1)

a) \(\dfrac{18ab}{27bc}=\dfrac{2a}{3c}\)

b) \(\dfrac{-21b^2y^2}{-28by}=\dfrac{3by}{4}\)

c) \(\dfrac{-49a^3}{14b^3}=\dfrac{-7a^3}{2b^3}\)

d) \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{2x^2}{3y^3}\)

2)

a) \(\dfrac{a^3\left(a-5\right)}{a-5}=a^3\)

b) \(\dfrac{3\left(b+7\right)^4}{8\left(b+7\right)^6}=\dfrac{3}{8\left(b+7\right)^2}\)

c) \(\dfrac{15x\left(x+5\right)^2}{20x^2\left(x+5\right)}=\dfrac{3\left(x+5\right)}{4x}\)

d) \(\dfrac{x^3-4x^2}{y\left(x-4\right)}=\dfrac{x^2\left(x-4\right)}{y\left(x-4\right)}=\dfrac{x^2}{y}\)

e) \(\dfrac{5\left(a-2c\right)^2}{2a^2-4ac}=\dfrac{5\left(a-2c\right)^2}{2a\left(a-2c\right)}=\dfrac{5\left(a-2c\right)}{2a}\)

3)

a) \(\dfrac{ax-3a}{bx-3b}=\dfrac{a\left(x-3\right)}{b\left(x-3\right)}=\dfrac{a}{b}\) (câu này mình sửa lại đề)

b) \(\dfrac{5x+20y}{15x+60y}=\dfrac{5\left(x+4y\right)}{15\left(x+4y\right)}=\dfrac{1}{3}\)

c) \(\dfrac{3b-9c}{5b^2-15bc}=\dfrac{3\left(b-3c\right)}{5b\left(b-3c\right)}=\dfrac{3}{5b}\)

d) \(\dfrac{8a^2+40ab}{ab+5b^2}=\dfrac{8a\left(a+5b\right)}{b\left(a+5b\right)}=\dfrac{8a}{b}\)

4)

a) \(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)

\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

b) \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

5)

a) \(\dfrac{45x\left(3-x\right)}{15\left(x-3\right)^3}=\dfrac{-45x\left(x-3\right)}{15\left(x-3\right)^3}=\dfrac{-3x}{\left(x-3\right)^2}\)

b) \(\dfrac{36\left(x-2\right)^3}{32-16x}=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=\dfrac{-9\left(x-2\right)^2}{4}\)

c) \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}=\dfrac{-x}{5y}\)

d) \(\dfrac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\dfrac{-\left(y+x\right)\left(x-y\right)}{\left(x-y\right)^3}=\dfrac{-x-y}{\left(x-y\right)^2}\)

1.

a, \(\dfrac{18ab}{27bc}=\dfrac{18ab:9b}{27bc:9b}=\dfrac{2a}{3c}\)

b, \(\dfrac{-21b^2y^2}{-28by}=\dfrac{-21b^2y^2:\left(-7\right)by}{-28by:\left(-7\right)by}=\dfrac{3by}{4}\)

c, \(\dfrac{-49a^3}{14b^3}=\dfrac{-49a^3:7}{14b^3:7}=\dfrac{-7a^3}{2b^3}\)

d, \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{6xy^2\cdot2x^2}{6xy^2\cdot3y^3}=\dfrac{2x^2}{3y^3}\)

2.

a,\(\dfrac{a^3\cdot\left(a-5\right)}{a-5}=\dfrac{a^3}{1}=a^3\)

b,\(\dfrac{3\cdot\left(b+7\right)^4}{8\cdot\left(b+7\right)^6}=\dfrac{3}{8\cdot\left(b+7\right)^2}\)

c,\(\dfrac{15x\cdot\left(x+5\right)^2}{20x^2\cdot\left(x+5\right)}=\dfrac{3\cdot\left(x+5\right)}{4x}\)

d,\(\dfrac{x^3-4x^2}{y\cdot\left(x-4\right)}=\dfrac{x^2}{y}\)

e,\(\dfrac{5\cdot\left(a-2x\right)^2}{2a^2-4ac}=\dfrac{5\cdot\left(a-2x\right)}{2a}\)

3.

a,\(\dfrac{ax-3a}{bx-3b}=\dfrac{a\cdot\left(x-3\right)}{b\cdot\left(x-3\right)}=\dfrac{a}{b}\)

b, \(\dfrac{5x+20y}{15x+60y}=\dfrac{5\cdot\left(x+4y\right)}{15\cdot\left(x+4y\right)}=\dfrac{5}{15}=\dfrac{1}{3}\)

c, \(\dfrac{3b-9c}{5b^2-15bc}=\dfrac{3\cdot\left(b-3c\right)}{5b\cdot\left(b-3c\right)}=\dfrac{3}{5b}\)

d, \(\dfrac{8a^2+40ab}{ab+5b^2}=\dfrac{8a\cdot\left(a+5b\right)}{b\cdot\left(a+5b\right)}=\dfrac{8a}{b}\)

4.

a,\(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\cdot\left(x^2-4x+4\right)}{x\cdot\left(x^3-8\right)}=\dfrac{3\cdot\left(x-2\right)^2}{x\cdot\left(x-2\right)\cdot\left(x^2+2x+4\right)}=\dfrac{3\cdot\left(x-2\right)}{x\cdot\left(x^2+2x+4\right)}=\dfrac{3\cdot\left(x-2\right)}{x\cdot\left(x+2\right)^2}\)

b, \(\dfrac{7x^2+14x+7}{3x^2+3x}=\dfrac{7\cdot\left(x^2+2x+1\right)}{3x\cdot\left(x+1\right)}=\dfrac{7\cdot\left(x+1\right)^2}{3x\cdot\left(x+1\right)}=\dfrac{7\cdot\left(x+1\right)}{3x}\)

5.

a, \(\dfrac{45x\cdot\left(3-x\right)}{15x\cdot\left(x-3\right)^3}=\dfrac{3\cdot\left(3-x\right)}{\left(x-3\right)^3}=\dfrac{-3\cdot\left(x-3\right)}{\left(x-3\right)^3}=\dfrac{-3}{\left(x-3\right)^2}\)

b, \(\dfrac{36\cdot\left(x-2\right)^3}{36-16x}=\dfrac{36\cdot\left(x-2\right)^3}{16\cdot\left(2-x\right)}=\dfrac{36\cdot\left(-\left(x-2\right)\right)^3}{16\cdot\left(2-x\right)}=\dfrac{-36\cdot\left(2-x\right)^3}{16\cdot\left(2-x\right)}=\dfrac{-9\cdot\left(2-x\right)^2}{4}\)

c, \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\cdot\left(x-y\right)}{5y\cdot\left(y-x\right)}=\dfrac{-x\cdot\left(y-x\right)}{5y\cdot\left(y-x\right)}=\dfrac{-x}{5y}\)

d, \(\dfrac{y^2-x^2}{x^3-3x^2y+3xy^2+y^3}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{\left(x-y\right)^3}=\dfrac{-\left(x+y\right)\cdot\left(y-x\right)}{\left(x-y\right)^3}=\dfrac{-\left(x+y\right)}{\left(x-y\right)^2}\)

1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0

Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)

b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)

B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)

B = \(\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

B =  \(\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}\)

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

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

\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)

\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)

ai làm hộ mình đi

Mình mới lớp 7 thui, mình ko bít lớp 8, xin lỗi, tha lỗi cho mình nha.

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

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

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

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

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

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)

a)\(\frac{12x^5y^2}{8x^3y^5}=\frac{3x^2}{2y^3}\)

b)\(\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)

a/ \(\frac{3x^2}{2y^3}\)

b/ \(\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)

a)\(\frac{12x^5y^2}{8x^3y^5}\)

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

b)\(\frac{x^2+2x+1}{5x^2+5x}\)

=\(\frac{\left(x+1\right)^2}{5x\left(x+1\right)}\)

=\(\frac{x+1}{5x}\)

t tôi nhé bn

\(a,\frac{12x^5y^2}{8x^3y^5}=\frac{4x^3y^2.3x^2}{4x^3y^2.2y^3}=\frac{3x^2}{2y^3}\)

\(b,\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{\left(x+1\right)\left(x+1\right)}{5x\left(x+1\right)}=\frac{x+1}{5x}\)

Xong rồi nhé! Chúc bạn học giỏi.

a, \(\frac{12x^5y^2}{8x^3y^5}=\frac{3x^2}{2y^3}\)

b, \(\frac{x^2+2x+1}{5x^2+5x}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)

a)\(\frac{12x^5y^2}{8x^3y^5}=\frac{4\cdot3\cdot x^3x^2y^2}{4\cdot2\cdot x^3y^2y^3}=\frac{3x}{2y^3}\)

b)\(\frac{x^2+2x+1}{5x^2+5x}=\frac{x^2+2\cdot x\cdot1+1^2}{5x\left(x+1\right)}=\frac{\left(x+1\right)^2}{5x\left(x+1\right)}=\frac{x+1}{5x}\)