Tính nhanh
a) A=4/1.5+4/5.9+...+4/2001.2005
b) B=3/10.12+3/1214+...+3/998.1000
Ai lm đc nhớ nghi lời, giải đầy đủ~
Cảm ơn!!!
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\(B\left(8\right)\in\left\{0;8;16;24;32\right\}\)
\(Ư\left(12\right)\in\left\{1;2\right\}\)
Sửa đề:
\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)
\(A=4.\dfrac{33}{68}\)
\(A=\dfrac{33}{17}\)
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
99 - 3.(y + 1) = 45
3.(y + 1) = 99 - 45
3.(y + 1) = 54
y + 1 = 54: 3
y + 1 = 18
y = 18 - 1
y = 17
\(99-3\times\left(y+1\right)=45\\ \Rightarrow3x\left(y+1\right)=54\\ \Rightarrow y+1=18\\ \Rightarrow y=17.\)
\(\left(y+1\right)+\left(y+3\right)+\left(y+5\right)+..+\left(y+19\right)\\ 18y\left(1+3+5+...+19\right)=250\)
Đến đây bn tính tổng rồi tìm x nhé.
Số học sinh trung bình:
\(48\cdot\dfrac{1}{6}=8\left(hs\right)\)
Số học sinh khá:
\(8:50\%=16\left(hs\right)\)
Số học sinh giỏi:
\(48-8-16=24\left(hs\right)\)
Số học sinh trung bình là :
\(48.\dfrac{1}{6}=8\left(hsinh\right)\)
Số học khá là :
\(8:50\%=16\left(hsinh\right)\)
Số học giỏi là :
\(48-\left(8+16\right)=24\left(hsinh\right)\)
Đáp số...
Bài 1 :
a) \(...=45\left(56+44\right)-500=45.100-500=4500-500=4000\)
b) \(...=3^2.6-6^{18}+1=45+1-6^{18}=46-6^{18}\)
c) \(...=756\left(66+48-14\right)=756.100=75600\)
d) \(...=8.15-\left(149-7^2\right)=120-\left(149-49\right)=120-100=20\)
Bài 2 :
a) \(x-21=46\Rightarrow x=21+46\Rightarrow x=87\)
b) \(2^x.4=128\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
c) \(3x+27=162\Rightarrow3x=162-27\Rightarrow3x=135\Rightarrow x=45\)
d) \(4x-8=4^{2023}.4^{2021}\)
\(\Rightarrow4x-8=4^{2023+2021}\)
\(\Rightarrow4x-8=4^{4044}\)
\(\Rightarrow4x=4^{4044}+8\)
\(\Rightarrow x=\left(4^{4044}+8\right):4\)
\(\Rightarrow x=4^{4043}+2\)
\(\left(2^2\right)^8.2^{20}=2^{16}.2^{20}=2^{36};\left(3^2\right)^{12}.\left(3^3\right)^5.\left(3^4\right)^4\)\(=3^{24}.3^{15}.3^{16}=3^{55}\)
\(64^3.4^5.16^2=\left(4^3\right)^3.4^5.\left(4^2\right)^2=4^9.4^5.4^4=4^{18}\)
Sửa đề:
\(A=\dfrac{3}{5}+\dfrac{3}{20}+\dfrac{3}{44}+\dfrac{3}{77}\)
\(A=2.\left(\dfrac{3}{5}+\dfrac{3}{20}+\dfrac{3}{44}+\dfrac{3}{77}\right)\)
\(A=\dfrac{6}{10}+\dfrac{6}{40}+\dfrac{6}{88}+\dfrac{6}{154}\)
\(A=6.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}\right)\)
\(A=6.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)\)
\(A=6.\left(\dfrac{1}{2}-\dfrac{1}{14}\right)\)
\(A=6.\dfrac{6}{14}\)
\(A=\dfrac{36}{14}=\dfrac{18}{7}\)
\(=\dfrac{1}{5}.3+\dfrac{1}{5}.\dfrac{3}{4}+\dfrac{1}{11}.\dfrac{3}{4}+\dfrac{1}{11}.\dfrac{3}{7}\)
\(=\dfrac{1}{5}.\left(3+\dfrac{3}{4}\right)+\dfrac{1}{11}.\left(\dfrac{3}{4}+\dfrac{3}{7}\right)\)
\(=\dfrac{1}{5}.\dfrac{15}{4}+\dfrac{1}{11}.\dfrac{33}{28}=\dfrac{3}{4}+\dfrac{3}{28}=\dfrac{6}{7}\)
\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)
\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)
\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)
\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)
\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)
\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)
\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2005}\)
\(\Rightarrow A=\dfrac{2004}{2005}\)