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Ta có: x chia hết cho 60
=> x thuộc B(60)={ 0, 60, 120, 180, 240, 360, 420, 480,540, 600, 660, 720,780,...}
Vì 750>x>200
Nên x thuộc {240, 360, 420, 480, 540,600, 660, 720}
Ta có: x chia hết cho 60
=> x = B(60)
Mà B(60) = {0;60;120;180;240;300;360;420;480;540;600;660;720;780;...}
Nhưng 750 > x > 200
Vậy x = { 240;300;360;420;480;540;600;660;720}
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\(\Rightarrow2^{3n-n}=16=2^4\Rightarrow2n=4\Rightarrow n=2\)
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Ta có : \(\left\{{}\begin{matrix}2x=3y\\4y=3z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8}\end{matrix}\right.\)
`=> x/9 =y/6 =z/8=>x/9 =y/6 = (2z)/16` và `x-y+2z=57`
ADTC dãy tỉ số bằng nhau ta có :
`x/9 =y/6 = (2z)/16 = (x-y+2z)/(9-6+16) = 57/19=3`
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\Rightarrow x=3\cdot9=27\\\dfrac{y}{6}=3\Rightarrow y=3\cdot6=18\\\dfrac{z}{8}=3\Rightarrow z=3\cdot8=24\end{matrix}\right.\)
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\(\left(2x-3\right)^4-\left(2x-3\right)^2=0\)
\(\Rightarrow\left(2x-3\right)^2\left[\left(2x-3\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-3\right)^2=0\\\left(2x-3\right)^2=1\end{cases}}\)
Từ đó tìm được \(x=\frac{3}{2},x=2,x=1\)
(2x-3)4-(2x-3)2=0
suy ra có 2 TH
TH1 (2x-3)4=0
(2x-3)4=04
2x-3=0
2x=0+3
2x=3
x=3:2
x=1,5
tH 2
(2X-3)2=0
(2X-3)2=02
2X-3=0
2x=0+3
2x=3
x=3:2
x=1.5
vậy x \(\in\){1,5}
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\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{2}{5}=\dfrac{4}{5}\\3x-\dfrac{2}{5}=-\dfrac{4}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{6}{5}\\3x=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
\(\left(\dfrac{2}{5}-3x\right)^2=\left(\dfrac{-4}{5}\right)^2\)
\(\text{Vậy}\dfrac{2}{5}-3x=\left(\dfrac{-4}{5}\right)\)
\(3x=\dfrac{2}{5}+\dfrac{4}{5}=\dfrac{6}{5}\)
\(x=\dfrac{6}{5}:3=\dfrac{2}{5}\)
\(\text{hoặc}\dfrac{2}{5}-3x=\dfrac{4}{5}\)
\(3x=\dfrac{2}{5}+\left(\dfrac{-4}{5}\right)=\left(\dfrac{-2}{5}\right)\)
\(x=\dfrac{-2}{5}:3=\dfrac{-2}{15}\)
\(\Rightarrow x\in\left\{\dfrac{2}{5};\dfrac{-2}{15}\right\}\)
\(\text{Hok tốt!}\)
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Bạn tham khảo link này nha ! Có lời giải đó :
http://olm.vn/hoi-dap/detail/26954556179.html
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Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{12x-15y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=\frac{0}{7+9+11}=0\)
\(\Rightarrow\left\{\begin{matrix}\frac{12x-15y}{7}=0\\\frac{20z-12x}{9}=0\\\frac{15y-20z}{11}=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}12x-15y=0\\20z-12x=0\\15y-20z=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)
\(\Rightarrow12x=15y=20z\)
\(\Rightarrow\frac{12x}{60}=\frac{15y}{60}=\frac{20z}{60}\)
\(\Rightarrow\frac{x}{5}=\frac{y}{4}=\frac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=\frac{x+y+z}{5+4+3}=\frac{48}{12}=4\)
\(\left\{\begin{matrix}\frac{x}{5}=4\\\frac{y}{4}=4\\\frac{z}{3}=4\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=20\\y=16\\z=12\end{matrix}\right.\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(20;16;12\right)\)