Help me
\(4.\left(5+5^1+5^2+5^3+......+5^{100}\right)+5=5^n\)
Giải chi tiết dùm mk nha!
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\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)
Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5
\(-3^2-5.\left(x-1\right)=4x+7\)
\(\Rightarrow-9-5x+5=4x+7\)
\(\Rightarrow-9+5-7=5x+4x\)
\(\Rightarrow9x=-11\)
\(\Rightarrow x=\frac{-11}{9}\)
Vậy....
\(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+..+\dfrac{1}{44.49}\right)\left(\dfrac{1-3-5-7-..-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+..+\dfrac{5}{44.49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\\ A=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
\(A=\dfrac{9}{196}\left(\dfrac{1-3-5-7-...-49}{89}\right)\)
Ta đặt: \(P=1-3-5-7-...-49\\ =1-\left(3+5+7+..+49\right)\\ =1-624\\ =-623\\ \Rightarrow\dfrac{9}{196}.-\dfrac{623}{89}=-\dfrac{9}{28}.\)
(x-1/2)=19/4:5/3
(x-1/2)=19/4x3/5
x-1/2=57/20
x=57/20+1/2
x=67/20
tk cho mk nha
\(\frac{19}{4}:\left(x-\frac{1}{2}\right)=\frac{5}{3}\)
\(x-\frac{1}{2}=\frac{19}{4}:\frac{5}{3}=\frac{57}{20}\)
\(x=\frac{57}{20}+\frac{1}{2}\)
\(x=\frac{67}{20}\)
\(B=4\cdot\left(-\frac{1}{2}\right)^3:\left(\frac{4}{5}\right)^0\cdot\frac{1}{2}-\frac{\frac{3}{5}-\frac{3}{9}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{13}}\)
\(=4\cdot\frac{-1}{8}:1\cdot\frac{1}{2}-\frac{3\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}\)
\(=-\frac{1}{4}-\frac{3}{7}=-\frac{19}{28}\)