Ta có: \(\dfrac{x}{6}\) = \(\dfrac{y}{12}\)
⇒\(\left(\dfrac{x}{6}\right)^2\) = \(\left(\dfrac{y}{12}\right)^2\) =\(\dfrac{xy}{6.12}\)= \(\dfrac{648}{72}\) = \(9\)
⇒\(\dfrac{x^2}{36}\) = \(9\) ⇒ \(x^2\) = \(324\)

    \(\dfrac{y^2}{144}=9\) ⇒ \(y^2=1296\)

⇒ \(x=\pm18\); \(y=\pm36\)
Vậy cặp số \(\left(x;y\right)\in\left\{\left(18;36\right);\left(-18;-36\right)\right\}\)
    

Đặt \(\dfrac{x}{6}=\dfrac{y}{12}=k\Rightarrow x=6k;y=12k\)

Ta có: \(xy=648\)

\(\Rightarrow6k.12k=648\)

\(\Rightarrow72k^2=648\)

\(\Rightarrow k^2=648:72\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3\)

* Với \(k=1\Rightarrow x=6.1=6;y=12.1=12\)

* Với \(k=-1\Rightarrow x=6.\left(-1\right)=-6;y=12.\left(-1\right)=-12\)

Vậy \(x=6;y=12\) hoặc \(x=-6;y=-12\)

\(#Nulc`\)

1. How many calories do people need a day to stay in shape?

2. When did you start working as a volunteer?

3. Who do you often share your hobby with?

4. Where does Lan buys 6 apples?

5. Who will meet his friends next Sunday?

6. What did your uncle drink last year?

\(#Nulc`\)

1. How many calories do people need a day to stay in shape?

2. When did you start working as a volunteer?

3. Who do you often share your hobby with?

4. Where does Lan buy six apples?

5. Who will meet his friends next Sunday?

6. What did your uncle drink last year?

Mong moi người giúp

Đúng mik tick

cái gì đấy khó thế

Lời giải:

Áp dụng BĐT Cô-si: 

$a+b+c\geq 3\sqrt[3]{abc}=3(1)$
Tiếp tục áp dụng BĐT Cô-si:

$a^3+a\geq 2a^2$

$b^3+b\geq 2b^2$

$c^3+c\geq 2c^2$

$\Rightarrow a^3+b^3+c^3\geq 2(a^2+b^2+c^2)-(a+b+c)$

Lại có:

$a^2+1\geq 2a$

$b^2+1\geq 2b$

$c^2+1\geq 2c$

$\Rightarrow a^2+b^2+c^2\geq 2(a+b+c)-3=(a+b+c)+(a+b+c)-3$

$\geq a+b+c+3-3=a+b+c(2)$

$\Rightarrow a^3+b^3+c^3\geq 2(a^2+b^2+c^2)-(a+b+c)\geq a^2+b^2+c^2(3)$

Từ $(1); (2); (3)$ ta có đpcm.

Lời giải:
Áp dụng BĐT Cô-si:
a^3+2b^3=a^3+b^3+b^3\geq 3\sqrt[3]{a^3b^6}=3ab^2$

$a^3+1+1\geq 3a$

$b^3+1+1\geq 3b$

Cộng theo vế các BĐT trên:

$a^3+2b^3+(a^3+2)+2(b^3+2)\geq 3ab^2+3a+6b$

$\Leftrightarrow 2(a^3+2b^3)+6\geq 3(ab^2+a+2b)=3.4=12$

$\Rightarrow a^3+2b^3\geq (12-6):2=3$

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=1$

chịu, mình mới học lớp 5 à.

mình cũng chỉ hoch lớp 5

Áp dụng BĐT trị tuyệt đối:

\(A=\left|x+3\right|+\left|5-x\right|+\left|x-2\right|\ge\left|x+3+5-x\right|+\left|x-2\right|\)

\(\Rightarrow A\ge8+\left|x-2\right|\)

Mà \(\left|x-2\right|\ge0;\forall x\)

\(\Rightarrow A\ge8\)

Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\left(x+3\right)\left(5-x\right)\ge0\\\left|x-2\right|=0\\\end{matrix}\right.\)

\(\Rightarrow x=2\)

Lời giải:

$3\text{VT}=\frac{3a}{3a+1}+\frac{3b}{3b+1}+\frac{3c}{3c+1}$

$=1-\frac{1}{3a+1}+1-\frac{1}{3b+1}+1-\frac{1}{3c+1}$

$=3-\left[\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\right]$
Áp dụng BĐT Cauchy-Schwarz:

$\frac{1}{3a+1}+\frac{1}{3b+1}+\frac{1}{3c+1}\geq \frac{9}{3a+1+3b+1+3c+1}=\frac{9}{3(a+b+c)+3}=\frac{9}{3.6+3}=\frac{3}{7}$

$\Rightarrow 3\text{VT}\leq 3-\frac{3}{7}=\frac{18}{7}$

$\Rightarrow \text{VT}\leq \frac{6}{7}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=2$