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1/ ĐÁP ÁN:
\(\frac{-9}{33}=\frac{3}{-11}\); \(\frac{15}{9}=\frac{5}{3}\); \(\frac{-12}{19}=\frac{60}{-95}\)
2/ ĐÁP ÁN:
\(\frac{-7}{20}=\frac{3}{-18}=\frac{-9}{54}\ne\frac{12}{18}=\frac{-10}{-15}\ne\frac{14}{20}\)
3/ ĐÁP ÁN:
\(\frac{2}{3}=\frac{40}{60}\); \(\frac{3}{4}=\frac{45}{60}\); \(\frac{4}{5}=\frac{48}{60}\); \(\frac{5}{6}=\frac{50}{60}\)
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
a) \(2x-\frac{2}{3}-7x=\frac{3}{2}-1\\ 2x-7x-\frac{2}{3}=\frac{1}{2}\\ -5x=\frac{1}{2}+\frac{2}{3}\\ -5x=\frac{7}{6}\\ x=\frac{7}{6}:\left(-5\right)\\ x=\frac{-7}{30}\)Vậy \(x=\frac{-7}{30}\)
b) \(\frac{3}{2}x-\frac{2}{5}=\frac{1}{3}x-\frac{1}{4}\\ \frac{3}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{1}{4}\\ \frac{7}{6}x=\frac{3}{20}\\ x=\frac{3}{20}:\frac{7}{6}\\ x=\frac{9}{70}\)Vậy \(x=\frac{9}{70}\)
c) \(\frac{2}{3}-\frac{5}{3}x=\frac{7}{10}x+\frac{5}{6}\\ \frac{2}{3}-\frac{5}{6}=\frac{7}{10}x+\frac{5}{3}x\\ \frac{-1}{6}=\frac{71}{30}x\\ x=\frac{-1}{6}:\frac{71}{30}\\ x=\frac{-5}{71}\)Vậy \(x=\frac{-5}{71}\)
d) \(2x-\frac{1}{4}=\frac{5}{6}-\frac{1}{2}x\\ 2x+\frac{1}{2}x=\frac{5}{6}+\frac{1}{4}\\ \frac{5}{2}x=\frac{13}{12}\\ x=\frac{13}{12}:\frac{5}{2}\\ x=\frac{13}{30}\)Vậy \(x=\frac{13}{30}\)
e) \(3x-\frac{5}{3}=x-\frac{1}{4}\\ 3x-x=\frac{5}{3}-\frac{1}{4}\\ 2x=\frac{17}{12}\\ x=\frac{17}{12}:2\\ x=\frac{17}{24}\)Vậy \(x=\frac{17}{24}\)
Èo, chăm thế? Chăm hơn cả mik cơ, gần 11 h rồi onl thì thấy bài được bạn HISI làm hết rồi :((
Trả lời :
\(a,\frac{1}{6}=\frac{x}{18}\Rightarrow\frac{3}{18}=\frac{x}{18}\Rightarrow x=3\)
\(b,\frac{x}{8}=\frac{-1}{4}\Rightarrow\frac{x}{8}=\frac{-2}{8}\Rightarrow x=-2\)
\(c,\frac{4}{5}=\frac{x}{10}\Rightarrow\frac{x}{10}=\frac{8}{10}\Rightarrow x=8\)
\(d,\frac{11}{5}=\frac{-22}{x}\Rightarrow\frac{-22}{x}=\frac{-22}{10}\Rightarrow x=10\)
a) 1/6 = x/18
1.18= 6.x
18 = 6.x
x= 18: 6
x= 3
b) x/8 = -1/4
x.4=-1.8
x.4= - 8
x=-8 :4
x = -4
c) 4/5= x/10
4 . 10= x.5
40 = x . 5
x= 40 : 5
x = 8
d)11/5 = -22/x
11.x = -22.5
11.x =-110
x = -110 : 11
x = -10
chúc bn học tốt !
Bài 1:
a; \(\dfrac{x}{3}\) = \(\dfrac{4}{y}\)
\(xy\) = 12
12 = 22.3; Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6;12}
Lập bảng ta có:
| \(x\) | -12 | -6 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 6 | 12 |
| y | -1 | -2 | -3 | -4 | -6 | -12 | 12 | 6 | 4 | 3 | 2 | 1 |
Theo bảng trên ta có các cặp \(x;y\) nguyên thỏa mãn đề bài là:
(\(x\)\(;y\)) =(-12; -1);(-6; -2);(-4; -3);(-2; -6);(-1; 12);(1; 12);(2;6);(3;4);(4;3);(6;2);(12;1)
b; \(\dfrac{x}{y}\) = \(\dfrac{2}{7}\)
\(x\) = \(\dfrac{2}{7}\).y
\(x\) \(\in\)z ⇔ y ⋮ 7
y = 7k;
\(x\) = 2k
Vậy \(\left\{{}\begin{matrix}x=2k\\y=7k;k\in z\end{matrix}\right.\)
\(\frac{4}{5}\)và \(\frac{-8}{-10}\)
\(\frac{-20}{24}\)và \(\frac{-5}{6}\)
\(\frac{7}{3}\)và \(\frac{-14}{-6}\)