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\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)
Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
Ở đây mình dùng tính chất phân phối tách :
x-3×(x+1)thành
x-3x-3×1 nhưng thầy giáo bạn viết gọn 3×1 thành 3 vì 3×1 vẫn bằng chính nó
a) \(3^x-2=5^2\)
\(\Rightarrow3^x-2=25\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
b) \(\left(x+1\right)^2=36\)
\(\Rightarrow\left(x+1\right)^2=6^2\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5:\left(2x-15\right)^3=1\)
\(\Rightarrow\left(2x-15\right)^2=1\)
\(\Rightarrow2x-15=1\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=16:2=8\)
Chúc em học tốt nhé!
a) \(3^x-2=5^2\)
\(\Rightarrow3^x-2=25\)
\(\Rightarrow3^x=27\)
\(\Rightarrow3^x=3^3\)
\(\Rightarrow x=3\)
b) \(\left(x+1\right)^2=36\)
\(\Rightarrow\left(x+1\right)^2=6^2\)
\(\Rightarrow x+1=6\)
\(\Rightarrow x=5\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5:\left(2x-15\right)^3=1\)
\(\Rightarrow\left(2x-15\right)^2=1\)
\(\Rightarrow\left(2x-15\right)^2=1^2\)
\(\Rightarrow2x-15=1\)
\(\Rightarrow2x=16\)
\(\Rightarrow x=8\)
Chúc em học tốt nhé!
Bài 1:tính
a)65.(-19)+19.(-35)
=65.(-19)+(-19).35
=(-19).(65+35)
=(-19).100
=-1900
b)85.(35-27)-35.(85-27)
=85.35-85.27-35.85+35.27
=(85.35-35.85)+(-85.27+35.27)
=27.(-85+35)
=27.(-50)
=1350
c)47.(45-15)-47.(45+15)
=47.[(45-15)-(45+15)]
=47.[30-60]
=47.(-30)
=-1410
Bài2: Tìm các số nguyên x biết
a)(-2).(x+6)+6.(x-10)=8
-2x-12+6x-60=8
4x-72=8
4x=72+8
4x=50
x=\(\frac{25}{2}\)
b)(-4).(2x+9)-(-8x+3)-(x+13)=0
-6x-36+8x-3-x-13=0
x-41=0
x=41
Bài 1: Tính
a) \(65.\left(-19\right)+19.\left(-35\right)\)
= \(-1235+-665\)
= \(-1900\)
b) \(85.\left(35-27\right)-35.\left(85-27\right)\)
= \(-1350\)
c) \(47.\left(45-15\right)-47.\left(45+15\right)\)
=\(-1410\)
Bài 2: Tìm x:
\(\left(-2\right).\left(x+6\right)+6.\left(x-10\right)=8\)
\(x=20\)
Bài giải chi tiết đây em nhé:
\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\)+...+ \(\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\)(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+ \(\dfrac{2}{7.9}\)+...+ \(\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) +... + \(\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\) ( 1 - \(\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)
1 - \(\dfrac{1}{2x+1}\) = \(\dfrac{9}{19}\) : \(\dfrac{1}{2}\)
1 - \(\dfrac{1}{2x+1}\) = \(\dfrac{18}{19}\)
\(\dfrac{1}{2x+1}\) = \(1-\dfrac{18}{19}\)
\(\dfrac{1}{2x+1}\) = \(\dfrac{1}{19}\)
\(2x+1\) = 19
2\(x\) = 19 - 1
2\(x\) = 18
\(x\) = 18: 2
\(x\) = 9