Hôm qua mình giải cho bạn rồi mà nhỉ?

\(\left(2x+3y\right)^2+2\left(2x+3y\right)+1\)

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

Đặt \(\dfrac{1}{315}=x,\dfrac{1}{651}=y\)

\(\Rightarrow A=\left(2+x\right)y-3x\left(4-y\right)-4xy+12x\)

\(=2y+xy-12x+3xy-4xy+12x\)

\(=2y\)

Thay \(y=\dfrac{1}{651}\Rightarrow A=\dfrac{2}{651}\)

Vậy...

Ta có: \(\sqrt{2018}>\sqrt{2017}\)

\(\sqrt{2018}>\sqrt{2015}\)

\(\Leftrightarrow2\sqrt{2018}>\sqrt{2017}+\sqrt{2015}\)

Vậy...

\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}\)

\(=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\)

\(=\dfrac{2\sqrt{5}+4-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{8}{5-4}=8\)

\(\Rightarrowđpcm\)

\(\left|17x-5\right|+\left|17x+5\right|=0\)

+) Xét \(x\ge\dfrac{5}{17}\) có:
\(17x-5+17x+5=0\)

\(\Rightarrow34x=0\Rightarrow x=0\) ( ko t/m )
+) Xét \(\dfrac{-5}{17}\le x< \dfrac{5}{17}\) có:

\(5-17x+17x+5=0\Rightarrow10=0\) ( vô lí )
+) Xét \(x< \dfrac{-5}{17}\) có:
\(5-17x-17x-5=0\)

\(\Rightarrow-34x=0\Rightarrow x=0\) ( ko t/m )

Vậy không có giá trị x thỏa mãn

17x + 5 hay 17 + 5?

a, \(x^4+x^3+x+1\)

\(=x^3\left(x+1\right)+\left(x+1\right)\)

\(=\left(x^3+1\right)\left(x+1\right)\)

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

b, \(x^2y+xy^2-x-y\)

\(=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)

c, \(5xy^2-10xyz+5xz^2\)

\(=5x\left(y^2-2yz+z^2\right)\)

\(=5x\left(y-z\right)^2\)

e, \(x^4+x^2+1\)

\(=x^4+2x^2+1-x^2\)

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

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

\(A=x^5-70x^4-70x^3-70x^2-70x+34\)

\(=x^5-71x^4+x^4-71x^3+x^3-71x^2+x^2-71x+x-71+105\)

\(=x^4\left(x-71\right)+x^3\left(x-71\right)+x^2\left(x-71\right)+x\left(x-71\right)+\left(x-71\right)+105\)

\(=\left(x^4+x^3+x^2+x+1\right)\left(x-71\right)+105\)

Thay x = 71\(\Rightarrow A=105\)

Vậy...

bàn tròn là bàn không méo 

bàn không méo là mèo không bán

**** mình nha