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\(\frac{4}{12}=\frac{4:4}{12:4}=\frac{1}{3}\)
\(\frac{24}{30}=\frac{24:6}{30:6}=\frac{4}{5}\)
\(\frac{25}{100}=\frac{25:25}{100:25}=\frac{1}{4}\)
\(\frac{60}{80}=\frac{60:20}{80:20}=\frac{3}{4}\)
\(\frac{9}{18}=\frac{9:9}{18:9}=\frac{1}{2}\)
\(\frac{60}{36}=\frac{60:12}{36:12}=\frac{5}{3}\)
\(\frac{72}{54}=\frac{72:18}{54:18}=\frac{4}{3}\)
\(\frac{35}{210}=\frac{35:35}{210:35}=\frac{1}{6}\)
4/12=1/3
24/30=4/5
25/100=1/4
60/80=3/4
9/18=1/2
60/36=5/3
72/54=4/3
35/210=1/6
2/9,7/9,1 nửa,2/2,
Còn lại bó tay bởi vì em mới lên lớp 3 thôi à!
Lời giải:
Xét tử số:
$\text{TS}=1+25^4+25^8+...+25^{28}$
$25^4.\text{TS}=25^4+25^8+...+25^{32}$
$\Rightarrow \text{TS}(25^4-1)=25^{32}-1$
$\text{TS}=\frac{25^{32}-1}{25^4-1}$
Xét mẫu số:
$\text{MS}=1+25^2+..+25^{30}$
$25^2.\text{MS}=25^2+25^4+...+25^{32}$
$\Rightarrow \text{MS}(25^2-1)=25^{32}-1$
$\Rightarrow \text{MS}=\frac{25^{32}-1}{25^2-1}$
Do đó:
$A=\frac{25^{32}-1}{25^4-1}:\frac{25^{32}-1}{25^2-1}=\frac{25^2-1}{25^4-1}$
$=\frac{25^2-1}{(25^2-1)(25^2+1)}=\frac{1}{25^2+1}$
a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
Đặt phân số trên là A
\(A=\frac{25^{28}+25^{24}+...+25^4+25^0}{\left(25^{28}+25^{24}+...+25^4+25^0\right)+\left(25^{30}+25^{26}+...+25^6+25^2\right)}\)
\(\frac{1}{A}=\frac{\left(25^{28}+25^{24}+...+25^4+25^0\right)+\left(25^{30}+25^{26}+...+25^6+25^2\right)}{25^{28}+25^{24}+...+25^4+25^0}\)
\(\frac{1}{A}=1+\frac{25^{30}+25^{26}+...+25^6+25^2}{25^{28}+25^{24}+...+25^4+25^0}\)
Đặt \(B=\frac{25^{30}+25^{26}+...+25^6+25^2}{25^{28}+25^{24}+...+25^4+25^0}\)
\(\frac{B}{25^2}=\frac{25^{30}+25^{26}+...+25^6+25^2}{25^{30}+25^{26}+...+25^6+25^2}=1\Rightarrow B=25^2\)
=> \(\frac{1}{A}=1+B=1+25^2\Rightarrow A=\frac{1}{1+25^2}\)
Bài 1:
a: 6/9=2/3
6/24=1/4
48/96=1/2
42/98=3/7
b: 24/36=2/3
18/30=3/5
15/120=1/8
80/240=1/3
c: 5/25=1/5
75/100=3/4
64/720=4/45
16/1000=2/125
Bài 2:
Các phân số bằng 2/3 là 34/51; 20/30; 84/126
Bài 1 :
a) 6/9 = 2/3 ; 6/24 = 1/4 ; 48/96 = 1/2 ; 42/98 = 3/7
b) 24/36 = 2/3 ; 18/30 = 3/5 ; 15/120 = 1/8 ; 80 / 240 = 1/3
c) 5 / 25 = 1/5 ; 75/100=3/4 ; 64/720=4/45 ; 16/1000=2/125
Bài 2 :
Các phân số bằng 2/3 là :
\(\text{34/51; 20/30; 84/126}\)
\(\frac{4}{12}\)=\(\frac{1}{3}\);\(\frac{24}{30}\)=\(\frac{12}{15}\);\(\frac{25}{100}\)=\(\frac{1}{4}\)
4 /12 = 1 / 3
24 / 30 = 12 / 15 = 4 / 5
25 / 100 = 1 / 4