Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)
\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)
\(x\) + 1 = 16
\(x\) = 16 - 1
\(x\) = 15
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)
\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)
\(\Leftrightarrow376x+752=375x+1125\)
\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)
\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)
\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)
\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)
\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)
\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}-\dfrac{3}{2}+1=\)
\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1=-\dfrac{3}{2}\)
= 4 . -1/8 - 2 . -1/4 + 3 . -1/2 + 1
= -1/2 - -1/2 + -3/2 + 1
= -1/2
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55
\(a.ĐK:x\ne3;1\)
\(\Rightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)+2\left(3x-10\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{7\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow x-1+6x-20=7\left(x^2-4x+3\right)\)
\(\Leftrightarrow7x-21=7x^2-28x+21\)
\(\Leftrightarrow7x^2-35x+42=0\)
\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
b.\(ĐK:x\ne2;4\)
\(\Rightarrow\dfrac{x-1}{x-2}-\dfrac{x+3}{4-x}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(4-x\right)}=\dfrac{2}{\left(x-2\right)\left(4-x\right)}\)
\(\Leftrightarrow\left(x-1\right)\left(4-x\right)-\left(x+3\right)\left(x-2\right)=2\)
\(\Leftrightarrow4x-x^2-4+x-x^2+2x-3x+6-2=0\)
\(\Leftrightarrow-2x^2+4x=0\)
\(\Leftrightarrow-2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{1}{2\left(x-3\right)}+\dfrac{3x-10}{\left(x-1\right)\left(x-3\right)}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1+2\left(3x-10\right)=7\left(x-1\right)\left(x-3\right)\)
\(\Leftrightarrow7\left(x^2-4x+3\right)=x-1+6x-20=7x-21\)
\(\Leftrightarrow\left(x-3\right)\left(7x-7\right)-7\left(x-3\right)=0\)
=>(x-3)(7x-14)=0
=>x=3(loại) hoặc x=2(nhận)
b: \(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)=-2\)
\(\Leftrightarrow x^2-5x+4+x^2+x-6=-2\)
\(\Leftrightarrow2x^2-4x=0\)
=>2x(x-2)=0
=>x=0(nhận) hoặc x=2(loại)
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)
\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)
\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)
Em giải như XYZ olm em nhé
Sau đó em thêm vào lập luận sau:
\(x\) = \(\dfrac{11}{18}\)
Vì \(\in\) N*
Vậy \(x\in\) \(\varnothing\)