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1.
`16 + (27 - 7.6 ) - (94 -7 - 27.99)`
`= 16+ 27 - 7.6 - 94 + 7 + 27.99`
`= 16 + 27(99 +1) - 7(6-1) - 94`
`= -78 + 27.100 - 7.5`
`= 2587`
2.
`A = 2/1.4 + 2/4.7 + 2/7.10 +...+ 2/97.100`
`A= 2(1/1.4 + 1/4.7 + 1/7.10 +...+1/97.100)`
`3A = 2 (3/1.4 + 3/4.7 + 3/7.10+...+ 3/97.100)`
`3/2 A = 1 - 1/4 + 1/4 - 1/7 +...+ 1/97 - 1/100`
`3/2A = 1 - 1/100`
`3/2 A= 99/100`
`A= 99/100 : 3/2`
`A=33/50`
Vậy `A= 33/50`
1.16+(27-7.6)-(94-7-27.99)=16+27-7.6-94+7+27.99
=(27+27.99)+(27+7-94)+16
=27.100-60+16
=2700-44=2656
2.A=\(\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{97.100}\)
=\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
a) \(16+\left(27-7\cdot6\right)-\left(94\cdot7-27\cdot99\right)\)
\(=16+27-7\cdot6-94\cdot7+27\cdot99\)
\(=16+27\left(1+99\right)-7\left(6+94\right)=16+2700-700=2016\)
b)\(A=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(=\frac{1}{3}\left(\frac{2}{1}-\frac{2}{4}+\frac{2}{4}-\frac{2}{7}+\frac{2}{7}-\frac{2}{10}+...+\frac{2}{97}-\frac{2}{100}\right)\)
\(=\frac{1}{3}\left(2-\frac{2}{100}\right)=\frac{1}{3}\cdot\frac{99}{50}=\frac{33}{50}\)
\(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{97.100}\)
\(A=2\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)\)
\(A=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=2\left(1-\frac{1}{100}\right)\)
\(A=2.\frac{99}{100}=..............\)
Tự làm nốt nha
Ta có : (12+13+14+....+110)×x=19+28+....+82+91(12+13+14+....+110)×x=19+28+....+82+91
Đặt A=19+28+....+82+91A=19+28+....+82+91
=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)
=1+102+103+...+109=102+103+....+109+1010=1+102+103+...+109=102+103+....+109+1010
=(12+13+14+....+110)×10=(12+13+14+....+110)×10 (1)
Từ (1), suy ra:
(12+13+14+...+110)×x=(12+13+14+....+110)×10(12+13+14+...+110)×x=(12+13+14+....+110)×10
⇒x=10⇒x=10
Vậy x=10x=10
~ Học tốt ~
Ta có : (12+13+14+....+110)×x=19+28+....+82+91(12+13+14+....+110)×x=19+28+....+82+91
Đặt A=19+28+....+82+91A=19+28+....+82+91
=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)=(9−1−1−....−1)+(82+1)+(73+1)+.....+(19+1)
=1+102+103+...+109=102+103+....+109+1010=1+102+103+...+109=102+103+....+109+1010
=(12+13+14+....+110)×10=(12+13+14+....+110)×10 (1)
Từ (1), suy ra:
(12+13+14+...+110)×x=(12+13+14+....+110)×10(12+13+14+...+110)×x=(12+13+14+....+110)×10
⇒x=10⇒x=10
Vậy x=10x=10
~ Học tốt ~
x/1.4+x/4.7+x/7.10+x/10.13+x/13.16=5/2
=>x/3(1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)=5/2
=>x/3.(1/4-1/16)=5/2
=>x/3.3/16=5/2
=>x/3=5/2:3/16
=>x/3=40/3
=>x=40
Vậy x=40
x/1.4 + x/4.7 + x/7.10 + x/10.13 + x/13.16 = 5/6
=> x.1/3.(3/1.4 + 3/4.7 + 3/7.10 + 3/10.13 + 3/13.16) = 5/6
=> x.1/3.(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16) = 5/6
=> x.1/3.(1 - 1/16) = 5/6
=> x.1/3.15/16 = 5/6
=> x.1/3 = 5/6 : 15/16
=> x.1/3 = 8/9
=> x = 8/9 : 1/3
=> x = 8/3
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{\left(3x-2\right)\left(3x+1\right)}=\frac{670}{2011}\)
\(\Rightarrow\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(3x-2\right)\left(3x+1\right)}\right)=\frac{670}{2011}\)
\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{3x-2}-\frac{1}{3x+1}=\frac{670}{2011}:\frac{1}{3}\)
\(\Rightarrow1-\frac{1}{3x+1}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{3x+1}=1-\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{3x+1}=\frac{1}{2011}\)
=>3x+1=2011
=>3x=2011-1
=>x=2010:3
=>x=670
vậy x=670
Dặt \(A=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{\left(3x-2\right).\left(3x+1\right)}\)
\(3A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{\left(3x-2\right)\left(3x+1\right)}\)
\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{\left(3x-2\right)}-\frac{1}{\left(3x+1\right)}\)
\(3A=1-\frac{1}{3x+1}\)
\(A=\left(1-\frac{1}{3x+1}\right):3=\frac{670}{2011}\)
\(1-\frac{1}{3x+1}=\frac{670}{2011}.3\)
\(1-\frac{1}{3x+1}=\frac{2010}{2011}\)
\(\frac{1}{3x+1}=1-\frac{2010}{2011}\)suy ra \(\frac{1}{3x+1}=\frac{1}{2011}\)
suy ra 3x+1=2011
3x=2000
x=2000/3
\(\Leftrightarrow2^x+16=16\Leftrightarrow2^x=0\\ \)
=> Không có giá trị nào của x để 2^x =0
Vậy pt vô nghiệm
nhé