Question

[Linh Nhật]

bài 1 :a) Cho abc=2. Rút gọn biểu thức : M=\(\dfrac{a}{ab+a+2}\)+\(\dfrac{b}{bc+b+1}\)+\(\dfrac{2c}{ac+2c+2}\)

b) Cho abc=1. Rút gọn biểu thức : N= \(\dfrac{a}{ab+a+1}\)+\(\dfrac{b}{bc+b+1}\)+\(\dfrac{c}{ac+c+1}\)
bài 2: Cho a+b+c=0 ( a,b,c cùng khác 0) . Rút gọn các biểu thức:

a) A=\(\dfrac{a^2}{bc}\)+\(\dfrac{b^2}{ca}\)+\(\dfrac{c^2}{ab}\)

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

Answer 1

[Thiên Hàn]

Bài 2:

a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)

\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)

\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)

\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)

Vì \(a+b+c=0\)

Nên a + b = -c (1)

Thay (1) vào A, ta được:

\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)

\(A=\dfrac{1}{abc}.3abc\)

\(A=3\)

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

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

Vì \(a+b+c=0\)

Nên b + c = -a

=> ( b + c )² = (-a)²

=> b² + c² + 2bc = a²

=> b² + c² = a² - 2bc (1)

Tương tự ta có: c² + a² = b² - 2ac (2)

a² + b² = c - 2ab (3)

Thay (1), (2) và (3) vào B, ta được:

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

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

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

\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)

\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)

Mà \(a^3+b^3+c^3=3abc\) ( câu a )

\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)

\(\Rightarrow B=\dfrac{3}{2}\)

Answer 2

[Thiên Hàn]

Bài 1:

a) GT: abc = 2

\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)

\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)

\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)

\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)

\(M=\dfrac{1+b+bc}{bc+b+1}\)

\(M=1\)

b) GT: abc = 1

\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)

\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)

\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)

\(N=\dfrac{1+b+bc}{bc+b+1}\)

\(N=1\)