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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 1 2024
Bài 1:
e; \(\dfrac{10}{21}\) - \(\dfrac{3}{8}\) : \(\dfrac{15}{4}\)
= \(\dfrac{10}{21}\) - \(\dfrac{3}{8}\) x \(\dfrac{4}{15}\)
= \(\dfrac{10}{21}\) - \(\dfrac{1}{10}\)
= \(\dfrac{100}{210}\) - \(\dfrac{21}{210}\)
= \(\dfrac{79}{210}\)
f; (\(\dfrac{2}{3}\) + \(\dfrac{3}{4}\)).(\(\dfrac{5}{7}\) + \(\dfrac{5}{14}\))
= (\(\dfrac{8}{12}\) + \(\dfrac{9}{12}\)).(\(\dfrac{10}{14}\) + \(\dfrac{5}{14}\))
= \(\dfrac{17}{12}\).\(\dfrac{15}{14}\)
= \(\dfrac{85}{56}\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 1 2024
\(\dfrac{1}{n\left(n+1\right)}=\dfrac{1+n-n}{n\left(n+1\right)}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 1 2024
Bài 2:
a; \(x\) - \(\dfrac{1}{2}\) = \(\dfrac{3}{10}\).\(\dfrac{5}{6}\)
\(x\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{4}\)
\(x\) = \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{3}{4}\)
Vậy \(x\) = \(\dfrac{3}{4}\)
b; \(\dfrac{x}{5}\) = \(\dfrac{-3}{14}\) \(\times\) \(\dfrac{7}{3}\)
\(\dfrac{x}{5}\) = \(\dfrac{-1}{2}\)
\(x\) = \(\dfrac{-1}{2}\) \(\times\) 5
\(x\) = \(\dfrac{-5}{2}\)
Vậy \(x\) = \(\dfrac{-5}{2}\);
c; \(x\) : \(\dfrac{4}{11}\) = \(\dfrac{11}{4}\) \(\times\) 2
\(x\) : \(\dfrac{4}{11}\) = \(\dfrac{11}{2}\)
\(x\) = \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{11}\)
\(x\) = 2
Vậy \(x\) = 2
d; \(x^2\) + \(\dfrac{9}{-25}\) = \(\dfrac{2}{5}\) : \(\dfrac{5}{8}\)
\(x^2\) - \(\dfrac{9}{25}\) = \(\dfrac{16}{25}\)
\(x^2\) = \(\dfrac{16}{25}\) + \(\dfrac{9}{25}\)
\(x^2\) = \(\dfrac{25}{25}\)
\(x^2\) = 1
\(\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Vậy \(x\)\(\in\) {-1; 1}
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
29 tháng 1 2024
Bài 3:
a; A = \(\dfrac{2}{13}\)\(\times\) \(\dfrac{5}{9}\)+ \(\dfrac{2}{13}\)\(\times\)\(\dfrac{4}{9}\) + \(\dfrac{11}{13}\)
A = \(\dfrac{2}{13}\) \(\times\)(\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{11}{13}\)
A = \(\dfrac{2}{13}\) \(\times\) \(\dfrac{9}{9}\) + \(\dfrac{11}{13}\)
A = \(\dfrac{2}{13}\) + \(\dfrac{11}{13}\)
A = 1
b; B = \(\dfrac{1}{10}\).\(\dfrac{4}{11}\) + \(\dfrac{1}{10}\).\(\dfrac{8}{11}\) - \(\dfrac{1}{10}\).\(\dfrac{1}{11}\)
B = \(\dfrac{1}{10}\) x (\(\dfrac{4}{11}\) + \(\dfrac{8}{11}\) - \(\dfrac{1}{11}\))
B = \(\dfrac{1}{10}\) x (\(\dfrac{12}{11}\) - \(\dfrac{1}{11}\))
B = \(\dfrac{1}{10}\) x \(\dfrac{11}{11}\)
B = \(\dfrac{1}{10}\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 2 2024
a) \(\dfrac{5}{11}\cdot\dfrac{5}{7}+\dfrac{5}{11}\cdot\dfrac{2}{7}+\dfrac{6}{11}=\dfrac{5}{11}\cdot\left(\dfrac{5}{7}+\dfrac{2}{7}\right)+\dfrac{6}{11}=\dfrac{5}{11}\cdot1+\dfrac{6}{11}=\dfrac{5}{11}+\dfrac{6}{11}=\dfrac{11}{11}=1\)
b) \(\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{9}{11}-\dfrac{3}{13}\cdot\dfrac{4}{11}=\dfrac{3}{13}\cdot\left(\dfrac{6}{11}+\dfrac{9}{11}-\dfrac{4}{11}\right)=\dfrac{3}{13}\cdot\dfrac{11}{11}=\dfrac{3}{13}\cdot1=\dfrac{3}{13}\)
c) \(\dfrac{-5}{6}\cdot\dfrac{4}{19}+\dfrac{7}{12}\cdot\dfrac{4}{-19}-\dfrac{40}{57}=\dfrac{-5}{6}\cdot\dfrac{4}{19}+\dfrac{-7}{12}\cdot\dfrac{4}{19}-\dfrac{40}{57}=\dfrac{4}{19}\cdot\left(\dfrac{-5}{6}+\dfrac{-7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{47}=\dfrac{-17}{57}-\dfrac{40}{57}=\dfrac{-57}{57}=-1\)
d) \(\left(\dfrac{11}{4}\cdot\dfrac{-5}{9}+\dfrac{4}{9}\cdot\dfrac{11}{-4}\right)\cdot\dfrac{8}{33}=\left(\dfrac{11}{4}\cdot\dfrac{-5}{9}+\dfrac{-4}{9}\cdot\dfrac{11}{4}\right)\cdot\dfrac{8}{33}=\dfrac{11}{4}\cdot\dfrac{8}{33}\cdot\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)\)
\(=\dfrac{11}{4}\cdot\dfrac{8}{33}\cdot1=\dfrac{11\cdot8}{4\cdot33}=\dfrac{2}{3}\)
e) \(\left(\dfrac{12}{61}-\dfrac{31}{22}+\dfrac{14}{91}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)=\left(\dfrac{12}{61}-\dfrac{31}{22}+\dfrac{14}{91}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{12}{61}-\dfrac{31}{22}+\dfrac{14}{91}\right)\cdot0=0\)
AH
Akai Haruma
Giáo viên
4 tháng 2 2024
Lời giải:
a.
$=\frac{3}{5}-\frac{7}{4}=\frac{12-35}{20}=\frac{-23}{20}$
b.
$=-(2+\frac{5}{8})=-\frac{21}{8}$
c.
$=-(\frac{1}{8}+\frac{5}{9})=-\frac{9+8.5}{8.9}=\frac{-49}{72}$
d.
$=\frac{6}{13}-\frac{14}{39}=\frac{18}{39}-\frac{14}{39}=\frac{4}{39}$
e.
$=\frac{-3}{4}+\frac{5}{7}=\frac{5}{7}-\frac{3}{4}$
$=\frac{20-21}{7.4}=\frac{-1}{28}$
Bài 1:
\(D=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}=\dfrac{49}{50}\)
Bài 2:
Gọi số học sinh khối 6 là x(bạn)
(Điều kiện: \(x\in Z^+\))
Nếu xếp hàng 10;12;15 đều dư 3 bạn nên \(x-3\in BC\left(10;12;15\right)\)
=>\(x-3\in B\left(60\right)\)
=>\(x-3\in\left\{60;120;180;240;300;360;420;...\right\}\)
=>\(x\in\left\{63;123;183;243;303;363;423\right\}\)
mà x<=400
nên \(x\in\left\{63;123;183;243;303;363\right\}\)(1)
Khi xếp hàng 11 thì vừa đủ nên \(x\in B\left(11\right)\left(2\right)\)
Từ (1),(2) suy ra x=363
vậy: Khối 6 có 363 bạn
Bài 1:
\(D=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ =1-\dfrac{1}{50}\\ =\dfrac{50}{50}-\dfrac{1}{50}\\ =\dfrac{49}{50}\)