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Câu 1:
a: x+2,57=14,25-6,3
=>x+2,57=7,95
=>x=5,38
b: 12,9-x=4,2+2,4
=>12,9-x=6,6
hay x=6,3
7 x 17 x 125 = (8 x 125) x 17 = 1000 x 17 = 17000
4 x 37 x 25 = (25 x 4 ) x 7 = 100 x 7 = 700
43 x 27 + 93 x 43 + 57 x 81 + 69 x 57 = 43 x (27 + 93) + 57 x (81 +69) = 43 x 120 + 57 x 120 + 57x30 = 120 x (43 + 57) + 1710 = 12000 + 1710 = 13710
\(e,112-45+5x=87\)
\(67+5x=87\)
\(5x=20\)
\(x=4\)
\(f,6^2+64:\left(x-1\right)=52\)
\(36+64:\left(x+1\right)=52\)
\(64:\left(x+1\right)=16\)
\(x+1=4\)
\(x=3\)
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
\(a,\) \(5.\left(4+6x\right)=290\)
\(4+6x=290:5\)
\(4+6x=58\)
\(6x=58-4\)
\(6x=54\)
\(x=54:6\)
\(x=9\left(tm\right)\)
Vậy ..................
\(b,x.3,7+x.6,3=120\)
\(x\left(3,7+6,3\right)=120\)
\(10x=120\)
\(x=120:10\)
\(x=12\left(tm\right)\)
Vậy ....................
c) \(\left(15.24-x\right):0,25=100:\dfrac{1}{4}\)
\(\left(360-x\right):\dfrac{1}{4}=400\)
\(360-x=400.\dfrac{1}{4}\)
\(360-x=100\)
\(x=360-100\)
\(x=260\left(tm\right)\)
Vậy ...................
a) 5. ( 4+6 . x ) = 290
( 4+6 . x ) = 290 : 5
4 + 6 . x = 58
6 .x = 58 - 4
6.x = 54
x = 54 : 6
x = 9
x + 2,57 = 14,25 -6,3
x + 2,57 = 7,95
x = 7,95 - 2,57
x = 5,38
b, 12,9-x = 4,2 +2,4
12,9-x = 6,6
x = 12,9 - 6,6
x= 6,3
c, x . 93 + x . 0,7 = 100
x . (93 + 0,7 ) = 100
x . 93,7 = 100
x= 100 : 93,7
x= 1000/ 937
d, \(\dfrac{x}{5}=\dfrac{4}{2}\) = \(\dfrac{x}{5}=2\)
x : 5 = 2
x= 2.5
x= 10
a) x + 2,57 = 14,25 - 6,3 ; b) 12,9 - x = 4,2 + 2,4
x = 7,95 - 2,57 x = 12,9 - 6,6
x = 5,38 x = 6,3
c) 93x + 0,7x = 100 ; d) x/5 = 4/2
(93 + 0,7)x = 100 x = 2 x 5
x = 100/93,7 x = 10