Because you always copy my solution,Tran Nhat Duong

I don't understand meaning of the post

The ratio of 2a to b is: \(\dfrac{7}{3}.2=\dfrac{14}{3}\)

I found the number that STAR present is \(\overline{STAR}=8712\)

Cause \(RATS\cdot4=STAR=2178\cdot4=8712\)

So the value of S+T+A+R is 8 + 7 + 1 + 2 = 18.

WE have

I found the number that STAR present is STAR=8712

Cause RATS⋅4=STAR=2178⋅4=8712

So the value of S+T+A+R is 8 + 7 + 1 + 2 = 18.

Answer:18

We have to prove \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

Let it be : \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)(1).

We have (1) <=> \(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\), true with all a,b > 0.

Done ! 

We have:

\(\left(a-b\right)^2\ge0\)

=> a² + b² \(\ge2ab\)

=> a² + b² + 2ab \(\ge2\left(2ab\right)=4ab\)

=> \(\left(a+b\right)^2\ge4ab\)

Because a > 0, b > 0 => a+b > 0

\(\Rightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a}{ab}+\dfrac{b}{ab}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

So, .......

i loved pokemon

Draw \(CH\perp AB\). 

\(\Delta ABC\) right at C has \(AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+9^2}=15\) 

\(AC^2=AH.AB\Rightarrow AH=\dfrac{12^2}{15}=9.6\Rightarrow DH=9.6-5=4.6\)

\(\Delta ACH\) right at H has \(CH=\sqrt{AC^2-AH^2}=\sqrt{12^2-\left(9.6\right)^2}=7.2\)

\(\Delta CDH\) right at H has

\(CD=\sqrt{CH^2+DH^2}=\sqrt{\left(7.2\right)^2+\left(4.6\right)^2}=\sqrt{73}\) (units)

Draw CH⊥ABCH⊥AB. 

ΔABCΔABC right at C has AB=√AC2+BC2=√122+92=15AB=AC2+BC2=122+92=15 

AC2=AH.AB⇒AH=12215=9.6⇒DH=9.6−5=4.6AC2=AH.AB⇒AH=12215=9.6⇒DH=9.6−5=4.6

ΔACHΔACH right at H has CH=√AC2−AH2=√122−(9.6)2=7.2CH=AC2−AH2=122−(9.6)2=7.2

ΔCDHΔCDH right at H has

CD=√CH2+DH2=√(7.2)2+(4.6)2=√73CD=CH2+DH2=(7.2)2+(4.6)2=73 (units)

my answer is :

225+16=241(cm2)

answer:241 cm2

We have: \(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^{\text{4}}\)

                    \(=\left[\left(x+y\right)\left(x+4y\right)\right]\left[\left(x+2y\right)\left(x+3y\right)\right]+y^4\)

                    \(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

                    \(=\left[\left(x^2+5xy+5y^2\right)-y^2\right]\left[\left(x^2+5xy+5y^2\right)+y^2\right]+y^4\)

                    \(=\left(x^2+5xy+5y^2\right)^2-y^4+y^4\)

                    \(=\left(x^2+5xy+5y^2\right)^2\)

So A is a square number.

We have: A=(x+y)(x+2y)(x+3y)(x+4y)+y4

                    =[(x+y)(x+4y)][(x+2y)(x+3y)]+y4

                    =(x2+5xy+4y2)(x2+5xy+6y2)+y4

                    =[(x2+5xy+5y2)−y2][(x2+5xy+5y2)+y2]+y4

                    =(x2+5xy+5y2)2−y4+y4

                    =(x2+5xy+5y2)2

So A is a square number.