\(7^{2018}+7^{2017}-7^{2016}\)

\(=7^{2016}\left(7^2+7-1\right)=7^{2016}.55⋮11\)

\(\Rightarrowđpcm\)

\(\left(\dfrac{1}{3}-x\right)^2+\dfrac{3}{2}=\dfrac{1}{2}\)

\(\Rightarrow\left(\dfrac{1}{3}-x\right)^2=-1\)

Mà \(\left(\dfrac{1}{3}-x\right)^2\ge0\)

\(\Rightarrow\)Không có giá trị x thỏa mãn

Vậy...

\(A=5x+\sqrt{x^2+6x+9}\)

\(=5x+\sqrt{\left(x+3\right)^2}=5x+\left|x+3\right|\)

+) Xét \(x\ge-3\) có:

\(A=5x+x+3=6x+3\)

+) Xét x < -3 có:
\(A=5x-x-3=4x-3\)

Vậy...

\(a^3-b^3-3a+3b\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2-3\right)\)

a, \(A\in Z\Rightarrow2x+3⋮x-3\)

\(\Rightarrow2x-6+9⋮x-3\)

\(\Rightarrow2\left(x-3\right)+9⋮x-3\)

\(\Rightarrow9⋮x-3\)

\(\Rightarrow x-3\in\left\{1;-1;3;-3;9;-9\right\}\)

\(\Rightarrow x\in\left\{4;2;6;0;12;-6\right\}\)

Vậy...

b, \(B\in Z\Rightarrow2x^2+x-5⋮2x+1\)

\(\Rightarrow x\left(2x+1\right)-5⋮2x+1\)

\(\Rightarrow5⋮2x+1\)

\(\Rightarrow2x+1\in\left\{1;-1;5;-5\right\}\)

\(\Rightarrow x\in\left\{0;-1;2;-3\right\}\)

Vậy...

Bài 1:
\(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)=15\)

\(\Leftrightarrow5\left(8x-5\right)=15\)

\(\Leftrightarrow8x=8\Leftrightarrow x=1\)

Vậy x = 1

Bài 2:

\(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)

\(=\left(7x+1-x-7\right)\left(7x+1+x+7\right)\)

\(=\left(6x-6\right)\left(8x+8\right)\)

\(=48\left(x-1\right)\left(x+1\right)\)

\(=48\left(x^2-1\right)=VP\)

\(\Rightarrowđpcm\)

\(A=-\left|2-3x\right|+\dfrac{1}{2}\le\dfrac{1}{2}\)

Dấu " = " khi \(-\left|2-3x\right|=0\Rightarrow x=\dfrac{2}{3}\)

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

\(\sqrt{x+2\sqrt{x-1}}=2\)

\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=2\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)

\(\Leftrightarrow\sqrt{x-1}+1=2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\)

Vậy x = 2

xin lỗi mk mới học lớp 6

\(4x^2-2x-3y-9y^2\)

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

\(=\left(2x-3y\right)\left(2x+3y\right)-\left(2x+3y\right)\)

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