- Nhân cả tử và mẫu phân thức thứ nhất với a

- Nhân cả tử và mẫu phân thức thứ 2 với ac

- Thay abc =2016 ta có mẫu số chung là :

3ac - 4032 +2016a

- Rút gọn => đáp án : -1

\(P=\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}-\frac{4032-3ac}{3ac-4032+2016a}\)

Ta rút gọn từng biểu thức

\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\)

\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\)

\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}\)

\(=-1+\frac{2016}{3c-2bc+2016}\)

\(\Rightarrow P=-1\)

Ta có:

\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\left(1\right)\)

\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\left(2\right)\)

\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}=-1+\frac{2016}{3c-2bc+2016}\left(3\right)\)

\(P=\left(1\right)+\left(2\right)+\left(3\right)=-1\)

Vậy .........

\(P=\left(1\right)-\left(2\right)+\left(3\right)=-1\)

\(P=\dfrac{2bc-2016}{3c-2bc+2016}-\dfrac{2b}{3-2b+ab}+\dfrac{4032-3ac}{3ac-4032+2016c}\)
\(=\dfrac{2bc-abc}{3c-2bc+abc}-\dfrac{2b}{3-2b+ab}+\dfrac{2abc-3ac}{3ac-2abc+a^2bc}\)
\(=\dfrac{2b-ab}{3-2b+ab}-\dfrac{2b}{3-2b+ab}+\dfrac{2b-3}{3-2b+ab}\)
\(=\dfrac{2b-ab-2b+2b-3}{3-2b+ab}\)
\(=\dfrac{-3+2b-ab}{3-2b+ab}=-1\).

a) \(\dfrac{3ac}{a^3b}=\dfrac{3c}{a^2b}\)

\(\dfrac{6c}{2a^2b}=\dfrac{3c}{a^2b}\)

\(\Rightarrow\dfrac{3ac}{a^3b}=\dfrac{6c}{2a^2b}\)

b) \(\dfrac{3ab-3b^2}{6b^2}=\dfrac{3b\left(a-b\right)}{6b^2}=\dfrac{a-b}{2b}\left(dpcm\right)\)

`a, (3ac)/(a^3b) = (3c)/(a^2b)`

`(6c)/(2a^2b) = (3c)/(a^2b)`

Vậy hai phân thức `=` nhau

`b, (3ab-3b^2)/(6b^2) = (3b(a-b))/(6b^2) = (a-b)/(2b)`

Vậy hai phân thức `=` nhau

a: a^3+b^3+c^3-3abc

=(a+b)^3+c^3-3ab(a+b)-3bac

=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)

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

b: Đề sai rồi bạn

c: 2(a+b+c)*(b/2+c/2-a/2)

=(a+b+c)(b+c-a)

=(b+c)^2-a^2

=c^2+2bc+c^2-a^2

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

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

\(=a^3+ab^2+ac^2+a^2b+b^3+c^2b+a^2c+b^2c+c^3-a^2b-abc-a^2c-ab^2-b^2c-abc-abc-bc^2-ac^2\)

\(=a^3+b^3+c^3-3abc\left(đpcm\right)\)

b) Bạn chỉ cần nhân bung cả 2 vế ra là được á .

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

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

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

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

\(=2bc+b^2+c^2-a^2\left(đpcm\right)\)

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c+2a-3c}{2b+3d+2b-3d}=\dfrac{a}{b}\)

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

Suy ra \(\dfrac{a}{b}=\dfrac{c}{d}\)

tớ không hiểu đầu bài

Đề bài chỉ cho a+b+c=0 và yêu cầu cm ab + 2bc + 3ac < hoặc = 0