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Bài 1: <Cho là câu a đi>:
a. \(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow1-\frac{1}{x+1}=\frac{49}{50}\)
\(\rightarrow\frac{1}{x+1}=1-\frac{49}{50}=\frac{1}{50}\)
\(\rightarrow x+1=50\rightarrow x=49\)
Vậy x = 49.
a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)
\(=\frac{5}{2}\cdot\frac{7}{1}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)
b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)
\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)
\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)
c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)
\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)
\(=\frac{1}{4}-\frac{1}{4}+\frac{3}{4}=\frac{3}{4}\)
Bài 2 Bạn tự làm nhé
1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{67}{4}\)
b,Các phép tính khác làm tương tự
Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ
c,tương tự
2.
a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{7}{12}\div x=\frac{-77}{20}\)
Đến đây dễ bạn tự làm
b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)
\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)
\(\frac{14}{5}x+50=-34\)
\(\frac{14}{5}x=-84\)
Tự làm tiếp
c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)
Bài 1:
\(\left(\frac{3}{5}x+8\right):20=1\)
\(\frac{3}{5}x+8=1.20\)
\(\frac{3}{5}x+8=20\)
\(\frac{3}{5}x=20-8\)
\(\frac{3}{5}x=12\)
\(x=12:\frac{3}{5}\)
\(x=20\)
\(\left(\frac{5}{2}x-3\right):15=\frac{3}{10}\)
\(\frac{5}{2}x-3=\frac{3}{10}.15\)
\(\frac{5}{2}x-3=\frac{9}{2}\)
\(\frac{5}{2}x=\frac{9}{2}+3\)
\(\frac{5}{2}x=\frac{15}{2}\)
\(x=\frac{15}{2}:\frac{5}{2}\)
\(x=3\)
để \(\frac{n-1}{n+3}\)là số nguyên thì n-1 chia hết cho n+3
ta có:n-1=n+3-4
để n-1 chia hết cho n+3
thì -4 chia hết cho n+3
=>n+3\(\in\)Ư(-4)
Ư(-4)={-1,-2,-4,4,2,1}
ta có bảng:
| n+3 | 1 | -1 | 2 | -2 | 4 | -4 |
| n | -2 | -4 | -1 | -5 | 1 | -7 |
vậy với n\(\in\){-7,-5,-4,-2,-1,1} thì \(\frac{n-1}{n+3}\)có giá trị nguyên
Câu 1a:
75% - 1 1/2 + 0,5 : 5/12 - (-1/2)^2
= 0,75 - 1,5 + 0,5 x 12/5 - 0,25
= (0,75 - 0,25) - 1,5 + 1,2
= 0,5 - 1,5 + 1,2
= - 1 + 1,2
= 0,2
Bài 1b:
(5/7.0,6 - 5: 3 1/2). (40% - 1,4).(-2)^3
= (3/7 - 5 x 2/7).(0,4 - 1,4).8
= - 1.(-1).8
= 8
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
x/3 = -12/9
=> x/3 = -4/3
=> x = -4
vậy_
1.Ta có: \(\frac{x}{3}=-\frac{12}{9}\)
=> \(\frac{3x}{9}=-\frac{12}{9}\)
=> 3x = -12
=> x = -12 : 3
=> x = -4
\(\frac{4}{5}x-\frac{8}{5}=-\frac{1}{2}\)
=> \(\frac{4}{5}x=-\frac{1}{2}+\frac{8}{5}\)
=> \(\frac{4}{5}x=\frac{11}{10}\)
=> \(x=\frac{11}{10}:\frac{4}{5}\)
=> \(x=\frac{11}{8}\)
bài 1: \(\frac{x}{3}=\frac{-12}{9}\)=> 9x=-36
=> x=-4
vậy x=-4
\(\frac{4}{5}x-\frac{8}{5}=\frac{-1}{2}\)=> \(\frac{4}{5}x=\frac{-1}{2}+\frac{8}{5}\)
=> \(\frac{4}{5}x=\frac{-5}{10}+\frac{16}{10}\)=\(\frac{11}{10}\)=> \(x=\frac{11}{10}:\frac{4}{5}\)=\(\frac{11}{10}.\frac{5}{4}\)=\(\frac{11}{8}\)
vậy x=\(\frac{11}{8}\)
\(\frac{1}{5}.\left|x\right|-1\frac{2}{5}=\frac{2}{5}\)=> \(\frac{1}{5}.\left|x\right|-\frac{7}{5}=\frac{2}{5}\)
=> \(\frac{1}{5}.\left|x\right|=\frac{2}{5}+\frac{7}{5}=\frac{9}{5}\)=> |x| =\(\frac{9}{5}:\frac{1}{5}\)=9
=> x=9 hoặc x=-9
vậy x=9 hoặc x=-9
1.\(\frac{1}{5}\left|x\right|-1\frac{2}{5}=\frac{2}{5}\)
=> \(\frac{1}{5}\left|x\right|=\frac{2}{5}+\frac{7}{5}\)
=> \(\frac{1}{5}\left|x\right|=\frac{9}{5}\)
=> \(\left|x\right|=\frac{9}{5}:\frac{1}{5}\)
=> \(\left|x\right|=9\)
=> \(\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
Bài 2: Ta có: \(\frac{n+1}{n-2}=\frac{\left(n-2\right)+3}{n-2}=1+\frac{3}{n-2}\)
Để \(\frac{n+1}{n-2}\)là số nguyên <=> 3 \(⋮\)n - 2
<=> n - 1 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng :
Vậy ...
Bài 1 :
\(\frac{x}{3}=\frac{-12}{9}\)
\(\Rightarrow\frac{3x}{9}=\frac{-12}{9}\)
\(\Rightarrow3x=-12\)
\(\Rightarrow x=-4\)
\(\frac{4}{5}x-\frac{8}{5}=-\frac{1}{2}\)
\(\Rightarrow\frac{4}{5}x=-\frac{1}{2}+\frac{8}{5}\)
\(\frac{4}{5}x=\frac{11}{10}\)
\(\Rightarrow x=\frac{11}{10}:\frac{4}{5}\)
\(x=\frac{11}{8}\)
\(\frac{1}{5}.|x|-1\frac{2}{5}=\frac{2}{5}\)
\(\Rightarrow\frac{1}{5}.|x|=\frac{2}{5}+\frac{7}{5}\)
\(\frac{1}{5}.|x|=\frac{9}{5}\)
\(\Rightarrow|x|=\frac{9}{5}:\frac{1}{5}=9\)
\(\Rightarrow\hept{\begin{cases}x=9\\x=-9\end{cases}}\)
kudo shinichi n - 2 chứ k phải n - 1 nhá =))
2.
Để \(\frac{n+1}{n-2}\in Z\) thì \(n\in Z\)
\(\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)
Để \(\frac{n+1}{n-2}\in Z\)thì \(\frac{3}{n-2}\in Z\)
\(\Rightarrow n-2\in\left\{-3;-1;1;3\right\}\)
\(\Rightarrow n\in\left\{-1;1;3;5\right\}\)