+2ca\)

Tham khảo:

Cho a≠b≠c, a+b≠c và c2+2ab-2ac-2bc=0 Hãy rút gọn \(B=\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\) - Hoc24

\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)

\(=\dfrac{a^2+a^2-2ac+c^2}{b^2+b^2-2bc+c^2}\)

\(=\dfrac{2a^2-2ac+c^2}{2b^2-2bc+c^2}\)

= a mũ 2+b mũ 2+c mũ 2-2ab-2bc+2ac - b mũ 2+2bc-c mũ 2+2ab -2ac

=a mũ 2

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

b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+ac+c^2-b^2}\)

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

Cái đầu ko rút gọn được

Cái sau:

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

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

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

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

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

\(=a^2\)

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

b ) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{a^2+2ab+b^2-c^2}{a^2+2ac+c^2-b^2}\)

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

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

b) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\frac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)

a) \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)

\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)

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

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

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

\(=a^2\)

Siêu sao bóng đá Lần sau nhớ gõ Latex nhé, tiêu đề bạn nên viết rõ ra như là Toán lớp 8 nhân đa thứ với đa thức chẳng hạn