a) (a+b)(a³+b³) \(\ge\)2(a⁴+b⁴) 

<=> a⁴ + ab³ + ba³ + b⁴ \(\ge\)2(a⁴ + b⁴)

<=> ab³ - b⁴ + ba³ - a⁴ \(\ge\)0

<=> (a - b)(b³ - a³)\(\ge\)0

<=> -  (a - b)² (a² + ab + b²) \(\ge\)0 (sai)

=> Xem lại đề

mk cg cảm thấy sai nhưng mà cậu xem lại nhé vì đề đúng 

alibaba

Ta có: \(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{\left(a+1\right)b^2}{b^2+1}\)

\(\ge\left(a+1\right)-\frac{\left(a+1\right)b^2}{2b}=a+1-\frac{ab+b}{2}\)

Tương tự ta có:\(\frac{b+1}{c^2+1}\ge b+1-\frac{bc+c}{2};\frac{c+1}{a^2+1}\ge c+1-\frac{ca+a}{2}\)

Cộng theo vế ta có: \(VT\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}=6-\frac{3+ab+bc+ca}{2}\)

Mà theo BĐT AM-GM: \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)

Suy ra \(VT\ge6-3=3\)(ĐPCM)

x(5 - 3) + 4x = 5x - 3 + 12

5x - 3x + 4x = 5x - 3 + 12

2x + 4x = 5x + 9

6x = 5x + 9

x = 9 

x( 5 - 3 ) + 4x = 5x - 3 + 12

5x - 3x + 4x = 5x - 3 + 12

2x + 4x = 5x + 9

6x = 5x + 9

\(\Leftrightarrow\)x = 9

\(\left(2x-5\right)^3-\left(3x-4\right)^3+\left(x+1\right)^3=0\)

\(\Leftrightarrow\left(2x-5-3x+4\right)\left[\left(2x-5\right)^2+\left(2x-5\right)\left(3x-4\right)+\left(3x-4\right)^2\right]+\left(x+1\right)^3=0\)

\(\Leftrightarrow-\left(x+1\right)\left[\left(2x-5\right)^2+\left(2x-5\right)\left(3x-4\right)+\left(3x-4\right)^2\right]+\left(x+1\right)^3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2-\left(2x-5\right)^2-\left(2x-5\right)\left(3x-4\right)-\left(3x-4\right)^2=0\left(1\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\left(x+1+2x-5\right)\left(x+1-2x+5\right)-\left(2x-5\right)\left(3x-4\right)-\left(3x-4\right)^2=0\)

\(\Leftrightarrow\left(3x-4\right)\left(6-x-2x+5-3x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\-6x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-5}{2}\end{cases}}}\)