a: =>4/9*a/b=5/27+3/27=8/27

=>a/bb=8/27:4/9=8/27*9/4=72/108=2/3

b: =>9/13*a/b=40/91+1/7=53/91

=>a/b=53/63

`a)4/9xxa/b-1/9=5/27`

`=>4/9xxa/b=5/27+1/9`

`=>4/9xxa/b=5/27+3/27`

`=>4/9xxa/b=8/27`

`=>a/b=8/27:4/9`

`=>a/b=8/27xx9/4`

`=>a/b=2/3`

Vậy `a=2` và `b=3`

__

`b)9/13xxa/b+1/7=40/91`

`=>9/13xxa/b=40/91-1/7`

`=>9/13xxa/b=40/91-13/91`

`=>9/13xxa/b=27/91`

`=>a/b=27/91:9/13`

`=>a/b=27/91xx13/9`

`=>a/b=39/91`

Vậy `a=39` và `b=91`

a) \(A=x\left(x-\dfrac{4}{9}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

b) \(A=x\left(x-\dfrac{4}{9}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{9}< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{9}>0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x< 0\\x>\dfrac{4}{9}\end{matrix}\right.\)

c) \(A=x\left(x-\dfrac{4}{9}\right)< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{9}< 0\end{matrix}\right.\)( do \(x>x-\dfrac{4}{9}\))

\(\Leftrightarrow\dfrac{4}{9}>x>0\)

a)   A=x.(x-4/9)=0

<=>X=0 và X=4/9

b).  A=x.(x-4/9)>0

<=>X>0 và X>4/9

c).   A=x.(x-4/9)<0

<=>X<0 và X<4/9

b: Ta có: A<0

nên \(0< x< \dfrac{4}{9}\)

c: Ta có: A>0

nên \(\left[{}\begin{matrix}x< 0\\x>\dfrac{4}{9}\end{matrix}\right.\)

`5/9 xx a/b -11/21 =4/21`

`=> 5/9 xx a/b = 4/21 + 11/21`

`=> 5/9 xx a/b=15/21`

`=> a/b= 15/21 : 5/9`

`=>a/b= 15/21 xx 9/5`

`=> a/b=135/105`

`=>a/b=9/7`

Vậy `a/b=9/7`

\(\dfrac{5}{9}\times\dfrac{a}{b}-\dfrac{11}{21}=\dfrac{4}{21}\)

\(\Leftrightarrow\dfrac{5}{9}\times\dfrac{a}{b}=\dfrac{15}{21}\)

\(\Leftrightarrow\dfrac{a}{b}\times\dfrac{5}{9}=\dfrac{15}{21}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{15}{21}:\dfrac{5}{9}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{135}{105}=\dfrac{9}{7}\)

vaayjj : \(\dfrac{a}{b}=\dfrac{9}{7}\)

\(\dfrac{5}{9}\times\dfrac{a}{b}-\dfrac{11}{21}=\dfrac{4}{21}\\ \Rightarrow\dfrac{5}{9}\times\dfrac{a}{b}=\dfrac{4}{21}+\dfrac{11}{21}\\ \Rightarrow\dfrac{5}{9}\times\dfrac{a}{b}=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{7}\div\dfrac{5}{9}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{7}\times\dfrac{9}{5}\\ \Rightarrow\dfrac{a}{b}=\dfrac{9}{7}\)

Vậy \(\dfrac{a}{b}=\dfrac{9}{7}\)

Ta có : A(0) = a.0² + b.0 + c = c = 4

=> c = 4

Lại có A(1) = a.1² + b.1 + c = a + b + c = 9

mà c = 4

=> a + b = 5 (1)

Mặt khác A(2) = a.2² + b.2 + c = 4a + 2b + c = 14

mà c = 4

=> 4a + 2b = 10

=> 2a + b = 5 (2)

Từ (1)(2) => a = 0 ; b = 5

=> A(x) = 5x + 4

Vậy a = 0 ; b = 5 ; c = 4

a) Ta có: \(\left|2a-4\right|+\left|a-2\right|=0\)

\(\Leftrightarrow2\left|a-2\right|+\left|a-2\right|=0\)

\(\Leftrightarrow3\left|a-2\right|=0\)

\(\Leftrightarrow\left|a-2\right|=0\)

\(\Leftrightarrow a-2=0\)

hay a=2

Vậy: a=2

abc:(a+b+c)=100

aba=(a+b+c)x100

abc=a x100+bx100+cx100

ax100+bx10+c=ax100+bx100+cx100

( đề có vẻ sai )

abc:(a+b+c)=100

aba=(a+b+c)x100

abc=a x100+bx100+cx100

ax100+bx10+c=ax100+bx100+cx100

( đề có vẻ sai ) Nếu bn cảm thấy đúng thì k cho mình nhé!Học Tốt

Theo bài ra. ta có: a x b = c => b = c : a (1)  thế vào b x c = 4 x a

Có: c : a  x c = 4 x a => c² = 4 x a² (2)

a x c = 9 x b => a x c = 9 x c : a => a² = 9 => a = 3 (a dương)

a = 3 thì (2) <=> c² = 4 x a² = 5 x 3² = 36 = 6² => c = 6 (c dương)

Thế vào (1) có: b = c : a = 6 : 3 = 2

Vậy a = 3 ; b = 6 và c = 2