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: \(\dfrac{xy^2}{xy-y}=\dfrac{y\cdot xy}{y\cdot\left(x-1\right)}=\dfrac{xy}{x-1}\)

=>Hai phân thức này bằng nhau

b: \(\dfrac{xy+y}{x}=\dfrac{y\left(x+1\right)}{x}\)

\(\dfrac{xy+x}{y}=\dfrac{x\left(y+1\right)}{y}\)

Vì \(\dfrac{y\left(x+1\right)}{x}\ne\dfrac{x\left(y+1\right)}{y}\)

nên hai phân thức này không bằng nhau

c: \(\dfrac{-6}{4y}=\dfrac{-6:2}{4y:2}=\dfrac{-3}{2y}\)

\(\dfrac{3y}{-2y^2}=\dfrac{-3y}{2y^2}=\dfrac{-3y}{y\cdot2y}=\dfrac{-3}{2y}\)

Do đó: \(\dfrac{-6}{4y}=\dfrac{3y}{-2y^2}\)

=>Hai phân thức này bằng nhau

`a, (xy^2)/(xy+y) = (xy^2)/(y(x+1))`

`=(xy)/(x+1)`

Vậy `2` cặp phân thức bằng nhau.

`b, (xy-y)/x = (y(x-1))/x = (y^2(x-1))/(xy)`

`(xy-x)/y = (x(y-1))/y = (x^2(y-1))/(xy)`

Vậy `2` đa thức không bằng nhau

`(a^2-b^2)/(a^2b + ab^2) = ((a-b)(a+b))/(ab(a+b)) = (a-b)/(ab)`.

`(a-b)/(ab) = ((a-b)(a+b))/(ab(a+b)) = (a^2-b^2)/(ab(a+b))`

d: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{A}\)

hay A=x-2

Biểu thức có giá trị bằng 2 thì:

\(1,\\ a,=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\\ b,=\left(x-7\right)\left(x+7\right):\left(x-7\right)=x+7\\ 2,\dfrac{1}{a^2}-ab=\dfrac{1-a^3b}{a^2};\dfrac{1}{a^2}\text{ giữ nguyên}\\ 3,=\dfrac{-7}{t}\\ 4,=\dfrac{1-x+1-y}{x-y}=\dfrac{2-x-y}{x-y}\)

Bài 1:

\(a,\left(16x^3y^2-24x^2y^3+20x^4\right):16x^2=16x^2\left(xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\right):16x^2=xy^2-\dfrac{3}{2}y^3+\dfrac{5}{4}x^2\)

\(b,\left(x^2-49\right):\left(x-7\right)=\left[\left(x-7\right)\left(x+7\right)\right]:\left(x-7\right)=x+7\)

Bài 2:

\(\dfrac{1}{a^2}-ab=\dfrac{1-a^2b}{a^2}\)

\(\dfrac{1}{a^2}\)

Bài 3:

\(\dfrac{7\left(t-z\right)}{t\left(z-t\right)}=\dfrac{-7\left(z-t\right)}{t\left(z-t\right)}=\dfrac{-7}{t}\)

Bài 4:

\(\dfrac{x-1}{y-x}+\dfrac{1-y}{x-y}=\dfrac{x-1}{y-x}-\dfrac{1-y}{y-x}=\dfrac{x-1-1+y}{y-x}=\dfrac{x+y-2}{y-x}\)

a) 2 phân số trên bằng nhau vì khi rút gọn \(\dfrac{6}{-27}\)với -3 ta được \(\dfrac{-2}{9}\)

=>\(\dfrac{-2}{9}=\dfrac{6}{-27}\)

b)-1/-5 và 4/25

-1/-5=-25/-125

4/25=-20/-125

=>-1/-5>4/25

\(a.\)

\(-\dfrac{2}{9}\)

\(\dfrac{-6}{27}=-\dfrac{2}{9}\)

\(\Rightarrow-\dfrac{2}{9}=-\dfrac{6}{27}\)

\(b.\)

\(\dfrac{-1}{-5}=\dfrac{1}{5}=\dfrac{5}{25}\)

\(\dfrac{4}{25}\)

\(\dfrac{5}{25}>\dfrac{4}{25}\)

\(\Rightarrow\dfrac{-1}{-5}>\dfrac{4}{25}\)

a.

\(\dfrac{2a^2-3a-2}{a^2-4}=2\)

\(\Leftrightarrow\dfrac{2a^2-4a+a-2}{\left(a-2\right)\left(a+2\right)}=2\)

\(\Leftrightarrow\dfrac{\left(2a^2-4a\right)+\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=2\)

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

\(\Leftrightarrow\dfrac{\left(2a+1\right)\left(a-2\right)}{\left(a-2\right)\left(a+1\right)}=2\)

\(\Leftrightarrow\dfrac{2a+1}{a+1}=2\)

\(\Leftrightarrow\dfrac{2a+1}{a+1}=\dfrac{2\left(a+1\right)}{a+1}\)

\(\Leftrightarrow2a+1=2a+2\)

Suy ra pt vô nghiệm

a) \(\dfrac{2a^{2^{ }}-3a-2}{a^2-4}\)=2

<=> \(\dfrac{2a^{2^{ }}-3a-2}{\left(a-2\right)\left(a+2\right)}\)=2 (1)

ĐKXĐ: a-2 #0 => a#2

a+2#0 -> a#-2

(1) <=> \(\dfrac{2a^{2^{ }}-3a-2}{\left(a-2\right)\left(a+2\right)}\)= \(\dfrac{2\left(a^{^2}-4\right)}{\left(a-2\right)\left(a+2\right)}\)

=> 2a² - 3a - 2 = 2a² - 8

<=> 2a² - 3a - 2 - 2a² + 8 = 0

<=> -3a + 6 = 0

<=> -3 ( a-2)

<=> -3 = 0 ( vô no )

a-2 = 0 => a = 2

Vậy với A=2 thì biểu thức có giá trị = 2

á t bị nhầm chút:

\(\Leftrightarrow\dfrac{2a+1}{a+2}=2\)

\(\Leftrightarrow\dfrac{2a+1}{a+2}=\dfrac{2\left(a+2\right)}{a+2}\)

\(\Leftrightarrow2a+1=2a+4\)

suy ra pt vô nghiệm