75%-1/3+-1/4.1/3-1 1/3
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1\frac{5}{7}\cdot\frac{3}{4}=\frac{12}{7}\cdot\frac{3}{4}=\frac{9}{7}\)
\(\frac{10}{11}:1\frac{1}{3}=\frac{10}{11}:\frac{4}{3}=\frac{15}{22}\)
\(3,57\cdot4,1+2,43\cdot4,1=(3,57+2,43)\cdot4,1\)
\(=6\cdot4,1=24,6\)
Cuối tự làm
~Chúc bn học tốt~
\(1\frac{5}{7}x\frac{3}{4}=\frac{12}{7}x\frac{3}{4}=\frac{9}{7}\)
\(\frac{10}{11}:1\frac{1}{3}=\frac{10}{11}:\frac{4}{3}=\frac{10}{11}x\frac{3}{4}=\frac{30}{44}=\frac{15}{22}\)
\(3,57x4,1+2,43x4,1=4,1x\left(3,57+2,43\right)=4,1x6=24,6\)
\(3,42:0,57x8,4-6,8=6x8,4-6,8=50,4-6,8=43,6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(D=(\frac{1}{2}-1)(\frac{1}{3}-1)(\frac{1}{4}-1)...(\frac{1}{50}-1)=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{1-50}{50}\)
\(=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-49}{50}=-\frac{1.2.3.4...49}{2.3.4...50}=\frac{-1}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=-1/2.3-1/3.4-1/4.5-1/5.6
=-(1/2.3+1/3.4+1/4.5+1/5.6)
=-(1/2-1/3+1/3-1/4+1/5-1/6)
=-(1/2-1/6)
=1/6-1/2=-1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
gọi 3 phần bị chia ra lần lượt là:x,y,z.
Ta có: x+y+z=900 và \(\frac{\frac{x}{1}}{3}=\frac{\frac{y}{1}}{4}=\frac{\frac{z}{1}}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau
\(\frac{x}{\frac{1}{3}}=\frac{y}{\frac{1}{4}}=\frac{z}{\frac{1}{5}}=\frac{x+y+z}{\frac{1}{3}+\frac{1}{4}+\frac{1}{6}}=\frac{900}{\frac{9}{12}}=1200\)
\(\frac{x}{\frac{1}{3}}=\frac{1}{3}.1200=400\)
\(\frac{y}{\frac{1}{4}}=\frac{1}{4}.1200=300\)
\(\frac{z}{\frac{1}{6}}=\frac{1}{6}.1200=200\)
Vậy 3 phần bị chia ra lần lượt là: 400, 300,200
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/2.1/3+1/3.1/4+1/4.1/5+1/5.1/6
=1/2.3+1/3.4+1/4.5+1/5.6
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1/2-1/6
=2/6=1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(75\%-\frac{1}{3}+\frac{-1}{4}.\frac{1}{3}-1\frac{1}{3}\)
\(=\frac{3}{4}-\frac{1}{3}-\frac{1}{4}.\frac{1}{3}-1-\frac{1}{3}\)
\(=\frac{3}{4}+\left(\frac{-1}{3}-\frac{1}{4}-1\right).\frac{1}{3}-1\)
\(=\frac{3}{4}+\frac{-19}{12}.\frac{1}{3}-1\)
\(=\frac{3}{4}+\frac{-19}{36}-1\)
\(=\frac{2}{9}-1=\frac{-7}{9}\)
\(75\%-\frac{1}{3}+\left(-\frac{1}{4}\right).\frac{1}{3}-1\frac{1}{3}\)
\(=\)\(\frac{3}{4}-\frac{1}{3}+-\frac{1}{12}-\frac{4}{3}\)
\(=\)\(\frac{5}{12}+-\frac{1}{12}-\frac{4}{3}\)
\(=\)\(\frac{4}{12}-\frac{4}{3}\)
\(=\)\(-1\)