Question

Bài 29. Cho \( a + b = 1 \).

Tính giá trị của các biểu thức sau: \( M = a^3 + b^3 + 3ab(a^2 + b^2) + 6a^2b^2(a + b) \).

Bài 30. Cho \( xyz = 2020 \)

Chứng minh rằng:
\[
\frac{2020x}{xy + 2020x + 2020} + \frac{y}{yz + y + 2020} + \frac{z}{xz + z + 1} = 1
\]

Bài 31. Cho \( a, b, c \) khác 0 và \( a + b + c = 0 \).

Rút gọn biểu thức:
\[
A = \frac{a^2}{a^2 - b^2 - c^2} + \frac{b^2}{b^2 - c^2 - a^2} + \frac{c^2}{c^2 - a^2 - b^2}
\]

Answer 1

Bài 29:

\(M=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\)

\(=1-3ab+3ab\left[1-2ab\right]+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)

=1

Bài 31:

a+b+c=0

=>b+c=-a; a+c=-b; a+b=-c

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

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

\(=\dfrac{a^2}{a^2-\left[\left(b+c\right)^2-2bc\right]}+\dfrac{b^2}{b^2-\left[\left(c+a\right)^2-2ac\right]}+\dfrac{c^2}{c^2-\left[\left(a+b\right)^2-2ab\right]}\)

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

\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}\)

\(=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2abc}=\dfrac{\left(-c\right)^3+c^3-3ab\left(-c\right)}{2abc}\)

\(=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)