Bài 1 :

Sửa để : \(N=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+....+\dfrac{1}{44.49}\right)\cdot\dfrac{1-3-5-7-..-49}{89}\)

\(N=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\cdot\dfrac{1-\left(3+5+7+..+49\right)}{89}\)

\(N=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\cdot\dfrac{1-624}{89}\)

\(N=\dfrac{1}{5}\cdot\dfrac{45}{196}\cdot\dfrac{-623}{89}\)

\(\Rightarrow N=\dfrac{9}{196}\cdot-7=\dfrac{-9}{28}\)

A)     A= 20

A=/2x+4,6/-2x+15,4

TH1:/2x+4,6/> hoặc =4,6

/2x+4,6/-2x+15,4

=(2x+4,6)-2x+15,4

=2x+4,6-2x+15,4

=2x+2x-4,6+15,4

=4x-20

Mình làm thế không biết đúng hay sai

\(\left|x-\frac{5}{7}\right|\ge0\Rightarrow\left|x-\frac{5}{7}\right|+\frac{2}{3}\ge\frac{2}{3}\Rightarrow A_{min}=\frac{2}{3}.\text{Dâu "=" xay ra}\Leftrightarrow x=\frac{5}{7}\)

\(\left|5-x\right|\ge0\Rightarrow\left|5-x\right|-8,5\ge-8,5\Rightarrow B_{min}=-8,5.\text{ Dâu "=" xay ra}\Leftrightarrow x=5\)

a/ \(A=\left(x-5\right)^2+\left(2x+1\right)^2-2\left(2x^2+8,5\right)\)

\(=\left(x^2-2.x.5+5^2\right)+\left[\left(2x\right)^2+2.2x.1+1^2\right]-2.2x^2-2.8,5\)

\(=\left(x^2-10x+25\right)+\left(4x^2+4x+1\right)-4x^2-17\)

\(=x^2-10x+25+4x^2+4x+1-4x^2-17\)

\(=x^2-6x+9\)

\(=x^2-2.3x+3^2\)

\(=\left(x-3\right)^2\)

b/ A = 0

\(\Rightarrow\left(x-3\right)^2=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=0+3=3\)

c/ A = 36

\(\Rightarrow\left(x-3\right)^2=36\)

\(\Rightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6+2=8\\x=-6+2=-4\end{matrix}\right.\)

Vậy................