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`@` `\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}`
a) \(15-5\left|x+4\right|=-12-3\)
\(\Leftrightarrow5\left|x+4\right|=30\)
\(\Leftrightarrow\left|x+4\right|=6\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)
b) \(\left(4x-8\right)\left(7-x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-8=0\\7-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}\)
c) \(\left(x^2-36\right)\left(x^2+5\right)=0\Rightarrow\left(x-6\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
d) \(-3\left(x+7\right)-11=2\left(x+5\right)\)
\(\Leftrightarrow-3x-32=2x+10\)
\(\Leftrightarrow5x=-42\Rightarrow x=-\frac{42}{5}\)
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5
Bài 1:
a) 2/19 + 2/10 + 2/22 + 17/19 + 2/11 + 4/5 + 8/11
=(2/19 +17/19) + 1/5 + 1/11 + 2/11 + 4/5 + 8/11
= 1 + (1/5 + 4/5) + (2/11 + 8/11 + 1/11)
= 1 + 1 + 1 = 3
b) 3/9 + 4/12 + 6/18 + 1/3 + 5/15 + 7/21
= 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3
= 1/3 x 6 = 2
c) 100 + (125x3-125x2-125) x (1 + 3 + 5 + 7 + ...+ 97 + 99)
= 100 + [125x(3-2-1)] x A
= 100 + (125x0) x A
= 100 + 0 x A
= 100 + 0
= 100
Bài 2:
Gọi số đó là ab
(a+b) x 6 = ab
a x 6 + b x 6= a x 10 + b
b x 5 = a x 4
suy ra a=5; b=4; ab=54
Bài 3:
Vì các số lẻ x 5 đều có tận cùng là 5 nên các tích đều có tận cùng là 5.
Mà 5x3=15 nên P có tận cùng là 5
Bài 1:
a) 43+45+47+...+565 [có (565-43)/2+1= 262 số hạng)
= [(565+43)*262]/2
= 79648
b) 21+24+27+...+318 [có (318-21)/3+1= 100 số hạng]
= [(318+21)*100]/2
= 16950
- Tính giá trị biểu thức:
a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145
b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7
c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11
- Tính giá trị biểu thức:
a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21
b) (5/17 x 21/32 x 47/24 x 0)
= 0
c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7
- Tìm x:
a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x
b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15
c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2
- Tìm x:
a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36
câu,b,không,đủ,thông,tin,nhan,bạn.
1a)6,28x18,24+18,24x3,72 d)0,9x95+1,8x2+0,9
=18,24x(6,28+3,72) = 0,9x95+0,9x4+0,9
=18,24x10 =0,9(95+4+1)
=182,4 =0,9x100=90
b) 35,7x99+35+0,7 e)0,25x611,7x40
=35,7x99+35,7 =(0,25x40)x611,7
=35,7x(99+1) =10x611,7
=35,7x100 =6117
=357 g)37,2x101-37-0,2
c)17,34x99+18-0,66 = 101x37,2-(37+0,2)
=17,34x99+17,34 =101x37,2-37,2
=17,34x(99+1) =37,2x(101-1)
=17,34x100 =37,2x100
=1734 =3720
Bạn ơi đăng từng bài thôi mk giải cho. Nhiều quá nhìn hoa mắt mặc dù bài rất dễ!!!
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
Bài 1
a) \(5\times72\times10\times2=\left(5\times2\times10\right)\times72=100\times72=7200\)
b) \(40\times125=5\times\left(8\times125\right)=5\times1000=5000\)
c) \(16\times6\times25=4\times4\times6\times25=\left(4\times6\right)\times\left(4\times25\right)=24\times100=2400\) Bài 2:
a) \(24\times57+43\times24=24\times\left(57+43\right)=24\times100=2400\)
b) \(12\times19+80\times12+12=12\times\left(19+80+1\right)=12\times100=1200\)
c) \(\left(36\times15\times169\right)\div\left(5\times18\times13\right)\)
\(=\left(18\times2\times3\times5\times13\times13\right)\div\left(5\times18\times13\right)\)
\(=\left(2\times3\times13\right)\times\left(18\times5\times13\right)\div\left(5\times18\times13\right)\)
\(=2\times3\times13\)
\(=78\)
d) \(\left(44\times52\times60\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times11\times4\times13\times4\times15\right)\div\left(11\times13\times15\right)\)
\(=\left(4\times4\times4\right)\times\left(11\times13\times15\right)\div\left(11\times13\times15\right)\)
\(=4\times4\times4\)
\(=64\)
Bài 3:
a) \(x-280\div35=5\times54\)
\(x-8=270\)
\(x=270+8\)
\(x=278\)
b) \(\left(x-280\right)\div35=54\div4\)
\(\left(x-280\right)\div35=\dfrac{27}{2}\)
\(x-280=\dfrac{27}{2}\times35\)
\(x-280=\dfrac{945}{2}\)
\(x=\dfrac{945}{2}+280\)
\(x=\dfrac{1505}{2}\)
c) \(\left(x-128+20\right)\div192=0\)
\(x-128+20=0\times192\)
\(x-128+20=0\)
\(x-128=0-20\)
\(x-128=-20\)
\(x=-20+128\)
\(x=108\)
d) \(4\times\left(x+200\right)=460+85\times4\)
\(4\times\left(x+200\right)=460+340\)
\(4\times\left(x+200\right)=800\)
\(x+200=800\div4\)
\(x+200=200\)
\(x=200-200\)
\(x=0\)
Bài 4:
a) \(\dfrac{7}{12}-\dfrac{5}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)
b) \(\dfrac{8}{11}+\dfrac{19}{11}=\dfrac{27}{11}\)
c) \(\dfrac{3}{8}+\dfrac{5}{12}=\dfrac{9}{24}+\dfrac{10}{24}=\dfrac{19}{24}\)
d) \(\dfrac{3}{4}+\dfrac{7}{12}=\dfrac{9}{12}+\dfrac{7}{12}=\dfrac{16}{12}=\dfrac{4}{3}\)
Bài 5:
a) \(x-\dfrac{6}{7}=\dfrac{5}{2}\)
\(x=\dfrac{5}{2}+\dfrac{6}{7}\)
\(x=\dfrac{47}{14}\)
b) \(\dfrac{12}{7}\div x+\dfrac{2}{3}=\dfrac{7}{5}\)
\(\dfrac{12}{7}\div x=\dfrac{7}{5}-\dfrac{2}{3}\)
\(\dfrac{12}{7}\div x=\dfrac{11}{15}\)
\(x=\dfrac{12}{7}\div\dfrac{11}{15}\)
\(x=\dfrac{180}{77}\)