1 tìm x biết
a, \(5^x+5^{x+2}=650\)
b \(\left(x+2\right)^2=81\)
c,\(1+5+9+13+17+...+x=5050\)
d,\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1978}=-4\)
e,\(\dfrac{1}{2016}:2015x=\dfrac{-1}{2015}\)
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\(\dfrac{45-x}{1963}+\dfrac{40-x}{1968}+\dfrac{35-x}{1973}+\dfrac{30-x}{1978}+4=0\)
\(\Rightarrow\dfrac{45-x}{1963}+\dfrac{40-x}{1968}+\dfrac{35-x}{1973}+\dfrac{30-x}{1978}=-4\)
\(\Rightarrow\left(\dfrac{45-x}{1963}+1\right)+\left(\dfrac{40-x}{1968}+1\right)+\left(\dfrac{35-x}{1973}+1\right)+\left(\dfrac{30-x}{1978}+1\right)=-4+1+1+1+1\)
\(\Rightarrow\dfrac{2008-x}{1963}+\dfrac{2008-x}{1968}+\dfrac{2008-x}{1973}+\dfrac{2008-x}{1978}=0\)
Vì \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\) nên 2008 - x = 0
\(\Rightarrow x=2008\)
Đề sai:
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1983}+1\right)=0\)
\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên \(2013-x=0\Leftrightarrow x=2013\)
\(\dfrac{55-x}{1963}\) + \(\dfrac{50-x}{1968}\) + \(\dfrac{45-x}{1973}\) + \(\dfrac{40-x}{1978}\) + 4 = 0
(1 + \(\dfrac{55-x}{1963}\) ) + ( 1 + \(\dfrac{50-x}{1968}\)) + (1+ \(\dfrac{45-x}{1973}\))+ (1 + \(\dfrac{40-x}{1978}\)) = 0
\(\dfrac{1963+55-x}{1963}\) + \(\dfrac{1968+50-x}{1968}\)+\(\dfrac{1973+45-x}{1973}\)+\(\dfrac{1978+40-x}{1978}\)=0
\(\dfrac{2018-x}{1963}\)+\(\dfrac{2018-x}{1968}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1978}\)=0
(2018 - \(x\))\(\times\)( \(\dfrac{1}{1963}\)+\(\dfrac{1}{1986}\)+\(\dfrac{1}{1973}\)+) =0
2018 \(-x\) = 0
\(x\) = 2018
\(\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+4=0\)
\(\Rightarrow\text{ }\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+1+1+1+1=0\)
\(\Rightarrow\text{ }\left(\dfrac{55-x}{1963}+1\right)+\left(\dfrac{50-x}{1968}+1\right)+\left(\dfrac{45-x}{1973}+1\right)+\left(\dfrac{40-x}{1978}+1\right)=0\)
\(\Rightarrow\text{ }\dfrac{2018-x}{1963}+\dfrac{2018-x}{1968}+\dfrac{2018-x}{1973}+\dfrac{2018-x}{1978}=0\)
\(\Rightarrow\text{ }\left(2018-x\right)\left(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\right)=0\)
Mà \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\)
\(\Rightarrow\text{ }2018-x=0\)
\(\Rightarrow\text{ }x=2018-0\)
\(\Rightarrow\text{ }x=2018\)
Vậy, \(x=2018.\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
Các câu dễ tự làm :v
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1981}=-4\) (sau khi đã sửa đề)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1981}+1\right)=0\)\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1981}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1981}\right)=0\)
\(\Rightarrow2013-x=0\Rightarrow x=2013\)
\(1+5+9+13+17+.....+x=5050\)
Số các số hạng là:
\(\dfrac{x-1}{4}+1=\dfrac{1}{4}x+\dfrac{3}{4}\)
Như vậy có :
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\) số hạng
Theo đề bài ta có:
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\left(x+1\right)=5050\)
\(\Rightarrow\left(\dfrac{1}{4}x+\dfrac{3}{4}\right)\left(x+1\right)=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+\dfrac{1}{4}x+\dfrac{3}{4}x+\dfrac{3}{4}=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+x+\dfrac{3}{4}=10100\)
Kiệt sức.đến đây ko nghĩ nổi nx
a,
\(5^x+5^{x+2}=650\\ 5^x\left(1+5^2\right)=650\\ 5^x\cdot26=650\\ 5^x=25\\ 5^x=5^2\\ \Rightarrow x=2\)
Vậy \(x=2\)
b,
\(\left(x+2\right)^2=81\\ \Rightarrow\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
Vậy \(x=7\) hoặc \(x=-11\)
d,
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\\ \dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}+4=0\\ \dfrac{45-x}{1968}+1+\dfrac{40-x}{1973}+1+\dfrac{35-x}{1978}+1+\dfrac{30-x}{1983}+1=0\\ \dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\\ \left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên
\(2013-x=0\\ x=2013\)
Vậy \(x=2013\)
e,
\(\dfrac{1}{2016}:2015x=\dfrac{-1}{2015}\\ 2015x=\dfrac{-2015}{2016}\\ x=\dfrac{-1}{2016}\)
Vậy \(x=\dfrac{-1}{2016}\)