a x 10 + b + b x 10 + a = 154

a x 11 + b x 11 = 154

11 x ( a + b ) = 154

a + b = 154 : 11

a + b = 14

=> a = ( 14 + 2 ) : 2 = 8

vậy a là 8

ab + ba = 154 và a - b = 2

10a + b + 10b + a = 154

(10a + a) + (10b + b) = 154

11a + 11b = 154

11(a + b) = 154

a + b = 154 : 11

a + b = 14

\(\Rightarrow\)a = (14 + 2) : 2 = 8

b = 8 - 2 = 6

Vậy ab = 86

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ab+ba=154

10a+b+10b+a=154

(10a+a)+(10b+b)=154

11a+11b=154

11(a+b)=154

a+b=154/11

a+b=14

=>a=(14+2)/2=8

ab+ba=154

10a+b+10b+a=154

(10a+a)+(10b+b)=154

11a+11b=154

11 x (a+b)=154

a+b=154/11

a+b=14

=>a=(14+2)/2=8

ab + ba = 154

=> 10a + b + 10b + a =154

=> 11a + 11b = 154

=> 11. ( a+b )= 154

=> a + b = 154 : 11 = 14

Mà a - b = 2

=> a = (14 + 2) : 2 = 8

=> b = 14 - 8 = 6

ab=86

ab + ba = 154

10a + b  +10b + a = 154

11a + 11b = 154

11(a + b) = 154

a +b = 154 : 11= 14

a = (14 + 2) : 2 = 8

b = 8 - 2 = 6

Vậy a=  8 ; b  = 6

Ta có: $\sqrt[]{ab+2c}=\sqrt[]{ab+(a+b+c)c}=\sqrt[]{ab+ac+bc+c^2}=\sqrt[]{(c+a)(c+b)}$ (do $a+b+c=2$)

Nên $\dfrac{ab}{\sqrt[]{ab+2c}}=\dfrac{ab}{\sqrt[]{(c+a).(c+b)}}=ab.\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}}$

Áp dụng bất đẳng thức Cauchy cho $\dfrac{1}{a+c};\dfrac{1}{b+c}>0$ có:

$\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}} \leq \dfrac{1}{2}.(\dfrac{1}{a+c}+\dfrac{1}{b+c})$

Nên $\dfrac{ab}{\sqrt[]{ab+2c}} \leq \dfrac{1}{2}.ab.(\dfrac{1}{a+c}+\dfrac{1}{b+c})= \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})$

Tương tự ta có: $\dfrac{bc}{\sqrt[]{bc+2a}} \leq \dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})$

$\dfrac{ca}{\sqrt[]{ca+2b}} \leq \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})$

Nên $Q \leq  \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})+\dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})+ \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}.(\dfrac{b(a+c)}{a+c}+\dfrac{a(b+c)}{b+c}+\dfrac{c(a+b)}{a+b}=\dfrac{1}{2}.(a+b+c)=1$ (do $a+b+c=2$)

Dấu $=$ xảy ra khi $a=b=c=\dfrac{2}{3}$

ab + ba = 154

10 x a + b  + 10 x b + a = 154

11 x a + 11 x b = 154

11 x ﴾ a+b﴿ = 154

a+b = 14 a =

﴾14 + 2﴿ : 2 = 8

Vậy a = 8

ab+ba=154

10a+b+10b+a=154

11a+11b=154

11(a+b)=154

a+b=154:11

a+b=14

Vì a+b=14;a-b=2

a là:

 (14+2):2=8

bn bấm vào câu hỏi tương tự nhé

a = 8

nhớ k cho mik nha

AB + BA = 154

10A + B + 10B + A = 154

(10A + A) + (10B + B) = 154

11A + 11B = 154

11(A + B) = 154

A + B = 154 : 11

A + B = 14

A = (14 + 2) : 2

A = 8

ab + ba=154

(10*a+b)+(10*b+a)=154

11*a+11*b=154

11*(a+b)=154

=> a+b=14

Vậy a=(14+2):2=8

a=8 đó

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