lộn na !!

7x - 7 - 5x - 35 = 0

2x + 28 = 0

2x = -28

x = -14 

Ta có : \(a^2\left(b+c\right)=b^2\left(a+c\right)\)

\(\Rightarrow a^2\left(b+c\right)-b^2\left(a+c\right)=0\)

\(\Rightarrow a^2b+a^2c-b^2a-b^2c=0\)

\(\Rightarrow a^2c-b^2c+a^2b-b^2a=0\)

\(\Rightarrow c\left(a^2-b^2\right)+ab\left(a-b\right)=0\)

\(\Rightarrow c\left[\left(a-b\right)\left(a+b\right)\right]+ab\left(a-b\right)=0\)

\(\Rightarrow\left(a-b\right)\left[c\left(a+b\right)+ab\right]=0\)

\(\Rightarrow\left(a-b\right)\left(ca+cb+ab\right)=0\)

Do a\(\ne b\)

\(\Rightarrow\left(ca+cb+ab\right)=0\)

Xét \(c^2\left(a+b\right)-a^2\left(b+c\right)\)

\(=c^2a+c^2b-a^2b-a^2c\)

\(=\left(c^2a-a^2c\right)+\left(c^2b-a^2b\right)\)

\(=ac\left(c-a\right)+b\left(c^2-a^2\right)\)

\(=ac\left(c-a\right)+b\left[\left(c-a\right)\left(c+a\right)\right]\)

\(=\left(c-a\right)\left[ac+b\left(c+a\right)\right]\)

\(=\left(c-a\right)\left(ac+bc+ba\right)\)

\(=0\)

\(\Rightarrow c^2\left(a+b\right)=a^2\left(b+c\right)=2016\)

Vậy...

đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Thay a và c vào VP và VT sẽ bằng nhau

x^3 + 6x^2 + 11x + 6 
= x^3 + x^2 + 5x^2 + 5x + 6x + 6 
= x^2(x + 1) + 5x(x + 1) + 6(x + 1) 
= (x + 1)(x^2 + 5x + 6) 
= (x + 1)(x^2 + 2x + 3x + 6) 
= (x + 1)[x(x + 2) + 3(x + 2) 
= (x + 1)(x + 2)(x + 3) 

\(3^{2017}-3^{2014}=3^{2014}.\left(3^3-1\right)=3^{2014}.26\)

\(3^{2016}=3^{2014}.3^2=3^{2014}.9\)

vì 26 > 9 => 3²⁰¹⁴.26 > 3²⁰¹⁴.9

Vậy \(3^{2017}-3^{2014}>3^{2016}\).