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\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)
= \(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)
= \(\dfrac{4\times9\times16\times25\times36}{3\times8\times15\times24\times35}\)
= \(\dfrac{1\times2\times2\times25\times36}{1\times2\times15\times24\times35}\)
= \(\dfrac{4\times25\times36}{30\times24\times35}\)
= \(\dfrac{1\times25\times36}{30\times6\times35}=\dfrac{1}{7}\)
Sai rồi nhé! Từ dấu bằng thứ 2 xuống dấu bằng thứ 3 bạn làm sao đc z?
\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\dfrac{25}{24}\cdot\dfrac{36}{35}=\dfrac{12}{7}\)
= 4/3*9/8*16/15*25/24*36/35
=2*2/1*3 * 3*3/2*4 *4*4/3*5 *5*5/4*6 * 6*6/5*7
= (2*3*4*5*6 / 1*2*3*4*5) * ( 2*3*4*5*6 / 3*4*5*6*7)
=6/1* 2/7
= 12/7
\(=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{17}{25}-\dfrac{8}{25}\right)+\left(\dfrac{13}{37}+\dfrac{24}{37}\right)=1\)
Giải:
\(9-3\times\left(x-9\right)=6\)
\(3\times\left(x-9\right)=9-6\)
\(3\times\left(x-9\right)=3\)
\(x-9=3:3\)
\(x-9=1\)
\(x=1+9\)
\(x=10\)
\(4+6\times\left(x+1\right)=70\)
\(6\times\left(x+1\right)=70-4\)
\(6\times\left(x+1\right)=66\)
\(x+1=66:6\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
\(\dfrac{x}{13}+\dfrac{15}{26}=\dfrac{46}{52}\)
\(\dfrac{x}{13}=\dfrac{23}{26}-\dfrac{15}{26}\)
\(\dfrac{x}{13}=\dfrac{4}{13}\)
\(\Rightarrow x=4\)
\(\dfrac{11}{14}-\dfrac{3}{x}=\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{11}{14}-\dfrac{5}{14}\)
\(\dfrac{3}{x}=\dfrac{3}{7}\)
\(\Rightarrow x=7\)
\(5\times\left(3+7\times x\right)=40\)
\(3+7\times x=40:5\)
\(3+7\times x=8\)
\(7\times x=8-3\)
\(7\times x=5\)
\(x=5:7\)
\(x=\dfrac{5}{7}\)
\(x\times6+12:3=120\)
\(x\times6+4=120\)
\(x\times6=120-4\)
\(x\times6=116\)
\(x=116:6\)
\(x=\dfrac{58}{3}\)
\(x\times3,7+x\times6,3=120\)
\(x\times\left(3,7+6,3\right)=120\)
\(x\times10=120\)
\(x=120:10\)
\(x=12\)
\(\left(15\times24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):0,25=400\)
\(360-x=400.0,25\)
\(360-x=100\)
\(x=360-100\)
\(x=260\)
\(71+65\times4=\dfrac{x+140}{x}+260\)
\(\left(x+140\right):x+260=71+260\)
\(x:x+140:x+260=331\)
\(1+140:x+260=331\)
\(140:x=331-1-260\)
\(140:x=70\)
\(x=140:70\)
\(x=2\)
\(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=155\)
\(10\times x+\left(1+4+7+...+28\right)=155\)
Số số hạng \(\left(1+4+7+...+28\right)\) :
\(\left(28-1\right):3+1=10\)
Tổng dãy \(\left(1+4+7+...+28\right)\) :
\(\left(1+28\right).10:2=145\)
\(\Rightarrow10\times x+145=155\)
\(10\times x=155-145\)
\(10\times x=10\)
\(x=10:10\)
\(x=1\)
Đều theo cách lớp 5 nha em!
\(\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{32}{40}+\dfrac{48}{56}+\dfrac{14}{21}\\ =\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{4}{5}+\dfrac{6}{7}+\dfrac{2}{3}\\ =\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{7}+\dfrac{6}{7}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\\ =1+1+1=3\)
Lời giải:
$\frac{1}{3}+\frac{1}{7}+\frac{1}{5}+\frac{32}{40}+\frac{48}{56}+\frac{14}{21}$
$=\frac{1}{3}+\frac{1}{7}+\frac{1}{5}+\frac{4}{5}+\frac{6}{7}+\frac{2}{3}$
$=(\frac{1}{3}+\frac{2}{3})+(\frac{1}{7}+\frac{6}{7})+(\frac{1}{5}+\frac{4}{5})$
$=\frac{3}{3}+\frac{7}{7}+\frac{5}{5}=1+1+1=3$
`1/8+1/24+1/48+1/80+1/120`
`=1/[2xx4]+1/[4xx6]+1/[6xx8]+1/[8xx10]+1/[10xx12]`
`=1/2xx(2/[2xx4]+2/[4xx6]+2/[6xx8]+2/[8xx10]+2/[10xx12])`
`=1/2xx(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)`
`=1/2xx(1/2-1/12)`
`=1/2xx(6/12-1/12)`
`=1/2xx5/12=5/24`
\(\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
=\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{10.12}\)
=\(\dfrac{1}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{10.12}\right)\)
=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
=\(\dfrac{1}{2}.\dfrac{5}{12}\)
=\(\dfrac{5}{24}\)
Dấu chấm(.)là nhân.