tìm x , biết
9^x-1=1
giúp mik nha , lời giải chi tiết giúp mik nhé
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Ta có: \(20^x:14^x=\dfrac{10}{7}x\)
\(\Leftrightarrow\left(\dfrac{10}{7}\right)^x=\dfrac{10}{7}x\)
\(\Leftrightarrow x=\left(\dfrac{10}{7}\right)^{x-1}\)
Đến đây mình bí rồi, xin lỗi bạn!
tham khảo
Z1= 60 răng
Z2= 30 răng
a) Ta có: i= Z1/Z2
=> i=60/30=2
b) đĩa líp quay nhanh hơn vì có số răng ít hơn đia xích
6\(^2\)+ 64 : ( x - 1 ) = 52
36 + 64 : ( x - 1 ) =52
64 ; ( x - 1 ) =64 : 52
x - 1 = \(\frac{16}{13}\)
x = \(\frac{16}{13}\)+1
x = \(\frac{29}{13}\)
HT
\(x^2=x^3\\ \Rightarrow x^2-x^3=0\\ \Rightarrow x^2\left(1-x\right)\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) 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
1. Tìm x
a) 1+2+3+...+x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x = 20
b) \(32.3^x=9.3^{10}+5.27^3\)
=>\(32.3^x=9.3^{10}+5.3^9\)(\(27^3=\left(3^3\right)^3=3^9\))
=>\(32.3^x=9.3.3^9+5.3^9\)
=>\(32.3^x=3^9\left(9.3+5\right)\)
=>\(32.3^x=3^9.32\)
=>x = 9
2.
Ta có 2A = 3A - A
=> 2A = \(3\left(1+3+3^2+3^3+....+3^{10}\right)\)\(-\)\(1-3-3^2-3^3-....-3^{10}\)
=> 2A = \(3+3^2+3^3+.....+3^{11}-\)\(1-3-3^2-3^3-...-3^{10}\)
=> 2A = \(3^{11}-1\)
=> 2A+1 = \(3^{11}-1+1\)=\(3^{11}\)
=> n = 11
Ta có : a)1 + 2 + 3 + ... + x = 210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 420
=> x(x + 1) = 20.21
=> x = 20
\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\left(ĐKXĐ:y\ne0\right)\)
\(\Rightarrow\dfrac{xy-27}{9y}=\dfrac{1}{18}\)
\(\Rightarrow18\left(xy-27\right)=9y\)
\(\Rightarrow2\left(xy-27\right)=y\)
\(\Rightarrow2xy-54=y\)
\(\Rightarrow2xy-y=54\Rightarrow y\left(2x-1\right)=54\)
\(\Rightarrow y=\dfrac{54}{2x-1}\)
- Suy ra 54 chia hết cho 2x - 1
\(\Rightarrow2x-1\inƯ\left(54\right)\)
\(\Rightarrow2x-1\in\left\{1;-1;2;-2;3;-3;9;-9;27;-27\right\}\)
Cho 2x - 1 bằng từng giá trị ở trên, ta tìm được :
\(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};2;-1;5;-4;14;-13\right\}\). Mà x không có giá trị ngoài tập số nguyên.
\(\Rightarrow x\in\left\{-13;-4;-1;0;1;2;5;14\right\}\)
Thay các giá trị x trên vừa tìm được vào y :
\(\Rightarrow y\in\left\{54;-54;18;-18;6;-6;2;-2\right\}\)
Vậy : Các số x và y thỏa mãn đề bài là : \(\left(x;y\right)\in\left\{\left(1;54\right),\left(0;-54\right),\left(2;18\right),\left(-1;-18\right),\left(5;6\right),\left(-4;-6\right),\left(14;2\right),\left(-13;-2\right)\right\}\)
\(\left(2x+1\right)\left(x^2-x\right)+x\left(5+x-2x^2\right)=3x+7\)
\(2x^3-2x^2+x^2-x+5x+x^2-2x^3=3x+7\)
\(5x-x=3x+7\)
\(4x-3x=7\)
\(x=7\)
(2x+1)(x^2-x)+x(-2x^2+x+5)=3x+7
=>2x^3-2x^2+x^2-x-2x^3+x^2+5x=3x+7
=>-x^2-x+x^2+5x=3x+7
=>4x=3x+7
=>x=7