a, 3¹⁰ + 3¹¹ = 3¹⁰ + 3¹⁰ . 3 = 3¹⁰( 1 + 3 ) = 3¹⁰ . 4 chia hết cho 4

b, 7⁶ + 7⁵ - 7⁴ = 7⁴ . 7² + 7⁴ . 7 - 7⁴ = 7⁴(7² + 7 - 1 ) = 7⁴ . 55 = 7⁴ . 11 . 5 chia hết cho 11

c, 10⁹ + 10⁸ + 10⁷ = 10⁷ . 10² + 10⁷ . 10 + 10⁷ = 10⁷(10² + 10 + 1 ) = 10⁷ . 111 chia hết cho 5 . 111 

Cạnh tăng 20%   thì  diện tích = (6/5)² =  36/25 =144/100 = 100%+ 44%

Vậy diện tích tăng 44%

a, \(\dfrac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}=\dfrac{x^2+10x+25+x^2-10x+25}{x^2+25}\)

\(=\dfrac{2\left(x^2+25\right)}{x^2+25}=2\forall x\)

\(\Rightarrowđpcm\)

b, \(\dfrac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}\)

\(=\dfrac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)

\(=\dfrac{29\left(x^2+1\right)}{x^2+1}=29\forall x\)

\(\Rightarrowđpcm\)

a, \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=x^4-\dfrac{4}{25}y^2\)

b, \(\left(3x-2y\right)\left(3x+2y\right)\left(9x^2+4y^2\right)\)

\(=\left(9x^2-4y^2\right)\left(9x^2+4y^2\right)\)

\(=81x^4-16y^4\)

Giải:

Ta có: \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{y-x}{3-4}=\dfrac{5}{-1}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-15\end{matrix}\right.\)

Vậy...

\(\left(2x-3\right)^{2015}=\left(2x-3\right)^{2013}\)

\(\Rightarrow\left(2x-3\right)^{2015}-\left(2x-3\right)^{2013}=0\)

\(\Rightarrow\left(2x-3\right)^{2013}\left[\left(2x-3\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-3\right)^{2013}=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\)

+) \(\left(2x-3\right)^{2013}=0\Rightarrow x=\dfrac{3}{2}\)

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

Vậy \(x=\dfrac{3}{2}\) hoặc x = 2 hoặc x = 1

Áp dụng bất đẳng thức Cauchy-Shwarz dạng Engel có:

\(a^2+b^2=\dfrac{a^2}{1}+\dfrac{b^2}{1}\ge\dfrac{\left(a+b\right)^2}{1+1}=\dfrac{1}{2}\)

Dấu " = " khi \(a=b=\dfrac{1}{2}\)

Vậy...

\(\left|3x-1\right|=1-3x\)

+) Xét \(x\ge\dfrac{1}{3}\) có:
\(3x-1=1-3x\Leftrightarrow6x=2\Leftrightarrow x=\dfrac{1}{3}\) ( t/m )

+) Xét \(x< \dfrac{1}{3}\) có:

\(1-3x=1-3x\)

\(\Leftrightarrow x\in R\forall x< \dfrac{1}{3}\)

Vậy \(x\le\dfrac{1}{3}\)