bài này khó quá bạn nhỉ, mặc dù tớ học lớp 6 nhưng toán nâng cao lớp 7 tớ ko biết làm, bạn thông cảm cho tớ nhá

to cung chiu

Sửa đề (1 + 2² + 3³ +...+2015²⁰¹⁵).(4.35 - 2.70).(1/2+1/6+1/12+1/30) 

=  (1 + 2² + 3³ +...+2015²⁰¹⁵).(140 - 140).(1/2+1/6+1/12+1/30) 

=  (1 + 2² + 3³ +...+2015²⁰¹⁵).0.(1/2+1/6+1/12+1/30) 

= 0

$A=\frac{3}{2^2}+\frac{8}{3^2}+\frac{15}{4^2}+....+\frac{{2023}^2-1}{{2023}^2}$

$A=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+....+1-\frac{1}{{2023}^2}$

$A=$

322+832+1542+....+20232-120232"" id="MathJax-Element-1-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-table; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=322+832+1542+....+20232−120232�=322+832+1542+....+20232-120232A=

1-122+1-132+1-142+....+1-120232"" id="MathJax-Element-2-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=1−122+1−132+1−1(2+....+1)120232�=1-122+1-132+1-142+....+1-1202321+12+13+...+122023−1

2022-(122+132+142+...+120232)"" id="MathJax-Element-3-Frame" role="presentation" tabindex="0" style="box-sizing: inherit; display: inline-block; line-height: 0; font-size: 18.08px; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;">A=2022−(122+132+142+...+120232)�=2022-(122+132+142+...+120232)A

\(B=\left(\frac{1}{2}\right)^{2015}.\left(\frac{2}{3}\right)^{2016}.\left(-3\right)^{2015}\)

\(B=\left(-3\right)^{2015}.\left(\frac{2}{3}\right)^{2016}.\frac{1}{2^{2015}}\)

\(B=\left(-3\right)^{2015}.\frac{1}{2^{2015}}.\frac{2^{2016}}{3^{2016}}\)

\(B=\left(-3^{2015}\right).\frac{1}{2^{2015}}.\frac{2^{2016}}{3^{2016}}\)

\(B=-\frac{1}{2^{2015}}.\frac{2^{2016}}{3^{2016}}.3^{2015}\)

\(B=-\frac{1.2^{2016}.3^{2015}}{2^{2015}.3^{2016}}\)

\(B=-\frac{2^{2016}.3^{2015}}{3^{2016}.2^{2015}}\)

\(B=\frac{3^{2015}.2}{3^{2016}}\)

\(B=-\frac{2}{3}\)

$\textbf{Câu 1}$

$A=\dfrac{|2x-13|-7}{4}.$

Vì $|2x-13|\ge0$ nên $|2x-13|-7\ge-7.$

Suy ra $A\ge-\dfrac74.$

Dấu ``='' xảy ra khi $2x-13=0$

$\Leftrightarrow x=\dfrac{13}{2}.$

Vậy giá trị nhỏ nhất của $A$ là $-\dfrac74.$

$\textbf{Câu 2}$

$B=|1-3x|+1.$

Vì $|1-3x|\ge0$ nên $B\ge1.$

Dấu ``='' xảy ra khi $1-3x=0$

$\Leftrightarrow x=\dfrac13.$

Vậy giá trị nhỏ nhất của $B$ là $1.$

có rảnh 

\(-\frac{1}{2016}\\ -1;0;2;3\\1 \)