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Đặt \(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}=k\)
\(\Rightarrow x=1998k;y=1999k;z=2000k\)
\(\left(x-z\right)^3=\left(2000k-1998k\right)^3=8k^3\)
\(8\left(x-y\right)^2\left(y-z\right)=8\left(1999k-1998k\right)^2.\left(1999k-2000k\right)\\ =8.k^2.k=8k^3\\ \Rightarrowđpcm\)
Sai đề kìa . Đề đúng đây :
\(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}\)
Đặt \(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}=k\left(k>0\right)\)
Ta có :
x = 1998k ; y = 1999k ; z =2000k
Ta có :
\(\left(x-z\right)^3=\left(1998k-2000k\right)^3=\left(-2k\right)^3=-8k\) (*)
\(8\left(x-y\right)^2\cdot\left(y-z\right)=8\left(1998k-1999k\right)^2\cdot\left(1999k-2000k\right)\)
\(=8\left(-1\right)^2\cdot\left(-1\right)=-8\) (**)
Từ (*) và (**) suy ra ĐPCM
\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)
\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)
\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)
\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)
\(x=\dfrac{21}{10}\)
\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
\(x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(x=\dfrac{2}{3}\)
\(c,\left(0,4x-1,3\right)^2=5,29\)
\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)
TH1: \(0,4x-1,3=2,3\)
\(0,4x=3,6\)
\(x=9\)
TH2: \(0,4x-1,3=-2,3\)
\(0,4x=-1\)
\(x=-\dfrac{5}{2}\)
=.= hok tốt!!
1) So sánh các lũy thừa
a.
4444\(^{3333}\) và 3333\(^{4444}\)
4444\(^{3333}\) =(4\(^3\)\()\) \(^{111}\)
3333\(^{4444}\) =\((\)3\(^4\)\()\) \(^{111}\)
\(\rightarrow\) (4\(^3\)\()\) \(^{111}\) =64\(^{111}\) ; \((\)3\(^4\)\()\) \(^{111}\) =81\(^{111}\)
\(\rightarrow\)64\(^{111}\) <81\(^{111}\)
\(\Rightarrow\) 4444\(^{3333}\) < 3333\(^{4444}\)
Lười làm quá,ý còn lại bn làm tương tự,có ý lấy số chung để so sánh,có ý lấy số mũ để so sánh,có ý như trên.
Bài 2:
\(\Leftrightarrow\left\{{}\begin{matrix}3^{3x-2x+y}=3^5\\5^{2x-x-y}=5^3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\x-y=3\end{matrix}\right.\)
=>x=1;y=-2
a) \(\dfrac{1}{5}:x=\dfrac{-1}{5}-\dfrac{-4}{5}\)
\(_{_{ }\Rightarrow\dfrac{-1}{5}:x=\dfrac{3}{5}}\)
\(\Rightarrow x=\dfrac{1}{5}:\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{1}{3}\)
Vậy \(x=\dfrac{1}{3}\)
b) \(-\left(x-2\right)^3=125\)
\(\Rightarrow\left(x-2\right)^3=-125\)
\(\Rightarrow x-2=-5\)
\(\Rightarrow x=-5+2\)
\(\Rightarrow x=-3\)
Vậy x=-3
\(\dfrac{4}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{4}{x}-\dfrac{2y}{6}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)
\(\Rightarrow\dfrac{4}{x}=\dfrac{1+2y}{6}\)
\(\Rightarrow24=x\left(1+2y\right)\)
\(\Rightarrow x;1+2y\inƯ\left(24\right)\)
\(Ư\left(24\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm8;\pm12;\pm24\right\}\)
Mà 1+2y lẻ nên:
\(\left\{{}\begin{matrix}1+2y=1\Rightarrow2y=0\Rightarrow y=0\\x=24\\1+2y=-1\Rightarrow2y=-2\Rightarrow y=-1\\x=-24\end{matrix}\right.\)
\(\left\{{}\begin{matrix}1+2y=3\Rightarrow2y=2\Rightarrow y=1\\x=8\\1+2y=-3\Rightarrow2y=-4\Rightarrow y=-2\\x=-8\end{matrix}\right.\)
1. Tìm x:
a) \(\left(x+36\right)^2=1936\Leftrightarrow x+36=\pm44.\) Vậy x = 8 hoặc x = -80
b) \(\left(\dfrac{3}{5}\right)^{x+2}=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}\right)^{x+2}=\left(\dfrac{3}{5}\right)^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
c) Xem lại đề
d) \(\left(\dfrac{9}{16}\right)^{x-5}=\left(\dfrac{4}{3}\right)^4\Leftrightarrow\left(\dfrac{3}{4}\right)^{2\left(x-5\right)}=\left(\dfrac{3}{4}\right)^{-4}\Leftrightarrow2\left(x-5\right)=-4\Leftrightarrow x=3\)
e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{27}\right)^x=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}.\dfrac{125}{27}\right)^x=\left(\dfrac{3}{5}\right)^4\Leftrightarrow\left(\dfrac{5}{3}\right)^{2x}=\left(\dfrac{5}{3}\right)^{-4}\Leftrightarrow2x=-4\) Vậy x = -2
3. Tính giá trị của biểu thức:
\(A=\left\{-\left[\left(\dfrac{1}{x}\right)^2\right]^3\right\}^5.\left\{-\left[\left(-x\right)^5\right]^2\right\}^3\) \(\left(x\notin0\right)\)
\(=\left\{-\left[-\dfrac{1}{x^2}\right]^3\right\}^5.\left\{-\left[-\left(-x\right)^5\right]^2\right\}^3=\left\{-\left[-\dfrac{1}{x^6}\right]\right\}^5.\left\{-\left[x^5\right]^2\right\}^3\)
\(=\left\{\dfrac{1}{x^6}\right\}^5.\left\{-x^{10}\right\}^3=\dfrac{1}{x^{30}}.\left(-x^{30}\right)=-1\)
a. VP: \(\left(x+y\right)^{1999}\cdot\left(x-y\right)^{1999}=\left[\left(x+y\right)\left(x-y\right)\right]^{1999}\)
\(=\left(x^2-xy+xy-y^2\right)^{1999}=\left(x^2-y^2\right)^{1999}=VT\)
--> đpcm
b. VT: \(\dfrac{\left(5^4-5^3\right)^3}{125^4}=\dfrac{500^3}{125^4}=\dfrac{125^3\cdot4^3}{125^4}=\dfrac{4^3}{125}=\dfrac{64}{125}=VP\)
--> đpcm