\(1\frac{4}{5}.2\frac{3}{6}.5\frac{1}{4}.\frac{12}{45}.\frac{20}{15}\)
Giải chi tiết nhé!
p/s: dấu chấm là dấu nhân nhé.
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\(\frac{17}{2}-\left|2x-\frac{5}{2}\right|=-\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{17}{2}-\frac{-7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{51}{6}+\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{29}{3}\)
\(2x-\frac{5}{2}=\frac{29}{3}\)hoặc \(2x-\frac{5}{2}=\frac{-29}{3}\)
Trường hợp 1:
\(2x-\frac{5}{2}=\frac{29}{3}\)
\(2x=\frac{29}{3}+\frac{5}{2}\)
\(2x=\frac{73}{6}\)
\(x=\frac{73}{6}:2\)
\(x=\frac{73}{12}\)
Trường hợp 2:
\(2x-\frac{5}{2}=\frac{-29}{3}\)
\(2x=\frac{-29}{3}+\frac{5}{2}\)
\(2x=\frac{-43}{6}\)
\(x=\frac{-43}{6}:2\)
\(x=\frac{-43}{12}\)
Vậy \(x=\frac{73}{12}\)hoặc \(x=\frac{-43}{12}\)
\(\frac{4,8\left(1994+3996+4010\right)}{\left(\frac{99-3}{4}+1\right)\left(\frac{99+3}{2}\right)-11.25}=\frac{48000}{25.51-11.25}=\frac{48000}{1000}=48\)
Ta có : \(5\cdot\left(\frac{1}{5}+\frac{1}{17}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{9}{15}+\frac{12}{68}\right)\)
\(=\) \(5\cdot\frac{1}{5}+5\cdot\frac{1}{17}-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)
\(=\) \(1+\frac{5}{17}-\left[\left(\frac{2}{5}+\frac{3}{5}\right)+\left(\frac{2}{17}+\frac{3}{17}\right)\right]\)
\(=\) \(1+\frac{5}{17}-\left(1+\frac{5}{17}\right)\)
\(=\) \(1+\frac{5}{17}-1-\frac{5}{17}\)
\(=\)\(0\)
Vậy ...
Tk ủng hộ mk nha các bn ❣❣ C.ơn nhiều ^^
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
a) 7/2.x - (7.(18+45+47))=3
7/2.x - 7.110 = 3
7/2.x - 770 =3 => 7/2.x = 3+770=773
=> x = 773 : 7/2 = 1546/7
b) 19/3 - ( 4x+6x+x)= 4/7 : 2/5
19/3 - 11x = 10/7
11x = 19/3 - 10/7 = 103/21
x = 103/21 : 11 = 103/231
a, x=-5/33
b, x=-30
c, (Nếu dấu của bạn là giá trị tuyệt đối) x=1/3 hoặc x=-1/3
Nếu cần mik giải rõ ràng cho
Ta có : \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{2\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}\right)}=\frac{1}{2}\)
Lại có : \(\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right).2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=\frac{0.2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=0\)
Khi đó \(B=\frac{1}{2}+0=\frac{1}{2}\)
c.\(=3\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\right)\)
\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=3\left(1-\frac{1}{101}\right)\)
\(=\frac{300}{101}\)