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a) x + 40*25 = 2000
x + 1000 = 2000
x = 2000 - 1000
x = 1000.
b) (x + 40)*25 = 2000
x + 40 = 2000 : 25
x + 40 = 80
x = 80 - 40
x = 40.
c) (x - 10)*5 = 100 - 20*4
(x - 10)*5 = 100 - 80
(x - 10)*5 = 20
x - 10 = 20 : 5
x - 10 = 4
x = 4 + 10
x = 14.
d) Các số hạng x + 2; x + 4; ... ; x + 1996 lập thành một dãy số cách đều với khoảng cách bằng 2.
Từ x + 2 đến x + 1996 có:
(1996 - 2) : 2 + 1 = 998 số hạng.
Tổng các số hạng ở vế trái là:
(x + 2) + (x + 4) + ... + (x + 1996) = (x + 1996 + x + 2)*998 : 2 = (2*x + 1998)*998 : 2
Vậy ta có:
(2*x + 1998)*998 : 2 = 998000
(2*x + 1998)*998 = 998000*2
2*x + 1998 = 998000*2 : 998
2*x + 1998 = 2000
2*x = 2000 - 1998
2*x = 2
x = 2 : 2
x = 1.
ủng hộ nha
Ta có: a) \(x+40.25=2000\)
\(\Rightarrow x+1000=2000\)
\(\Rightarrow x=2000-1000=1000\)
b) \(\left(x+40\right).25=2000\)
\(\Rightarrow x+40=2000:25=80\)
\(\Rightarrow x=80-40=40\)
c) \(\left(x-10\right).5=100-20.4\)
\(\Rightarrow\left(x-10\right).5=100-80\)
\(\Rightarrow\left(x-10\right).5=20\)
\(\Rightarrow x-10=20:5=4\)
\(\Rightarrow x=10+4=14\)
d) \(\left(x+2\right)+\left(x+4\right)+....+\left(x+1996\right)=998000\)
\(\Rightarrow\left(x+x+...+x\right)+\left(2+4+..+1996\right)=998000\)
\(\Rightarrow998x+997002=998000\)
\(\Rightarrow998x=998000-996002=998\)
\(\Rightarrow x=998:998=1\)
Ủng hộ nha m.n ^_^
a) x + 40*25 = 2000
x + 1000 = 2000
x = 2000 - 1000
x = 1000.
b) (x + 40)*25 = 2000
x + 40 = 2000 : 25
x + 40 = 80
x = 80 - 40
x = 40.
c) (x - 10)*5 = 100 - 20*4
(x - 10)*5 = 100 - 80
(x - 10)*5 = 20
x - 10 = 20 : 5
x - 10 = 4
x = 4 + 10
x = 14.
d) Các số hạng x + 2; x + 4; ... ; x + 1996 lập thành một dãy số cách đều với khoảng cách bằng 2.
Từ x + 2 đến x + 1996 có:
(1996 - 2) : 2 + 1 = 998 số hạng.
Tổng các số hạng ở vế trái là:
(x + 2) + (x + 4) + ... + (x + 1996) = (x + 1996 + x + 2)*998 : 2 = (2*x + 1998)*998 : 2
Vậy ta có:
(2*x + 1998)*998 : 2 = 998000
(2*x + 1998)*998 = 998000*2
2*x + 1998 = 998000*2 : 998
2*x + 1998 = 2000
2*x = 2000 - 1998
2*x = 2
x = 2 : 2
x = 1.
a) x + 40 x 25 = 2000
x + 1000 = 2000
x = 2000 - 1000
x = 1000
b) (\(x\) + 40) x 25 = 2000
x + 40 = 2000 : 25
x + 40 = 80
x = 80 - 40
x = 40
c) (x - 10) x 5 = 100 - 20 x 4
(x - 10) x 5 = 100 - 80
(x - 10) x 5 = 20
x - 10 = 20 : 5
x - 10 = 4
x = 4 + 10
x = 14.
d) Các số hạng x + 2; x + 4; ... ; x + 1996 lập thành một dãy số cách đều với khoảng cách bằng 2.
Từ \(x\) + 2 đến \(x\) + 1996 có:
(1996 - 2) : 2 + 1 = 998 số hạng.
Tổng các số hạng ở vế trái là:
(\(x\) + 2) + (\(x\) + 4) + ... + (\(x\) + 1996) = (\(x\) + 1996 + \(x\) + 2) x 998 : 2 = (2 x \(x\) + 1998) x 998 : 2
Vậy ta có:
(2 x \(x\) + 1998) x 998 : 2 = 998000
(2 x \(x\) + 1998) x 998 = 998000x 2
2 x \(x\) + 1998 = 998000 x 2 : 998
2 x \(x\) + 1998 = 2000
2 x \(x\) = 2000 - 1998
2 x \(x\) = 2
\(x\) = 2 : 2
x = 1
3 x 15 + 21 x 15 + 85 x 5
= 45 + 315 + 425
= 785
15 - 30 + 40
= 25
21 + 19 - 50 + 10
= 0
\(\dfrac{1}{5}-\dfrac{1}{4}+2\)
\(=-\dfrac{1}{20}+2\)
\(=\dfrac{39}{20}\)
\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{12}\times\dfrac{1}{4}\)
\(=\dfrac{5}{12}\times\dfrac{3}{12}\)
\(=\dfrac{5}{48}\)
\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)
\(=-\dfrac{9}{20}\)
\(3\times15+21\times15+85\times5\\ =15\times\left(3+21\right)+425\\ =15\times24+425\\ =360+425\\ =785\)
\(15-30+40\\ =\left(15+40\right)-30\\ =55-30\\ =25\)
\(21+19-50+10\\ =\left(21+19\right)-\left(50-10\right)\\ =40-40\\ =0\)
\(\dfrac{1}{5}-\dfrac{1}{4}+2\)
\(=\dfrac{4}{20}-\dfrac{5}{20}+\dfrac{40}{20}\)
\(=\dfrac{\left(4+40\right)}{20}-\dfrac{5}{20}\)
\(=\dfrac{44}{20}-\dfrac{5}{20}\)
\(=\dfrac{39}{20}\)
\(\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{12}\times\dfrac{1}{4}\)
\(=\dfrac{5}{48}\)
\(\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{3}{4}\)
\(=\dfrac{2}{20}+\dfrac{4}{20}-\dfrac{15}{20}\)
\(=\dfrac{6}{20}-\dfrac{15}{20}\)
\(=-\dfrac{9}{20}\)
đơn giản
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{39}{40}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{80}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}\)
\(\frac{1}{x+1}=\frac{1}{80}\)
\(\Rightarrow x+1=80\)
\(\Rightarrow x=80-1=79\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}\div2=\frac{39}{80}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}=\frac{1}{80}\)
\(\Leftrightarrow x+1=80\Rightarrow x=80-1=79\)
Vậy \(x=79\)
X : 0,125 + X : 0,5 + X x 10 = 40
x : ( 0,125 + 0,5 + 10 ) = 40
x : 10,625 = 40
x = 40 x 10,625
x = 425
`@` `\text {Ans}`
`\downarrow`
`1,`
`a)`
`5 \times 72 \times 10 \times 2`
`= 5 \times 2 \times 10 \times 72`
`= 10 \times 10 \times 72`
`= 100 \times 72`
`= 7200`
`b)`
`40 \times 125`
`= 4 \times 10 \times 25 \times 5`
`= (5 \times 10) \times (4 \times 25)`
`= 50 \times 100`
`= 5000`
`c)`
`4 \times 2021 \times 25`
`= (4 \times 25) \times 2021`
`= 100 \times 2021`
`= 202100`
`d)`
`16 \times 6 \times 25`
`= 4 \times 4 \times 6 \times 25`
`= (4 \times 25) \times 4 \times 6`
`= 100 \times 24`
`= 2400`
`2,`
`a)`
`24 \times 57 + 43 \times 24`
`= 24 \times (57+43)`
`= 24 \times 100`
`= 2400`
`b)`
`12 \times 19 + 80 \times 12 +12`
`= 12 \times (19 + 80 + 1)`
`= 12 \times 100`
`= 1200`
`c)`
`(36 \times 15 \times 169) \div (5 \times 18 \times 13)`
`= 36 \times 15 \times 169 \div 5 \div 18 \div 13`
`= 6 \times 6 \times 3 \times 5 \times 13 \times 13 \div 5 \div 3 \times 6 \div 13`
`= (6 \div 6) \times (3 \div 3) \times (5 \div 5) \times (13 \div 13) \times 6 \times 13`
`= 6 \times 13`
`= 78`
`d)`
`(44 \times 52 \times 60) \div ( 11 \times 13 \times 15)`
`= 44 \times 52 \times 60 \div 11 \div 13 \div 15`
`= 4 \times 11 \times 13 \times 4 \times 15 \times 4 \div 11 \div 13 \div 15`
`= (11 \div 11) \times (13 \div 13) \times (15 \div 15) \times 4 \times 4 \times`
`= 4 \times 4 \times 4`
`= 64`
`3,`
`a)`
`x - 280 \div 35 = 5 \times 54`
`x - 8 = 270`
`x = 270 + 8`
`x = 278`
`b)`
`(x - 280) \div 35 = 54 \div 4`
`(x - 280) \div 35 = 13,5`
`x - 280 = 13,5 \times 35`
`x - 280 = 472,5`
`x = 472,5 + 280`
`x = 752,5`
`c)`
`(x - 128 + 20) \div 192 = 0`
`x - 128 + 20 = 0 \times 192`
`x - 128 + 20 = 0`
`x - 108 = 0`
`x = 0 + 108`
`x = 108`
`d)`
`4 \times (x + 200) = 460 + 85 \times 4`
`4 \times (x+200) = 460 + 340`
`4 \times (x+200) = 800`
`x + 200 = 800 \div 4`
`x + 200 = 200`
`x = 200 - 200`
`x = 0`
`4,`
`a)`
`7/12 - 5/12`
`= (7 - 5)/12`
`= 2/12`
`= 1/6`
`b)`
`8/11 + 19/11`
`= (8+19)/11`
`= 27/11`
`c)`
`3/8 + 5/12`
`= 9/24 + 10/24`
`= 19/24`
`d)`
`3/4 + 7/12`
`= 9/12 + 7/12`
`= 16/12`
`= 4/3`
`5,`
`a)`
`x - 6/7 = 5/2`
`x = 5/2 + 6/7`
`x = 47/14`
`b)`
`12/7 \div x + 2/3 = 7/5`
`12/7 \div x = 7/5 - 2/3`
`12/7 \div x = 11/15`
`x = 12/7 \div 11/15`
`x = 180/77`
`@` `\text {Kaizuu lv uuu}`
x+10=80
x=80-10
x=70
x + 10 = 40 x 2
x + 10 = 80
x = 80 - 10
x = 70