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Bài 1:
\(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - \(\dfrac{2}{5}\)
=\(\dfrac{40}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{61}{60}\)
b; \(\dfrac{12}{13}\) x \(\dfrac{3}{4}\) - \(\dfrac{7}{13}\)
= \(\dfrac{9}{13}\) - \(\dfrac{7}{13}\)
= \(\dfrac{2}{13}\)
c; \(\dfrac{15}{17}\) : \(\dfrac{19}{34}\) - \(\dfrac{17}{19}\)
= \(\dfrac{15}{17}\) x \(\dfrac{34}{19}\) - \(\dfrac{17}{19}\)
= \(\dfrac{30}{19}\) - \(\dfrac{17}{19}\)
= \(\dfrac{13}{19}\)
Bài 2:
a; \(\dfrac{x}{5}\) x \(\dfrac{3}{7}\) = \(\dfrac{9}{35}\)
\(\dfrac{x}{5}\) = \(\dfrac{9}{35}\) : \(\dfrac{3}{7}\)
\(\dfrac{x}{5}\) = \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{3}{5}\) x 5
\(x\) = 3
\(\text{a. x + 567 + 15 = 2020 + 567 + 12 }\)
\(x=2020+567+12-15-567\)
\(x=2020+\left(567-567\right)+12-15\)
\(x=2020+0+12-15\)
\(x=2032-15\)
\(x=2017\)
\(\text{b. (63 + x) + 2010 = 63 + 2010}\)
\(63+x+2010=63+2010\)
\(x=63+2010-2910-63\)
\(x=\left(63-63\right)+\left(2010-2010\right)\)
\(x=0\)
học tốt
a) \(D=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...++\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
=> \(D=1-\frac{1}{1024}\)
b) \(Đ=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(=1-\frac{1}{20}=\frac{19}{20}\)
a) D=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{512}+\frac{1}{1024}.\)
\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\dots+\frac{1}{512}-\frac{1}{1024}\)
\(D=1-\frac{1}{1024}\)
\(D=\frac{1023}{1024}\)
\(Đ=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)
\(Đ=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)
\(Đ=1-\frac{1}{20}\)
\(Đ=\frac{19}{20}\)
Phần c như kiểu sai đề chỗ cuối hay sao ấy.
b, 3/5 + 4/7 + 2/8 + 10/25 + 9/21 + 28/16
= 3/5 + 4/7 + 2/8 + 2/5 + 3/7 + 14/8
= (3/5 + 2/5) + ( 4/7 + 3/7) + ( 2/8 + 14/8)
= 1 + 1 + 7/4
= 2 + 7/4 = 15/4
c , 8/7 + 7/6 + 5/8 + 10/12 + 24/28 + 6/16
= c , 8/7 + 7/6 + 5/8 + 5/6 + 6/7 + 1/2
= (8/7 + 6/7) + (7/6 + 5/6) + 5/8 + 1/2
= 14/7 + 12/6 + 5/8 + 1/2
= 2 + 2 + 5/8 + 1/2
= 4 + 9/8 = 41/8
1b 4/11+5/11=9/11 ;2/3+4/5=10/15+12/15=22/15 ;5/7+9/14=10/14+9/14=19/14 ;2+5/8=2/2+5/8=8/8+5/8=13/8
a) (865+135)+976=1000+976=1976
(891+109)+799=1000+799=1799
(2/5+3/5)+7/9=1+7/9=16/9
1) 1/2*1/6=1/12 ; 1/3*1/4=1/12. Ta có : 17/5+1/12*71/5-1/12*28/5=1/12*(17/5+71/5-28/5)=1/12*60/5=1/12*12=1 2)1/10+4/20+9/30+16/40+25/50 =1/10+2/10+3/10+4/10+5/10(rút gọn) =15/10=3/2 3)18/15*(2525/3636+1515/1818-2020/7272)=6/5*(25/36+5/6-5/18)(rút gọn)=6/5*25/36+6/5*5/6-6/5*5/18=5/6+1-1/3=9/6=3/2.
1) \(=\left(\frac{3}{5}+\frac{2}{5}\right).\frac{6}{11}\)
\(=1.\frac{6}{11}\)
\(=\frac{6}{11}\)
2)\(\frac{17}{25}.\left(\frac{11}{19}+\frac{6}{19}+\frac{2}{19}\right)\)
\(=\frac{17}{25}.1\)
\(=\frac{17}{25}\)
a, \(\frac{5}{11}\times\frac{7}{25}+\frac{15}{11}\times\frac{1}{5}\)
\(=\frac{5}{11}\times\frac{7}{25}+\frac{5}{11}\times\frac{3}{5}\)
\(=\frac{5}{11}\times\left(\frac{7}{25}+\frac{3}{5}\right)\)
\(=\frac{5}{11}\times\frac{22}{25}\)
\(=\frac{2}{5}\)
b, \(\frac{3}{7}\times\frac{25}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\frac{75}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\left(\frac{75}{19}-\frac{18}{19}\right)\)
\(=\frac{1}{7}\times3\)
\(=\frac{3}{7}\)
khó quá
a) 11/12 x 9/19 - 22/24 x 6/19 + 11/12 x 16/19.
= 11/12 x 9/19 - 11/12 x 6/19 + 11/12 x 16/19.
=11/12 x ( 9/19 -6/19 + 16/19)
=11/12x 1
=11/12
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