Áp dụng BĐT tam giác:

\(\left\{{}\begin{matrix}a< b+c\\b< a+c\\c< a+b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2< a\left(b+c\right)\\b^2< b\left(a+c\right)\\c^2< c\left(a+b\right)\end{matrix}\right.\)

Cộng vế:

\(\Rightarrow a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)

\(x+2y=3xy+3\)

\(\Leftrightarrow3xy-x-2y+3=0\)

\(\Leftrightarrow9xy-3x-6y+9=0\)

\(\Leftrightarrow3x\left(3y-1\right)-2\left(3y-1\right)+7=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3y-1\right)=-7\)

Vậy \(\left(x;y\right)=\left(1;-2\right);\left(3;0\right)\)

Ta có: \(f\left(-2\right)=4a-2b+c\)

 \(f\left(3\right)=9a+3b+c=13a+b+2c-4a+2b-c=-4a+2b-c\)

\(\Rightarrow f\left(-2\right).f\left(3\right)=\left(4a-2b+c\right)\left(-4a+2b-c\right)=-\left(4a-2b+c\right)^2\le0\) (đpcm)

`3x - 6/20=5x - 9 /30`

`=> 3x-5x= -9/30 + 6/20`

`=> -2x= 0`

`=>x=0:(-2)`

`=>x=0`

Câu đánh dấu đỏ của em đây nhé:

\(\dfrac{x}{y}\)  = \(\dfrac{2}{3}\) =>   \(\dfrac{x}{2}\) = \(\dfrac{y}{3}\) => \(\dfrac{x^2}{4}\) = \(\dfrac{x.y}{2.3}\) = \(\dfrac{54}{6}\) = 9

⇒ \(x^2\) = 9.4 = 36 ⇒ \(x^2\) = 6² ⇒ \(x\) = + - 6 

\(x\) = 6 ⇒ \(y\) = 6 : \(\dfrac{2}{3}\) = 9; 

\(x=-6\) ⇒ \(y=-6\) : \(\dfrac{2}{3}\) = -9

Gọi số tờ tiền mệnh giá 50 000 đồng; 20 000 đồng; 10 000 đồng lần lượt là : x, y, z ( x, y, z \(\in\) N^*) 

Theo bài ra ta có : 

50 000 \(x\) = 20 000 \(y\) = 10 000 \(z\);   \(x+y+z=85\)

              5\(x\) = 2 \(y\)  = \(z\)  ⇒ y = \(\dfrac{5}{2}\)\(x\);     \(z\)  =  5\(x\)

          ⇒ \(x+\dfrac{5}{2}x+5x\) = 85 ⇒ \(x.(1+\dfrac{5}{2}+5\)) = 85 

          ⇒ \(x\) . \(\dfrac{17}{2}\) = 85 ⇒ \(x\) = 85: \(\dfrac{17}{2}\) ⇒ \(x=10\)

          ⇒ \(y\) = 10 x \(\dfrac{5}{2}\) = 25;    \(z\)  = 10.5 = 50 

Kêt luận :....

`2^(x+2)-2^x =96`

`=>2^x (2^2-1)=96`

`=>2^x . 3=96`

`=>2^x=96:3`

`=>2^x=32`

`=>2^x=2^5`

`=>x=5`

`(2x-15)^5 =(2x-15)^3`

`=>(2x-15)^5 -(2x-15)^3=0`

`=> (2x-15)^3 [(2x-15)^2 -1]=0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=1\\2x-15=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)