ko ghi đề bài nha làm luôn

a) \(\frac{\left(2x+2y\right)+\left(5x+5y\right)}{\left(2x+2y\right)-\left(5x+5y\right)}=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}=\frac{\left(2+5\right)\left(x+y\right)}{\left(2-5\right)\left(x+y\right)}=\frac{-7}{3}\)

b)\(\frac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\frac{4x}{5x^2}=\frac{4}{5x}\)

a)ĐK: \(x\ne-y;x,y\ne0\)

\(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\frac{\left(x+y\right)\left(2+5\right)}{\left(x+y\right)\left(2-5\right)}=-\frac{7}{3}\)

b) ĐK: ...bạn tự xét...

\(\frac{4x^2-4xy}{5x^3-5x^2y}=\frac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\frac{4x}{5x^2}=\frac{4}{5x}\)

Vậy ...

Lời giải:

a) \(\frac{x^2-16}{4x-x^2}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)

b) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

c)

\(\frac{(x+y)^2-z^2}{x+y+z}=\frac{(x+y-z)(x+y+z)}{x+y+z}=x+y-z\)

d)

Biểu thức không rút gọn được

e)

\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b)^3-3ab(a+b)+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b)}{a^2+b^2+c^2-ab-bc-ac}\)

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

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

a) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}\)

\(=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)

b) \(\frac{x^2+4x+3}{2x+6}\)

\(=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)

\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

\(=\frac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}\)

\(=\frac{x+1}{2}\)

a) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+3x+x+3}{2\left(x+3\right)}=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}=\frac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\frac{x+1}{2}\)

a) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x+4\right)\left(x-4\right)}{x\left(4-x\right)}\)

\(=\frac{\left(x+4\right)\left(x-4\right)}{-x\left(x-4\right)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)

\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

\(=\frac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\frac{x+1}{2}\)

c) \(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}\)

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

\(=\frac{2x\left(x-2\right)}{x\left(x+2\right)}\)

\(=\frac{2x^2-4x}{x^2+2x}\)

d) \(\frac{x^3-x^2y+xy^2}{x^3+y^3}\)

\(=\frac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{x}{x+y}\)

a)

\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)

c)

\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)

d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)

f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)

g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)

\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)

h)

\(\frac{4x^2-4xy}{5x^3-5x^2y}=\frac{4x(x-y)}{5x^2(x-y)}=\frac{4}{5x}\)

i) \(\frac{(x+y)^2-z^2}{x+y+z}=\frac{(x+y-z)(x+y+z)}{x+y+z}=x+y-z\)

k) \(\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{(x^3)^2+2.x^3.y^3+(y^3)^2}{x(x^6-y^6)}\)

\(=\frac{(x^3+y^3)^2}{x(x^3-y^3)(x^3+y^3)}=\frac{x^3+y^3}{x(x^3-y^3)}\)

a) \(=\frac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)

\(=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=\frac{-7}{3}\)

b)\(=\frac{3x\left(x+y\right)}{y}\)

c) \(\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)

\(=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)

a) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7\left(x+y\right)}{-3\left(x+y\right)}=-\frac{7}{3}.\)

b) \(\frac{15x\left(x+y\right)^3}{5y\left(x+y\right)^2}=\frac{3x\left(x+y\right)}{y}=\frac{3x^2+3xy}{y}\)

c) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\frac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}=\frac{8\left(x-y\right)}{10\left(x-y\right)}=\frac{4}{5}\)

d) \(\frac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\frac{x-z}{2}\)

h) \(\frac{3x\left(1-x\right)}{2\left(x-1\right)}=-\frac{3x\left(x-1\right)}{2\left(x-1\right)}=\frac{-3x}{2}\)

j) \(\frac{6x^2y^2}{8xy^5}=\frac{3x}{4y^3}\)

Câu b) bạn xem lại nhé.

Học tốt ^3^

b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)