\(1\frac{1}{3}x2\frac{3}{4}+42,7x0,8\)
X là nhân nha
ai nhanh cho 3 tick
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1. \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}\)
\(=\frac{42}{43}\)
2. Đặt \(A=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{90}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(1-\frac{1}{10}\right)\)
\(=2.\frac{9}{10}\)
\(=\frac{9}{5}\)
Ủng hộ mk nha !!! ^_^
1) 3/1×4 + 3/4×7 + 3/7×10 + ... + 3/40×43
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/40 - 1/43
= 1 - 1/43
= 42/43
2) 2/2 + 2/6 + 2/12 + ... + 2/90
= 2 × (1/2 + 1/6 + 1/12 + ... + 1/90)
= 2 × (1/1×2 + 1/2×3 + 1/3×4 + ... + 1/9×10)
= 2 × (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10)
= 2 × (1 - 1/10)
= 2 × 9/10
= 9/5
1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)
\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)
\(\Leftrightarrow\frac{38}{x+3}=16\)
\(\Leftrightarrow x+3=2,375\)
\(\Leftrightarrow x=-0,625\)
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)
\(\Leftrightarrow\frac{38}{x+3}=14\)
\(\Leftrightarrow\left(x+3\right)14=38\)
\(\Leftrightarrow14x+42=38\)
\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)
Vậy \(x=-\frac{2}{7}\)
\(\frac{4}{5}-|x-\frac{1}{6}|=\frac{2}{3}\)
\(\Rightarrow|x-\frac{1}{6}|=\frac{2}{15}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{6}=\frac{2}{15}\\x-\frac{1}{6}=-\frac{2}{15}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{1}{30}\end{cases}}\)
Vậy.....
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
\(2.x=\frac{1+2+3+...+9}{1-2+3-4+5-6+7-8+9}+\frac{25.150-60.5+20.75}{1+2+3+...+99}\)
\(2.x=\frac{\left(9+1\right).9:2}{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+9}+\frac{2.3.5^2.\left(5^2-2+2.5\right)}{\left(1+99\right).99:2}\)
\(2.x=\frac{45}{\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+9}+\frac{2.3.5^2.33}{100.99.\frac{1}{2}}\)
\(2x=\frac{45}{5}+\frac{50.99}{50.2.99.\frac{1}{2}}=9+\frac{1}{2.\frac{1}{2}}=9+1=10\)
=> 2x = 10
x = 5
\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow94< x< 92\)
mà x là số tựu nhiên => \(x\in\varnothing\)
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