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A=\(\frac{3.7+9.28+15.42+21.56}{6.14+12.35+18.63+30.77}\)
\(\Rightarrow A=\frac{3.7.\left(1+3.4+5.6+7.8\right)}{6.7.\left(2+2.5+3.9+5.11\right)}\)
\(\Rightarrow A=\frac{1+3.4+5.6+7.8}{2.\left(2+2.5+3.9+5.11\right)}\)
\(\Rightarrow A=\frac{1+12+30+56}{2.\left(2+10+27+55\right)}\)
\(\Rightarrow A=\frac{99}{2.94}\)
\(\Rightarrow A=\frac{99}{188}\)
Ta có :
\(A=\dfrac{10^{50}+2}{10^{50}-1}=\dfrac{10^{50}-1+3}{10^{50}-1}=\dfrac{10^{50}-1}{10^{50}-1}+\dfrac{3}{10^{50}-1}=1+\dfrac{3}{10^{50}-1}\)
\(B=\dfrac{10^{50}}{10^{50}-3}=\dfrac{10^{50}-3+3}{10^{50}-3}=\dfrac{10^{50}-3}{10^{50}-3}+\dfrac{3}{10^{50}-3}=1+\dfrac{3}{10^{50}-3}\)
Vì \(1+\dfrac{3}{10^{50}-1}< 1+\dfrac{3}{10^{50}-3}\Rightarrow A< B\)
a) \(x+\dfrac{3}{2}=\dfrac{9}{2}\)
\(x=\dfrac{9}{2}-\dfrac{3}{2}\)
\(x=3\)
Vậy x = 3 là giá trị cần tìm
b) \(\dfrac{1}{3}x+\dfrac{3}{4}x-75\%=-5\dfrac{1}{4}\)
\(x\left(\dfrac{1}{3}+\dfrac{3}{4}\right)-\dfrac{75}{100}=\dfrac{-19}{4}\)
\(x\left(\dfrac{4}{12}+\dfrac{9}{12}\right)-\dfrac{3}{4}=\dfrac{-19}{4}\)
\(x.\dfrac{13}{12}-\dfrac{3}{4}=\dfrac{-19}{4}\)
\(x.\dfrac{13}{12}=\dfrac{-19}{4}+\dfrac{3}{4}\)
\(x.\dfrac{13}{12}=-4\)
\(x=-4:\dfrac{13}{12}\)
\(x=-\dfrac{48}{13}\)
Vậy..................................
a) x + \(\dfrac{3}{2}\)= \(\dfrac{9}{2}\)
=> x= \(\dfrac{9}{2}\)- \(\dfrac{3}{2}\)
=> x=3
Vậy x=3
b) \(\dfrac{1}{3}\)x + \(\dfrac{3}{4}\)x - 75%= \(-5\dfrac{1}{4}\)
=> x.(\(\dfrac{1}{3}\)+ \(\dfrac{3}{4}\)) - \(\dfrac{3}{4}\)= \(\dfrac{-21}{4}\)
=> x. \(\dfrac{13}{12}\)- \(\dfrac{3}{4}\)=\(\dfrac{-21}{4}\)
=> x. \(\dfrac{13}{12}\)= \(\dfrac{-21}{4}\)+ \(\dfrac{3}{4}\)
=> x. \(\dfrac{13}{12}\)= \(\dfrac{-9}{2}\)
=> x= \(\dfrac{-9}{2}\): \(\dfrac{13}{12}\)= \(\dfrac{-9}{2}\). \(\dfrac{12}{13}\)
=> x= \(\dfrac{-54}{13}\)
Vậy x= \(\dfrac{-54}{13}\)
Bai 2: b)
\(B=\dfrac{\left(\dfrac{8^2}{128}+\dfrac{3}{4}\right):1\dfrac{3}{16}}{1\dfrac{11}{19}}\\ =\dfrac{\left(\dfrac{\left(2^3\right)^2}{2^7}+\dfrac{3}{4}\right):\dfrac{19}{16}}{\dfrac{30}{19}}=\dfrac{\left(\dfrac{1}{2}+\dfrac{3}{4}\right).\dfrac{16}{19}}{\dfrac{30}{19}}\\ =\dfrac{\dfrac{5}{4}.\dfrac{16}{19}}{\dfrac{30}{19}}=\dfrac{\dfrac{20}{19}}{\dfrac{30}{19}}=\dfrac{2}{3}\)
a) \(A=\left(-1,5\right)^2.2\dfrac{2}{3}.\dfrac{1}{6}+\left(\dfrac{4}{7}-\dfrac{2}{5}\right):1\dfrac{1}{35}\\ =2,25.\dfrac{8}{3}.\dfrac{1}{6}+\dfrac{6}{35}:\dfrac{36}{35}\\ =\dfrac{9}{4}.\dfrac{8}{3}.\dfrac{1}{6}+\dfrac{6}{35}.\dfrac{35}{36}\\ =1+\dfrac{1}{6}=1\dfrac{1}{6}\)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{50^2}\)
\(A=\frac{1}{1^2}+\frac{1}{2\times2}+\frac{1}{3\times3}+\frac{1}{4\times4}+.....+\frac{1}{50\times50}\)
\(A< \frac{1}{1}+\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+.....+\frac{1}{49\times50}\)
\(A< 1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{49}-\frac{1}{50}\)
\(A< 2-\frac{1}{50}\)
\(2-\frac{1}{50}< 2\)
\(\Rightarrow A< 2\)
Chúc bạn học tốt
ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4};...;\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)
=>
cho nó ta đều được thương là các số tự nhiên.
Gọi phân số cần tìm là:\(\dfrac{a}{b}\) với\(\left(a,b\in N^{\otimes}\right)\)
Theo đề, ta có:
\(\dfrac{28}{15}:\dfrac{a}{b}=\dfrac{28}{15}.\dfrac{b}{a}\Rightarrow28⋮a\)và \(b⋮15\)
Tương tự ta cũng có:\(21⋮a\)và \(b⋮10\)
\(49⋮a\)và \(b⋮84\)
\(\Rightarrow a\in\)ƯCLN\(\left(28;21;49\right)=7\)
Và b\(\in BCNN\left(15;10;84\right)=420\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{7}{420}=\dfrac{1}{60}\)
Vậy phân số cần tìm là:\(\dfrac{1}{60}\)
Gọi \(ƯC\left(2a+3;a+2\right)\) là \(d\)
\(\Rightarrow2a+3⋮d\) ; \(a+2⋮d\)
\(\Rightarrow2a+3⋮d\) ; \(2\left(a+2\right)⋮d\)
\(\Rightarrow2a+3⋮d\) ; \(2a+4⋮d\)
\(\Rightarrow2a+4-2a-3⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=\pm1\)
Vậy phân số có dạng \(\dfrac{2a+3}{a+2}\) là phân số tối giản.
Bn ơi sao2(a+2) lại chia hết cho d?????