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a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)
\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)
\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1\)
\(=-\dfrac{3}{2}\)
b) \(8.\sqrt{9}-\sqrt{64}\)
\(=8.3-8\)
\(=24-8\)
\(=16\)
c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)
\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)
\(=-1+\dfrac{5}{2}\)
\(=\dfrac{3}{2}\)
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-1\)
Ta đặt: \(P=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
=> \(2P=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
=> \(2P-P=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
=> \(P=1-\frac{1}{64}\)
Mà \(S=P-1\)
=> \(S=1-\frac{1}{64}-1=-\frac{1}{64}\)
Vậy \(S=-\frac{1}{64}\)
N = 3/2 + 5/4 + 9/8 + 17/16 + 33/32 + 65/64 - 7
N = (1 + 1/2) + (1 + 1/4) + (1 + 1/8) + (1 + 1/16) + (1 + 1/32) + (1 + 1/64) - 7
N = (1 + 1 + 1 + 1 + 1 + 1 ) + (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) - 7
N = 6 - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) - 7
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
A = 1 - 1/64
N = 6 - (1 - 1/64) - 7
N = 6 - 1 + 1/64 - 7
N = 5 + 1/64 - 7
N = -2 + 1/64
N = -128/64 + 1/64
N = -127/64
bạn soyen_Tiểu bàng giải sao dòng 3 đang cộng đến dòng 9 lại chuyển thành trừ vậy
a
(4x+7)2 = 25/36
(4x+7)2 = (5/6)2
==> 4x+7=5/6
4x=5/6-7=-37/6
x=-37/6:4=-37/24
b
(5+2x)3 = 8
(5+2x)3 = 23
==> 5+2x=2
2x=2-5=-3
x=-3:2=-1,5
c
3x.9=243
3x.32=35
3x=35:32
3x=33
3x=27 ==> x=27:3=9
d
42x = (-64).(-4)
42x = 43.41 = 44
==> 2x = 4
x = 4:2=2
Tick giúp mình nhe ahihi ^_^
Ta có quy luật:
1+1=12 = 1
2+2=22 = 4
3+3 = 32 = 9
..................
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9 + 9 = 92 = 81
a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)
b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)
\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)
\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)
c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)
d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)
\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)
\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)
e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
b: \(\left(17x-25\right):8+65=9^5:9^3\)
\(\Leftrightarrow\left(17x-25\right):8=9^2-65=81-65=16\)
\(\Leftrightarrow17x-25=128\)
hay x=9
b: (17x−25):8+65=95:93(17x−25):8+65=95:93
⇔(17x−25):8=92−65=81−65=16⇔(17x−25):8=92−65=81−65=16
⇔17x−25=128⇔17x−25=128
hay x=9