\(\frac{a-b}{1+ab}+\frac{b-c}{1+bc}+\frac{c-a}{1+ac}\)

\(=\frac{a-b}{1+ab}+\frac{b-a+a-c}{1+bc}+\frac{c-a}{1+ac}\)

\(=\frac{a-b}{1+ab}+\frac{b-a}{1+bc}+\frac{a-c}{1+bc}+\frac{c-a}{1+ac}\)

\(=\frac{b-a}{1+bc}-\frac{b-a}{1+ab}-\frac{c-a}{1+bc}+\frac{c-a}{1+ac}\)

\(=\left(b-a\right)\left(\frac{1}{1+bc}-\frac{1}{1+ab}\right)-\left(c-a\right)\left(\frac{1}{1+bc}-\frac{1}{1+ac}\right)\)

\(=\left(b-a\right)\left(\frac{1+ab-1-bc}{\left(1+ab\right)\left(1+bc\right)}\right)-\left(c-a\right)\left(\frac{1+ac-1-bc}{\left(1+bc\right)\left(1+ac\right)}\right)\)

\(=\left(b-a\right)\frac{b\left(a-c\right)}{\left(1+ab\right)\left(1+bc\right)}-\left(c-a\right)\frac{c\left(a-b\right)}{\left(1+bc\right)\left(1+ac\right)}\)

Quy đồng:

\(=\frac{\left(b-a\right)b\left(a-c\right)\left(1+ac\right)-\left(c-a\right)c\left(a-b\right)\left(1+ab\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)

\(=\frac{\left(b-a\right)b\left(a-c\right)\left(1+ac\right)-\left(a-c\right)c\left(b-a\right)\left(1+ab\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)

\(=\frac{\left(b-a\right)\left(a-c\right)\left(b\left(1+ac\right)-c\left(1+ab\right)\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)

\(=\frac{\left(b-a\right)\left(a-c\right)\left(b+abc-c-abc\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)

\(=\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(1+ab\right)\left(1+bc\right)\left(1+ac\right)}\)là tích của chúng.

a) \(x^3+y^3+z^3-3xyz\)

\(=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^3-3xyz\)

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

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

câu b đâu

bài 1.a. điều kiện xác định của phân thức là \(x^3-8\ne0\Leftrightarrow x\ne2\)

b .ta có \(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x+2}\)

bài 2.

\(A=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)

\(A=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right).\frac{\left(x+1\right)^2}{2x+1}\)

\(A=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x+1\right)}\right).\frac{\left(x+1\right)^2}{2x+1}\)

\(\Leftrightarrow A=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)

khi \(x=\frac{1}{2}\Rightarrow A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=-3\)

Ta có \(\dfrac{a-b}{ab+1}+\dfrac{b-c}{bc+1}+\dfrac{c-a}{ca+1}=\dfrac{\left(a-b\right)\left(bc+1\right)\left(ca+1\right)+\left(b-c\right)\left(ca+1\right)\left(ab+1\right)+\left(a-b\right)\left(bc+1\right)\left(ca+1\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\).

a. ĐK: a, b, c khác 0.

 \(\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}=1\)

\(\Leftrightarrow\left[\frac{a^2+b^2-c^2}{2ab}-1\right]+\left[\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2-\left(a^2-b^2\right)}{b}+\frac{c^2+\left(a^2-b^2\right)}{a}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2\left(a+b\right)-\left(a^2-b^2\right)\left(a-b\right)}{ab}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{\left(a+b\right)\left(c^2-\left(a-b\right)^2\right)}{2abc}=0\)

\(\Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left(1-\frac{a+b}{c}\right)=0\)

\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)\left(c-a-b\right)=0\)

\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\).

b) Không mất tính tổng quả. G/s: a = b + c

Khi đó ta có:

\(\frac{a^2+b^2-c^2}{2ab}=\frac{\left(b+c\right)^2+b^2-c^2}{2\left(b+c\right)b}=1\)

\(\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-\left(b+c\right)^2}{2bc}=-1\)

\(\frac{c^2+a^2-b^2}{2ca}=\frac{c^2+\left(b+c\right)^2-b^2}{2\left(b+c\right)c}=1\)

=> Điều phải chứng minh.

đây ko phải toán lớp 1 toán lớp 1 làm gì mà khó thế

b/ không mất tính tổng quát ta giả sử: a = b + c thì

\(\frac{a^2+b^2-c^2}{2ab}=\frac{b^2+2bc+c^2-c^2}{2\left(b+c\right)b}=\frac{2b^2+2bc}{2b^2+2bc}=1\)

Tương tự

\(\frac{c^2+a^2-b^2}{2ac}=\frac{2c^2+2ac}{2c^2+2ac}=1\)

\(\frac{b^2+c^2-a^2}{2bc}=\frac{-2bc}{2bc}=-1\)

Vậy trong ba số luôn có 2 số = 1 và 1 số = - 1

\(\frac{a^2+b^2-c^2}{2ab}+\frac{-a^2+b^2+c^2}{2bc}+\frac{a^2-b^2+c^2}{2ca}=1\)

\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+c^2a+c^2b-2abc-a^3-b^3-c^3=0\)

\(\Leftrightarrow\left(a+b-c\right)\left(a+c-b\right)\left(b+c-a\right)=0\)

\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\)

Vậy trong 3 số có 1 số bẳng tổng 2 số kia