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1996*1995-1996*1994=1996
(1996*1995-996)-(1996*1994)=1996-996=1000
(1996*1995-996)-(1996*1994+1000)=1996-996-1000=0
=>(1996*1995-996)/(1996*1994+1000)=1( vì tử số bằng mẫu số)
\(\frac{1996.1995-996}{1996.1994+1000}=\frac{\left(1000+996\right).1995-996}{\left(1000+996\right).1994+1000}\)
\(=\frac{1000.1995+996.1995-996}{996.1994+1000.1994+1000}=\frac{1000.1995+996\left(1995-1\right)}{996.1994+1000\left(1994+1\right)}\)
\(=\frac{1000.1995+996.1994}{996.1994+1000.1995}\)=1
a)4,25+3,15+3,75+5,85=(4,25+3,75)+(3,15+5,85)=8+9=17.
b)(3/5+4/7)+2/5=(3/5+2/5)+4/7=1+4/7=11/7.
c)6,24+5,36+3,76=(6,24+3,76)+5,36=10+5,36=15,36.
d)(19/15+3/5)+11/15=(19/15+11/15)+3/5=2+3/5=13/5.
ủng hộ nha có j kb vs mk
Tính bằng cách thuận tiện nhất:
a ) 4,25 + 3,15 + 3,75 + 5,85
= ( 4,25 + 3,75 ) + ( 3,15 + 5, 85 )
= 8 + 9
= 17
tks giúp mk vs ạ!
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{y\times\left(y+1\right)}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{y}-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow\frac{1}{y+1}=1-\frac{996}{997}=\frac{1}{997}\)
\(\Leftrightarrow y+1=997\Leftrightarrow y=996\)
Vậy y = 996
a,Đặt \(A=\frac{1}{1\times4}+\frac{1}{4\times7}+...+\frac{1}{97\times100}\)
\(\Rightarrow3A=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{97\times100}\)
\(\Rightarrow3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3A=1-\frac{1}{100}=\frac{99}{100}\)
\(\Rightarrow A=\frac{99}{300}\)
b, \(\frac{1}{2}\times\frac{2}{3}\times...\times\frac{99}{100}=\frac{1\times2\times...\times99}{2\times3\times...\times1000}=\frac{1}{100}\)
c, \(\frac{3}{4}\times\frac{8}{9}\times...\times\frac{99}{100}=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times...\times\frac{9.11}{10.10}=\frac{1.2.....9}{2.3.....10}\times\frac{3.4.....11}{2.3.....10}=\frac{1}{10}\times\frac{11}{2}=\frac{11}{20}\) (dấu . là dấu nhân)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{1}-\frac{1}{8}\)
\(=\frac{7}{8}\)'
a,
\(x+\frac{7}{15}=1\frac{1}{20}\)
\(x+\frac{7}{15}=\frac{21}{20}\)
\(x=\frac{21}{20}-\frac{7}{15}=\frac{63}{60}-\frac{28}{60}\)
\(x=\frac{35}{60}=\frac{7}{12}\)
b,
\(\left[3\frac{1}{2}-x\right]\cdot1\frac{1}{4}=\frac{15}{16}\)
\(\left[\frac{7}{2}-x\right]\cdot\frac{5}{4}=\frac{15}{16}\)
\(\frac{7}{2}-x=\frac{15}{16}:\frac{5}{4}=\frac{3}{4}\)
\(\frac{7}{2}-x=\frac{3}{4}\Rightarrow x=\frac{7}{2}-\frac{3}{4}\)
\(x=\frac{11}{4}\)
c,
\(1\frac{1}{5}x+\frac{2}{3}x=\frac{56}{125}\Leftrightarrow\frac{6}{5}x+\frac{2}{3}x=\frac{56}{125}\)
\(\frac{28}{15}x=\frac{56}{125}\Rightarrow x=\frac{6}{25}\)
d,
\(60\%x+0,4x+x:3=2\)
\(\frac{3}{5}x+\frac{2}{5}x+\frac{1}{3}x=2\)
\(\frac{4}{3}x=2\Rightarrow x=\frac{3}{2}\)
Nguyễn Anh Thiện
a)
x + \(\frac{7}{15}\) = \(1\frac{1}{20}\)
X + \(\frac{7}{15}=\frac{21}{20}\)
X \(=\frac{21}{20}-\frac{7}{15}\)
X \(=\frac{63}{60}-\frac{28}{60}=\frac{35}{60}=\frac{7}{12}\)
^^ Học tốt !
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
\(a,\frac{125\cdot81\cdot7}{49\cdot75\cdot27}=\frac{25\cdot5\cdot9\cdot9\cdot7}{7\cdot7\cdot15\cdot5\cdot3\cdot9}=\frac{25\cdot9}{7\cdot15\cdot3}=\frac{5\cdot5\cdot3\cdot3}{7\cdot3\cdot5\cdot3}=\frac{5}{7}\)
\(b,\frac{48\cdot25-48\cdot11-48}{16\cdot23-16\cdot10}=\frac{1200-528-48}{368-160}=\frac{624}{208}=3\)
\(c,\frac{15\cdot\left[4\cdot17+17\cdot5\right]}{18\cdot25\cdot34}=\frac{15\cdot\left[17\cdot\left(4+5\right)\right]}{15300}=\frac{15\cdot153}{15300}\)
\(=\frac{2295}{15300}=\frac{3}{20}\)
\(d,\frac{1996\cdot1995-996}{1000+1996\cdot1994}=\frac{3982020-996}{1000+3980024}=\frac{3981024}{3981024}=1\)