tính a:
\(\frac{1}{33}\)x (\(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\)
Tìm x
1) \(\frac{-7}{4}x\cdot\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)
2) \(\frac{1}{2016}:2015\cdot x=\frac{-1}{2015}\)
a) 74x.(3312+33332020+333333303030+3333333342424242)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247x.(1233+20203333+303030333333+4242424233333333)=32
74x.(3312+3320+3330+3342)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247x.(1233+2033+3033+4233)=32
74x.(333.4+334.5+335.6+336.7)=32\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247x.(3.433+4.533+5.633+6.733)=32
74x.33.(13−14+14−15+15−16+16−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247x.33.(31−41+41−51+51−61+61−71)=32
74x.33.(13−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247x.33.(31−71)=32
74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247x.33⋅214=32
b) 13+16+110+115+...+2x.(x−1)=20072009\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}31+61+101+151+...+x.(x−1)2=20092007
26+212+220+230+...+2(x−1).x=20072009\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}62+122+202+302+...+(x−1).x2=20092007
22.3+23.4+24.5+25.6+...+2(x−1).x=20072009\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}2.32+3.42+4.52+5.62+...+(x−1).x2=20092007
2.(12−13+13−14+14−15+15−16+...+1x−1−1x)=200720092.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−31+31−41+41−51+51−61+...+x−11−x1)=20092007
2.(12−1x)=200720092.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−x1)=20092007
1−2x=200720091-\frac{2}{x}=\frac{2007}{2009}1−x2=20092007
2x=22009\frac{2}{x}=\frac{2}{2009}x2=20092
=> x = 2009
a) \frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247x.(1233+20203333+303030333333+4242424233333333)=32
\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247x.(1233+2033+3033+4233)=32
\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247x.(3.433+4.533+5.633+6.733)=32
\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247x.33.(31−41+41−51+51−61+61−71)=32
\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247x.33.(31−71)=32
\frac{7}{4}x.33\cdot\frac{4}{21}=3247x.33⋅214=32
đến đây thì bn tự lm đk r
\(\frac{7}{4}\).(\(\frac{33}{12}\)+\(\frac{3333}{2020}\)+\(\frac{333333}{303030}\)+\(\frac{33333333}{42424242}\))
\(\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{12}+\frac{3333\div101}{2020\div101}+\frac{333333\div10101}{303030\div10101}+\frac{33333333\div1010101}{42424242\div1010101}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
= 7/4 . 44/7
= 11
\(\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33.101}{20.101}+\frac{33.10101}{30.10101}+\frac{33.1010101}{42.1010101}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{7}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{7}{4}.33.\frac{4}{21}\)
\(=11\)
Tham khảo nhé~
\(\frac{7}{4x.}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=22\)
\(\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right).x=1\)
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 :\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow4A=4.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)\)
\(\Rightarrow4A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}\)
\(\Rightarrow4A=\frac{32+16+8+4+2+1}{64}=\frac{63}{64}\)
\(\Rightarrow A=\frac{63}{64}:4=\frac{63}{256}\)
\(\Rightarrow\frac{63}{256}.x=1\)
\(\Leftrightarrow x=1:\frac{63}{256}=\frac{256}{63}\)
1.Tìm y
a) y3 + 3y = 12 x 11
b) 64 x y + 225 = 700 - 75 - 36 x y
2.Tính
a)\(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)
b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
bài 2
a] = 3 x \(\frac{4343}{7171}\)= \(\frac{17372}{7171}\)= \(\frac{172}{71}\)
b] = \(\frac{1}{33}\)x \(\frac{44}{7}\)= \(\frac{1}{3}\)x \(\frac{4}{7}\)=\(\frac{4}{21}\)
bài 1
a] y là 9
b] <=> 64y + 36y = 700 - 75 - 225
<=> 100y = 400
<=> y = 4
trên lớp cô sửa rồi nên mình giải luôn:
1) Tìm y
a) y3 + 3y = 12 x 11
y3 + 3y = 132
y x 10 + 3 + 3 x 10 + y = 132
( y x 10 + y ) + ( 3 x 10 + 3 ) = 132
11 x y + 33 = 132
11 x y = 132 - 33
11 x y = 99
y = 99 : 11
y = 9
b) 64 x y + 225 = 700 - 75 - 36 x y
64 x y + 225 = 625 - 36 x y
64 x y + 36 x y = 625 -225
64 x y + 36 x y = 400
( 64 + 36 ) x y = 400
100 x y = 400
y = 400 : 100
y = 4
2) Tính
a) \(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)
\(=\frac{4343}{7171}\times4\)
\(=\frac{43}{71}\times4\)
\(=\frac{172}{71}\)
b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
Ta có:
\(\frac{3333}{2020}=\frac{3333:101}{2020:101}=\frac{33}{20}\)
\(\frac{333333}{303030}=\frac{333333:10101}{303030:10101}=\frac{33}{30}\)
\(\frac{33333333}{42424242}=\frac{33333333:1010101}{42424242:1010101}=\frac{33}{42}\)
A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
A = \(\frac{1}{33}\times33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{7}{21}-\frac{3}{21}\right)\)
A = 1 x \(\frac{4}{21}\)
A = \(\frac{4}{21}\)
\(\frac{7}{4}+\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
bài 1:
\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\cdot\frac{4}{21}=11\)
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)
= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)
= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)
= \(\frac{31}{34}:2\)
= \(\frac{31}{68}\)
đề đúng
\(\frac{7}{4}\cdot\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
Tính :
\(B=\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(C=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
\(B=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{11}{4}+\dfrac{33}{20}+\dfrac{11}{10}+\dfrac{11}{14}\right)\)
\(=\dfrac{7}{4}\cdot\dfrac{11\cdot35+33\cdot7+11\cdot14+11\cdot10}{140}\)
\(=\dfrac{880}{20\cdot4}=11\)
\(C=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right)\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-41}{21}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{-100}{41}}\)
\(=\dfrac{\dfrac{5751+187\cdot27}{108}}{\dfrac{-100}{41}}=100\cdot\dfrac{-41}{100}=-41\)
Tính nhanh : P= -7/4 x (33/12 + 3333/2020 + 333333/303030 + 33333333/42424242)
\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{33.101}{20.101}+\frac{33x10101}{30x10101}+\frac{33x1010101}{42x1010101}\right)\)
\(P=-\frac{7}{4}.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(P=-\frac{7}{4}.\left(\frac{11}{4}+\frac{33}{20}+\frac{11}{10}+\frac{11}{14}\right)\)
\(P=-\frac{77}{8}.\left(\frac{1}{6}+\frac{3}{10}+\frac{1}{15}+\frac{1}{21}\right)=-\frac{77}{8}.\frac{35+63+14+10}{210}=-\frac{11x122}{8x30}\)
\(P=-\frac{671}{120}\)
P = -7/4 x (33/12 + 3333/2020 + 333333/303030 + 33333333/42424242)
= -7/4 x (33/12 + 33/20 + 33/30 + 33/42)
= -7/4 x [33 x (1/12 + 1/20 + 1/30 + 1/42)]
= -7/4 x [33 x (35/420 + 21/420 + 14/420 + 10/420)]
= -7/4 x (33 x 80/420)
= -7/4 x 33 x 4/21
= -7/4 x 4/21 x 33 (= -7x4 / 4x21 x33)
= -7/21 x 33 (= -7x33 / 21 = -1x7x3x11 / 3x7)
= -11/1
= -11
Đáp số P = -11
Các số gạch đi là do rút gọn phân số nhé!!!
7/4.( 33/12 + 3333/2020 + 333333/303030 + 33333333/42424242) = ?
\(\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
=\(\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
=\(\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
=\(\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
=\(\frac{231}{4}.\frac{4}{21}\)
= 11
-7/4 .(33/12+3333/2020+333333/303030+33333333/42424242)
\(\dfrac{-7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33.101}{20.101}.\dfrac{33.10101}{30.10101}.\dfrac{33.1010101}{42.1010101}\right)\)
\(=\dfrac{-7}{4}.\left(\dfrac{33}{12}.\dfrac{33}{20}.\dfrac{33}{30}.\dfrac{33}{42}\right)\)
\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=\dfrac{-7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=\dfrac{-7}{4}.33.\dfrac{4}{21}\)
\(=33.\dfrac{-1}{3}\)
\(=11\)
xl nhé kết quả là -11 mk viết thiếu dấu -