Cô Hoài ơi, giúp con với: 3/13 : -11/-6 + -3/13 : 11/-5 - 2/13
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2.22 + 3.23 + 4.24 + ... + n.2n = 2n+11
Đặt vế trái là A ta có:
A = 2.22 + 3.23 + 4.24 + ... (n -1).2n-1+ n.2n
2A = 2.23 + 3.24 + 4.25 +....+ (n- 1).2n + n.2n+1
2A-A = [2.23+3.24 + 4.25 +...+(n-1).2n+n.2n+1] - [2.22 + 3.23+...+n.2n]
A = -2.22+ (2.33-3.23) + (3.24 - 4.24) +...+ [(n-1).2n - n.2n ] + n.2n+1
A = -2.22 - 23 - 24 -...- 2n + n.2n+1
Đặt B = -2.22 - 23 - 24 - ... - 2n
2B = -2.23 - 24 - 25 -...-2n+1
2B - B = (-2.23 - 24 - 25 -..-2n+1) - (-2.22-23-24-..-2n)
B = -24 -24 - 25 -..2n-2n+1 + 23 + 23 + 24+ 25+ 2n
B = (-24 + 23) + (- 2n+1 + 23) +(-24+24)+(-25+25)+(-2n+2n)
B = -16 + 8 - 2n+1 + 8
B = (-16 + 8 + 8 ) - 2n+1
B = - 2n+1
A = n.2n+1 - 2n+1
Theo bài ra ta có:
n.2n+1 - 2n+1 = 2n+11
n.2n+1 - 2n+1 - 2n+11 = 0
2n+1.(n - 1 - 210) = 0
Vì n là số tự nhiên nên 2n+1 > 0
Vậy 2n+1.(n - 1- 210) = 0 ⇔ n - 1 - 210 = 0 ⇒ n = 1 + 210 ⇒ n = 1025
Vậy n = 1025
B = (1 - \(\dfrac{1}{2^2}\)).(1 - \(\dfrac{1}{3^2}\)).(1 - \(\dfrac{1}{4^2}\))...(1 - \(\dfrac{1}{201^2}\))
B = \(\dfrac{2^2-1}{2^2}\).\(\dfrac{3^2-1}{3^2}\).\(\dfrac{4^2-1}{4^2}\)...\(\dfrac{201^2-1}{201^2}\)
B = \(\dfrac{4-1}{2^2}\).\(\dfrac{9-1}{3^2}\).\(\dfrac{16-1}{4^2}\)...\(\dfrac{40401-1}{201^2}\)
B = \(\dfrac{3}{2^2}\).\(\dfrac{8}{3^2}\).\(\dfrac{15}{4^2}\)....\(\dfrac{40400}{201^2}\)
B = \(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\).\(\dfrac{3.5}{4.4}\).\(\dfrac{4.6}{5.5}\)...\(\dfrac{200.202}{201.201}\)
B = \(\dfrac{1}{2}\).\(\dfrac{202}{201}\)
B = \(\dfrac{101}{201}\)
Lời giải:
a.
$\frac{8}{23}.\frac{46}{24}-\frac{2}{5}x=\frac{1}{3}$
$\frac{2}{3}-\frac{2}{5}x=\frac{1}{3}$
$\frac{2}{5}x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$
$x=\frac{1}{3}: \frac{2}{5}=\frac{5}{6}$
b.
$\frac{10}{12}: \frac{2}{3}x=\frac{28}{9}.\frac{3}{56}$
$\frac{5}{4}x=\frac{1}{6}$
$x=\frac{1}{6}: \frac{5}{4}=\frac{2}{15}$
c.
$\frac{x-1}{24}=\frac{2}{x+1}$
$(x-1)(x+1)=2.24$
$x^2-1=48$
$x^2=49=7^2=(-7)^2$
$\Rightarrow x=7$ hoặc $x=-7$
d.
$(\frac{3}{4}x+\frac{1}{4}-\frac{1}{3}): (2+\frac{1}{6}-\frac{1}{4})=\frac{7}{46}$
$(\frac{3}{4}x-\frac{1}{12}):\frac{23}{12}=\frac{7}{46}$
$\frac{3}{4}x-\frac{1}{12}=\frac{7}{46}.\frac{23}{12}=\frac{7}{24}$
$\frac{3}{4}x=\frac{7}{24}+\frac{1}{12}=\frac{3}{8}$
$x=\frac{3}{8}: \frac{3}{4}=\frac{1}{2}$
e.
$2\frac{1}{2}x+0,5x=2\frac{1}{4}$
$2,5x+0,5x=2,25$
$x(2,5+0,5)=2,25$
$3x=2,25$
$x=2,25:3=0,75$
f.
$\frac{1}{3}x+\frac{2}{5}(x-1)=0$
$\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0$
$x(\frac{1}{3}+\frac{2}{5})=\frac{2}{5}$
$x.\frac{11}{15}=\frac{2}{5}$
$x=\frac{2}{5}: \frac{11}{15}=\frac{6}{11}$
g.
$x-3\frac{1}{2}x=-2\frac{6}{7}$
$x(1-3\frac{1}{2})=\frac{-20}{7}$
$x.\frac{-5}{2}=\frac{-20}{7}$
$x=\frac{-20}{7}: \frac{-5}{2}=\frac{8}{7}$
h.
$2(\frac{1}{2}x-\frac{1}{3})-\frac{3}{2}=\frac{1}{4}$
$2(\frac{1}{2}x-\frac{1}{3})=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}$
$\frac{1}{2}x-\frac{1}{3}=\frac{7}{4}:2=\frac{7}{8}$
$\frac{1}{2}x=\frac{7}{8}+\frac{1}{3}=\frac{29}{24}$
$x=\frac{29}{24}: \frac{1}{2}=\frac{29}{12}$
i.
$-2\frac{1}{3}x+1\frac{3}{4}x+3\frac{2}{3}=3\frac{1}{2}$
$x(-2\frac{1}{3}+1\frac{3}{4})=3\frac{1}{2}-3\frac{2}{3}$
$x.\frac{-7}{12}=\frac{-1}{6}$
$x=\frac{-1}{6}: \frac{-7}{12}=\frac{2}{7}$
\(\left(\dfrac{1997}{2024}+\dfrac{2021}{1999}\right)-\left(\dfrac{1997}{2024}+\dfrac{22}{1999}\right)\)
\(=\dfrac{1997}{2024}+\dfrac{2021}{1999}-\dfrac{1997}{2024}-\dfrac{22}{1999}\)
\(=\dfrac{2021}{1999}-\dfrac{22}{1999}=\dfrac{1999}{1999}=1\)
Số tiền phải trả cho cái thứ nhất là \(180000\left(1-30\%\right)=126000\left(đồng\right)\)
Số tiền phải trả cho cái thứ hai là:
\(126000\left(1-5\%\right)=119700\left(đồng\right)\)
Tổng số tiền phải trả là:
\(126000+119700=245700\simeq246000\left(đồng\right)\)
\(\dfrac{1}{2}\cdot\dfrac{5}{6}+\dfrac{2}{3}\cdot75\%\)
\(=\dfrac{1}{2}\cdot\dfrac{5}{6}+\dfrac{2}{3}\cdot\dfrac{3}{4}\)
\(=\dfrac{1\cdot5+2\cdot3}{12}=\dfrac{5+6}{12}=\dfrac{11}{12}\)
\(\dfrac{3}{13}:\dfrac{-11}{-6}+\dfrac{-3}{13}:\dfrac{11}{-5}-\dfrac{2}{13}\)
\(=\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{5}{11}-\dfrac{2}{13}\)
\(=\dfrac{3}{13}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{2}{13}=\dfrac{3}{13}-\dfrac{2}{13}=\dfrac{1}{13}\)
Đây là lớp 6 á