Bài 1. (3 điểm) Thực hiện phép tính (một cách hợp lý nếu có thể):
a) \( 625 : 5^2 - 24 \cdot 8 + (2025 - 2024)^{2026} \)
b) \( B = \frac{1}{3} + \frac{1}{2 \cdot 4} + \frac{1}{3 \cdot 5} + \frac{1}{4 \cdot 6} + \ldots + \frac{1}{198 \cdot 200} + \frac{1}{199 \cdot 201} + \frac{1}{2 \cdot 201} \)
Bài 2. (3 điểm) Tìm x, biết:
a) \( 2^{2x+4} + 2^{2x} = 272 \)
b) \( \frac{(x-1) + (x-2) + (x-3) + \ldots + (x-50)}{1 + 2 + 3 + \ldots + 50} = -3 \)
bài 1:
\(a;625:5^2-24:8+\left(2025-2024\right)^{2026}\\ =5^4:5^2-24:8+1^{2026}\\ =5^2-3+1=25-3+1=23\)
\(b;\dfrac{1}{3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{198\cdot200}+\dfrac{1}{199\cdot201}+\dfrac{1}{2\cdot201}\\ =\dfrac{1}{3}+\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{198\cdot200}\right)+\left(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{199\cdot201}\right)+\dfrac{1}{2\cdot201}\\ =\dfrac{1}{3}+\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{198}-\dfrac{1}{200}\right)+\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)+\dfrac{1}{2\cdot201}\\ =\dfrac{1}{3}+\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{200}\right)+\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{201}\right)+\dfrac{1}{2\cdot201}\\ =\dfrac{1}{3}+\dfrac{1}{2}\cdot\dfrac{99}{200}+\dfrac{1}{2}\cdot\dfrac{22}{67}+\dfrac{1}{2}\cdot\dfrac{1}{201}\\ =\dfrac{1}{3}+\dfrac{1}{2}\cdot\left(\dfrac{99}{200}+\dfrac{22}{67}+\dfrac{1}{201}\right)\\ =\dfrac{1}{3}+\dfrac{1}{2}\cdot\dfrac{497}{600}=\dfrac{299}{400}\)
bài 2:
\(a;2^{2x+4}+2^{2x}=272\\ ⇒\:2^{2x}\cdot2^4+2^{2x}=272\\ ⇒\:2^{2x}\cdot\left(2^4+1\right)=272\\ ⇒\:2^{2x}=16\\ ⇒\:x=2\)
\(\dfrac{\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+...+\left(x-50\right)}{1+2+3+..+50}=-3\\ \dfrac{x-1+x-2+...+x-50}{1+2+3+...+50}=-3\\ \dfrac{\left(x+x+...+x\right)-\left(1+2+3+....+50\right)}{1+2+3+....+50}=-3\\ \dfrac{50x-1275}{1275}=-3\\ 50x-1275=-3825\\ 50x=-2550\\ x=-51\)
1.
a.
\(625:5^2-24:8+\left(2025-2024\right)^{2026}\)
\(=625:25-3+1^{2026}\)
\(=25-3+1=23\)
b.
\(B=\dfrac{1}{3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{198.200}+\dfrac{1}{199.201}+\dfrac{1}{2.201}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{199.201}\right)+\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{198.200}\right)+\dfrac{1}{2.201}\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)+\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{198}-\dfrac{1}{200}\right)+\dfrac{1}{2.201}\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{201}\right)+\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{200}\right)+\dfrac{1}{2.201}\)
\(=\dfrac{1}{2}-\dfrac{1}{2.201}+\dfrac{1}{2}.\dfrac{99}{200}+\dfrac{2}{2.201}\)
\(=\dfrac{1}{2}.\left(1+\dfrac{99}{200}\right)=\dfrac{299}{400}\)
2.a.
\(2^{2x+4}+2^{2x}=272\)
\(\Leftrightarrow2^{2x}.2^4+2^{2x}=272\)
\(\Leftrightarrow2^{2x}.\left(2^4+1\right)=272\)
\(\Leftrightarrow2^{2x}.17=272\)
\(\Leftrightarrow2^{2x}=16\)
\(\Leftrightarrow2^{2x}=2^4\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
2b.
Ta có: \(1+2+3+...+50=\dfrac{50.\left(50+1\right)}{2}=1275\)
Do đó:
\(\dfrac{\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+...+\left(x-50\right)}{1+2+3+...+50}=-3\)
\(\Leftrightarrow\dfrac{\left(x+x+x+...+x\right)-\left(1+2+3+...+50\right)}{1+2+3+...+50}=-3\)
\(\Leftrightarrow\dfrac{50x-1275}{1275}=-3\)
\(\Leftrightarrow50x-1275=-3825\)
\(\Leftrightarrow50x=-2550\)
\(\Leftrightarrow x=-51\)
