cái này chỉ là thử sức thôi nha chỉ dành cho lớp 5 ai giải được là thông minh
1.TÍNH
a 5454 / 5757 - 171717 / 191919
b 6/5 X 7/6 X 8/7 X 9/8 X ... 2020/2019 X 2021/2020
2.SO SÁNH
1/3 + 1/6 +1/10 + 1/15 + 1/21+ 1/28 + 1/36+ 1/45 và 1
1/a-1 và 1/a+1
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a: 7/8<8/9<9/10<10/11<11/12
b: 4/5<5/6<6/7<7/8
c: 1/3<3/5<7/9<9/11<11/13
Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
a)= 2021.2021-2020.(2021+1)
= 2021.(2020+1)-2020.(2021+1)
= (2021.2020)+2021-(2020.2021)-2020
= 1
b) B= (1+2-3-4)+(5+6-7-8)+(9+10-11-12)...........+(2017+2018-2019-2020)+2021
B= -4+(-4)+....................(-4)+2021
B= -4x505+2021
B= -2020 + 2021
B = 1
1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)
=\(0\)
a,\(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\) (1)
<=> \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
<=> \(\left(x+1\right)\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
=> x+1=0 (vì \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\ne0\))
<=> x=-1
Vậy pt (1) có tập nghiệm S\(=\left\{-1\right\}\)
b, \(\frac{x+6}{2015}+\frac{x+5}{2016}+\frac{x+4}{2017}=\frac{x+3}{2018}+\frac{x+2}{2019}+\frac{x+1}{2010}\)(2)
<=> \(\frac{x+6}{2015}+1+\frac{x+5}{2016}+1+\frac{x+4}{2017}+1=\frac{x+3}{2018}+1+\frac{x+2}{2019}+1+\frac{x+1}{2020}+1\)
<=> \(\frac{x+2021}{2015}+\frac{x+2021}{2016}+\frac{x+2021}{2017}-\frac{x+2021}{2018}-\frac{x+2021}{2019}-\frac{x+2021}{2020}=0\)
<=> \(\left(x+2021\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x+2021=0(vì \(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=-2021
Vậy pt (2) có tập nghiệm S=\(\left\{-2021\right\}\)
c,\(\frac{x+6}{2016}+\frac{x+7}{2017}+\frac{x+8}{2018}=\frac{x+9}{2019}+\frac{x+10}{2020}+1\) (3)
<=> \(\frac{x+6}{2016}-1+\frac{x+7}{2017}-1+\frac{x+8}{2018}-1=\frac{x+9}{2019}-1+\frac{x+10}{2020}-1+1-1\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}=\frac{x-2010}{2019}+\frac{x-2010}{2020}\)
<=> \(\frac{x-2010}{2016}+\frac{x-2010}{2017}+\frac{x-2010}{2018}-\frac{x-2010}{2019}-\frac{x-2010}{2020}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
=> x-2010=0 (vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\))
<=> x=2010
Vậy pt (3) có tập nghiệm S=\(\left\{2010\right\}\)
d, \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\) (4)
<=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=15-1-2-3-4-5\)
<=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
<=> (x-100)(\(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\))=0
=> x -100=0(vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\))
<=> x=100
Vậy pt (4) có tập nghiệm S=\(\left\{100\right\}\)
a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy \(x=-1.\)
Mình chỉ làm câu a) thôi nhé.
Chúc bạn học tốt!
\(\frac{1}{3}:x-\frac{2}{3}=\frac{5}{6}+\frac{7}{8}\)
\(\frac{1}{3}:x-\frac{2}{3}=\frac{41}{24}\)
\(\frac{1}{3}:x=\frac{19}{8}\)
\(x=\frac{8}{57}\)
\(\frac{2}{5}-x:\frac{3}{5}=\frac{6}{7}\)
\(x:\frac{3}{5}=\frac{-16}{35}\)
\(x=\frac{-48}{175}\)
\(\frac{4}{5}-\frac{26}{m}=\frac{-8}{15}\)
\(\frac{26}{m}=\frac{4}{3}\)
=> 26 x 3 = m x 4
78 = m x 4
m = 19,5
Câu 1 :\(\frac{1}{3}\div x-\frac{2}{3}=\frac{5}{6}+\frac{7}{8}\)
\(\Leftrightarrow\frac{1}{3x}=\frac{5}{6}+\frac{7}{8}+\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{3x}=\frac{4\cdot5}{24}+\frac{7\cdot3}{24}+\frac{2\cdot8}{24}\)
\(\Leftrightarrow\frac{1}{3x}=\frac{20}{24}+\frac{21}{24}+\frac{16}{24}\)
\(\Leftrightarrow\frac{1}{3x}=\frac{57}{24}\)
\(\Leftrightarrow24=57\cdot3x\)
\(\Leftrightarrow24=171x\)
\(\Leftrightarrow x=\frac{24}{171}=\frac{8}{57}\)
Câu 2: \(\frac{2}{5}-x\div\frac{3}{5}=\frac{6}{7}\)
\(\Leftrightarrow\frac{-5x}{3}=\frac{6}{7}-\frac{2}{5}\)
\(\Leftrightarrow\frac{-5x}{3}=\frac{5\cdot6}{35}-\frac{2\cdot7}{35}\)
\(\Leftrightarrow\frac{-5x}{3}=\frac{30}{35}-\frac{14}{35}\)
\(\Leftrightarrow-\frac{5x}{3}=\frac{16}{35}\)
\(\Leftrightarrow-175x=48\)
\(\Leftrightarrow x=-\frac{48}{175}\)
Câu 3: \(\frac{4}{5}-\frac{26}{m}=-\frac{8}{15}\)
\(\Leftrightarrow\frac{26}{m}=\frac{4}{5}-\left(-\frac{8}{15}\right)\)
\(\Leftrightarrow\frac{26}{m}=\frac{4}{5}+\frac{8}{15}\)
\(\Leftrightarrow\frac{26}{m}=\frac{12}{15}+\frac{8}{15}\)
\(\Leftrightarrow\frac{26}{m}=\frac{20}{15}\)
\(\Leftrightarrow20m=26\cdot15=390\)
\(\Leftrightarrow m=\frac{390}{20}=\frac{39}{2}\)
1) a) \(\frac{5454}{5757}-\frac{171717}{191919}=\frac{18\times3\times101}{19\times3\times101}-\frac{17\times10101}{19\times10101}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\)
b) \(\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times....\times\frac{2021}{2020}=\frac{6\times7\times8\times...\times2021}{5\times6\times7\times...\times2020}=\frac{2021}{5}\)
2) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}=2\times\frac{1}{6}+2\times\frac{1}{12}+2\times\frac{1}{20}+...+2\times\frac{1}{90}\)
\(=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\)
\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=2\times\left(\frac{1}{2}-\frac{1}{10}\right)=2\times\frac{2}{5}=\frac{4}{5}\)
b)Vì \(a-1< a+1\)
=> \(\frac{1}{a-1}>\frac{1}{a+1}\)
1/ a x b = 1/a - 1/b