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a) \(15\times\left(\frac{212121}{434343}+\frac{333333}{353535}\right)=15\times\left(\frac{21\times10101}{43\times10101}+\frac{33\times10101}{35\times10101}\right)\)
\(=15\times\left(\frac{21}{43}+\frac{33}{35}\right)=\frac{6462}{301}\)
b) \(\frac{639\times721721}{721\times639639}=\frac{639\times721\times1001}{721\times639\times1001}=1\)
c) \(\frac{327\times412+400}{328\times412-12}=\frac{\left(328-1\right)\times412+400}{328\times412-12}=\frac{328\times412-412+400}{328\times412-12}\)
\(=\frac{328\times412-12}{328\times412-12}=1\)
d) \(9\times\left(\frac{151515}{171717}+\frac{131313}{181818}\right)=9\times\left(\frac{15\times10101}{17\times10101}+\frac{13\times10101}{18\times10101}\right)=9\times\left(\frac{15}{17}+\frac{13}{18}\right)\)
\(=9\times\frac{491}{306}=\frac{491}{34}\)
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a, (x+2)+(x+4)+(x+6)+...+(x+100)=6000
(x+x+x+...+x)+(2+4+6+...+100)=6000
50.x+2550=6000
50.x=6000-2550
50.x=3450
x=3450:50
x=69
b, 1+2+3+4+...+x=15
10+...+x=15
x=15-10
x=5
Nho **** cho minh nha
Ta có: (x+x+x+...+x) + (2+4+6+...+100) = 6000
Ta thấy vế phải có: (100-2):2+1=50(số hạng)
Tổng của vế phải: [(2+100)*50]:2=2550
\(\Rightarrow\)có 50 số x
\(\Rightarrow\)50*x + 2550 = 6000
\(\Rightarrow\)50*x=6000-2550
\(\Rightarrow\)50*x=3450
\(\Rightarrow\)x=3450:50
\(\Rightarrow\)x=69
Vậy x=69
Mình đúng nè, nhớ k nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{23}{12}\)
\(\frac{314}{105}\)
\(\frac{59}{60}\)
\(\frac{199}{90}\)
\(\frac{1}{18}\)
\(\frac{13}{36}\)
\(\frac{4}{221}\)
\(\frac{4}{85}\)
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a, \(\frac{7}{4x}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)
\(\frac{7}{4x}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=22\)
\(\frac{7}{4x}\left[33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]=22\)
\(\frac{7}{4x}\left[33.\left(\frac{35}{420}+\frac{21}{420}+\frac{14}{420}+\frac{10}{420}\right)\right]=22\)
\(\frac{7}{4x}\left[33.\frac{4}{21}\right]=22\)
\(\frac{7}{4x}.\frac{44}{7}\)=22
\(\frac{11}{x}=22\)
x=11:22
x=\(\frac{1}{2}\)
b,\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)
Đặt A\(=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Ta có :\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow4A=4.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)
\(\Rightarrow4A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)
\(\Rightarrow4A=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)
\(\Rightarrow A=\frac{63}{64}:4=\frac{63}{256}\)
\(\Rightarrow\frac{63}{256}.x=1\)
\(\Leftrightarrow x=1:\frac{63}{256}=\frac{256}{63}\)
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\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
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\(8\frac{7}{10}+2\frac{3}{4}=\frac{87}{10}+\frac{11}{4}=\frac{174}{20}+\frac{55}{20}=\frac{229}{20}\)
Bạn chỉ cần đưa về phân số xong tính bình thường. Muốn đổi từ hỗn số sang phân số, ta chỉ cần lấy phần nguyên nhân cho mẫu rồi cộng với tử là xong. Chứ bạn cứ hỏi mấy bài dễ như thế này thì k giỏi đc đâu!!!
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A=1999/2000
B=199/200
C=511/512
hok tốt
Đáp án
mình lười trình bày cách làm lém, để đáp án thui nha
A = \(\frac{1999}{2000}\)
B = \(\frac{199}{200}\)
C = \(\frac{511}{512}\)
\(15\times\left(\frac{212121}{434343}+\frac{333333}{555555}\right)=15\times\left(\frac{21}{43}+\frac{3}{5}\right)=15\times\frac{234}{215}=\frac{702}{43}\)