\(2^x+2+2^x+1+2^x=224\)

\(\Leftrightarrow3\left(x^2\right)+3=224\)

\(\Leftrightarrow3\left(x^2+1\right)=224\)

\(\Leftrightarrow x^2+1=\frac{224}{3}\)

\(\Leftrightarrow x^2=\frac{221}{3}\)

\(\Leftrightarrow x=\sqrt{\frac{221}{3}}=\frac{\sqrt{663}}{3}\)

tổng của 3 số nguyên tố là 3

=> có ít nhất 1 số nguyên tố là chẵn

=> số cần tìm là 2

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-4x^3+2x^2-5x^3+10x^2-5x+2x^2-4x+2=0\)

\(\Leftrightarrow2x^2\left(x^2-2x+1\right)-5x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-5x+2\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(2x^2-x-4x+2\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)^2\left(2x-1\right)=0\)

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

\(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow2x^3\cdot\left(x-1\right)-7x^2\cdot\left(x-1\right)+7x\cdot\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x^3-1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x-1\right)\cdot\left(x^2+x+1\right)-7x\cdot\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[2\left(x^2+x+1\right)-7x\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2+2x+2-7x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-5x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-x-4x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[x\cdot\left(2x-1\right)-2\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)

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

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

Vậy \(x_1=\dfrac{1}{2};x_2=1;x_3=2\)

a) n+2 \(\in\)B(3)={0;3;6;9;12;15;18;21;...}

\(\Rightarrow\)n=1;4;7;10;13;16;19;....

b) 4n-5 \(\in\)B(13)={0;13;26;39;42;.....}

\(\Rightarrow\)n=5;18;31;44;47;...

c) 5n-1 \(\in\)B(7)={0;7;14;21;28;35;42;...}

\(\Rightarrow\)n=3

d) 25n+3 \(\in\)B(57)={0;57;114;171;228;285...}

\(\Rightarrow\)n=9

Tất cả các biểu thức này đều ko tồn tại max mà chỉ tồn tại min

\(B=\frac{x}{2}+\frac{x}{2}+\frac{4}{x^2}\ge3\sqrt[3]{\frac{4x^2}{4x^2}}=3\)

Dấu "=" xảy ra khi \(\frac{x}{2}=\frac{4}{x^2}\Leftrightarrow x=2\)

\(C=x^2+\frac{1}{x}+\frac{1}{x}\ge3\sqrt[3]{\frac{x^2}{x^2}}=3\)

Dấu "=" xảy ra khi \(x^2=\frac{1}{x}\Leftrightarrow x=1\)

\(D=9x^2+\frac{2}{3x}+\frac{2}{3x}\ge3\sqrt[3]{\frac{36x^2}{9x^2}}=3\sqrt[3]{4}\)

Dấu "=" xảy ra khi \(9x^2=\frac{2}{3x}\Leftrightarrow x=\frac{\sqrt[3]{2}}{3}\)

ơ hay , tự hỏi tự trả lời là sao ???

9x - 9x + 2x²= 2x²