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a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
a, \(120\cdot15+12\cdot470+12\cdot38\cdot10\)
\(=12\cdot10\cdot15+12\cdot47\cdot10+12\cdot38\cdot10\)
\(=120\left(15+47+38\right)=120\cdot100=12000\)
f,\(\left(2+4+6+...+100\right)\cdot\left(36\cdot333-111\cdot108\right)\)
\(=\left(2+4+6+...+100\right)\cdot\left(36\cdot111\cdot3-111\cdot108\right)\)
\(=\left(2+4+6+...+100\right)\cdot111\cdot\left(36\cdot3-108\right)\)
\(=\left(2+4+6+...+100\right)\cdot111\cdot\left(108-108\right)\)
\(=\left(2+4+6+...+100\right)\cdot111\cdot0=0\)
a)
120.15+12.470+12.38.10
=120.15+120.47+120.38=120(15+47+38)=120.100=12000
b)
13.58.4+32.26.2+52.10
=52.58+52.26+52.10=52.(58+26+10)=52.100=5200
c)
3,2.990+32=32.99+32=32.(99+1)=32.100=3200
d)
45.(13+78)+9.(87+22).5
=45.91+45.109=45.(91+109)=45.200=9000
e)
(24+72+36+60):12=12.(2+6+3+5):12=16
f)(2+4+6+...+100)(36.333−108.111)
=(2+4+6+...+100)(36.3.111−36.3.111)=(2+4+6+...+100).0=0
1,
a,-3/5
b,-1/2
c,19/39
d,1/4
e,-39/40
f,-59/56
2,
a,=
b,<
c,>
d,<
k cho mình nha
Ta thấy:
\(2^2=2.2>1.2\Rightarrow\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(3^2=3.3>2.3\Rightarrow\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.................
\(9^2=9.9>8.9\Rightarrow\dfrac{1}{9^2}< \dfrac{1}{8.9}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)
=> Đpcm
Ta thấy:
22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2
32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3
.................
92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9
⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9
⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89
=> ...(tự viết)
Ta thấy:
22=2.2>1.2⇒122<11.222=2.2>1.2⇒122<11.2
32=3.3>2.3⇒132<12.332=3.3>2.3⇒132<12.3
.................
92=9.9>8.9⇒192<18.992=9.9>8.9⇒192<18.9
⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9⇒122+132+142+...+192<11.2+12.3+13.4+...+18.9
⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89⇔122+132+142+...+192>1−12+12−13+13−14+...+18−19=1−19=89
=> 11111111111111111111110101010110000
HACK
Tính các tổng sau:
1, S=1-2+3_4+..+25-26
S =-1+3-5+7-...-53+55 ( có 28 số hạng )
= (-1+3)+(-5+7)+...+(-53+55) ( có 28:2=14 nhóm )
= 2+2+...+2
= 2 . 14
= 28
Giải:
A=1/22+1/32+1/42+...+1/92
Ta có:
1/22<1/1.2
1/32<1/2.3
1/42<1/3.4
...
1/92<1/8.9
⇒A<1/1.2+1/2.3+1/3.4+...+1/8.9
A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
A<1/1-1/9
A<8/9
Ta có:
1/22>1/2.3
1/32>1/3.4
1/42>1/4.5
...
1/92>1/9.10
⇒A>1/2.3+1/3.4+1/4.5+...+1/9.10
A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
A>1/2-1/10
A>2/5
Vậy 2/5<A<8/9 (đpcm)
Chúc bạn học tốt!
5.42 – 18 : 32
= 5.16 – 18 : 9 = 80 – 2 = 78
Phân tích ra thừa số nguyên tố: 78 = 2.39 = 2.3.13
mk bấm nhầm phép đè bài đây giải guips mk nha
43.32 và 152.183.72