Mn giải giúp em với (5*4^15*9^9-4*3^20*8^9) / (5*2^29*7*2^29*27^6)
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\(a+b=c+d\Leftrightarrow\left(a+b\right)^3=\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3+d^3+3cd\left(c+d\right)\)
\(\Leftrightarrow ab\left(a+b\right)=cd\left(c+d\right)\)
\(\Leftrightarrow ab=cd\)
\(\Leftrightarrow\dfrac{a}{c}=\dfrac{d}{b}=t\Leftrightarrow a=ct,d=bt\)
\(\left(a+b\right)^2\left(a^3+b^3\right)=a^5+b^5+2ab\left(a^3+b^3\right)+a^2b^2\left(a+b\right)\)
\(\left(c+d\right)^2\left(c^3+d^3\right)=c^5+d^5+2cd\left(c^3+d^3\right)+c^2d^2\left(c+d\right)\)
mà \(a+b=c+d,a^3+b^3=c^3+d^3,ab=cd\)
suy ra \(a^5+b^5=c^5+d^5\).
Các số lập được: 271, 217, 721, 712, 127, 172
271+217= 488
721+712= 1433
172+127= 299
488+ 299+ 1433 = 500+300+1500 - (12+1+67) = 2300 - 80= 2220
các số có 3 chữ số khác nhau được lập từ các chữ số 2; 7; 1
127; 172; 217; 271; 712; 721
các chữ số 1; 2; 7 xuất hiện ở hàng đơn vị, hàng chục, hàng trăm số lần là 2 lần
tổng các số vừa được lập là
(1 + 2 + 7 ) x ( 1 + 10 + 100) x 2 = 2220
9 ≤ 3n < 1000
32 ≤ 3n < 103
32 ≤ 3n ≤ 93 < 103
32 ≤ 3n ≤ 36 < 103
2 ≤ n ≤ 6
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^3=0\)
\(\Rightarrow\left(a+b\right)^3+3\left(a+b\right)c\left(a+b+c\right)+c^3=0\)
\(\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)+3ab\left(a+b\right)=0\)
\(\Rightarrow3\left(a+b\right)\left(ac+bc+c^2+ab\right)=0\)
\(\Rightarrow3\left(a+b\right)\left[c\left(b+c\right)+a\left(b+c\right)\right]=0\)
\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}c=0\\a=0\\b=0\end{matrix}\right.\)
*\(a=-b;c=0\Rightarrow P=a^{2021}-a^{2021}+0^{2021}=0\)
*\(b=-c;a=0\Rightarrow P=0^{2021}+b^{2021}-b^{2021}=0\)
*\(c=-a;b=0\Rightarrow P=a^{2021}+0^{2021}-a^{2021}=0\)
Vậy \(P=0\)
Đặt \(x^3=a^3;27y^3=b^3;8z^3=c^3\)
\(\Rightarrow a^3-b^3-c^3=3abc\)
\(\Rightarrow a^3-b^3-c^3-3abc=0\)
\(\Rightarrow a^3-\left(b+c\right)^3+3bc\left(b+c\right)-3abc=0\)
\(\Rightarrow\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2\right]-3bc\left(a-b-c\right)=0\)
\(\Rightarrow\left(a-b-c\right)\left(a^2+ab+ac+b^2+2bc+c^2-3bc\right)=0\)
\(\Rightarrow\left(a-b-c\right)\left(a^2+b^2+c^2+ab-bc+ca\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\a^2+b^2+c^2+ab-bc+ca=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c+a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\a=-b=-c\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3y-2z=0\\x=-3y=-2z\end{matrix}\right.\)
*\(x-3y-2z=0\) :
\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}=\dfrac{2z.x.3y}{6xyz}=1\)
*\(x=-3y=-2z\) :
\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}\dfrac{\left(x+x\right)\left(3y+3y\right)\left(-2z-2z\right)}{6xyz}=\dfrac{2x.6y.\left(-4\right)z}{6xyz}=-8\)
Mk sửa lại biểu thức P :\(P=\dfrac{\left(x-3y\right)\left(3y+2z\right)\left(x-2z\right)}{6xyz}\)
Ta có : x3 - 27y3 - 8z3 = 18xyz
<=> (x - 3y)3 + 9xy(x - 3y) - 8z3 = 18xyz
<=> [(x - 3y)3 - (2z)3] + 9xy(x - 3y - 2z) = 0
<=> (x - 3y - 2z)[(x - 3y)2 + (x - 3y).2z + 4z2] + 9xy(x - 3y - 2z) = 0
<=> (x - 3y - 2z)[(x - 3y)2 + (x - 3y).2z + 4z2 + 9zy] = 0
<=> \(\left(x-3y-2z\right)\left\{\left[\dfrac{1}{4}\left(x-3y\right)^2+\left(x-3y\right).2z+4z^2\right]+\dfrac{3}{4}\left(x-3y\right)^2+9xy\right\}=0\)
<=> \(\left(x-3y-2z\right)\left\{\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2\right\}=0\)
<=> \(\left[{}\begin{matrix}x-3y-2z=0\\\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2=0\end{matrix}\right.\)
THI1 x - 3y - 2z = 0
<=> x = 3y + 2z
Khi đó \(P=\dfrac{2z.x.3y}{6xyz}=1\)
TH2 \(\left[\dfrac{1}{2}\left(x-3y\right)+2z\right]^2+\dfrac{3}{4}\left(x+3y\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}\left(x-3y\right)+2z=0\\x+3y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2z=3y\\x=-3y\end{matrix}\right.\Leftrightarrow x=-3y=-2z\)
Khi đó P = \(\dfrac{\left(-6y\right).\left(-2x\right).\left(-4z\right)}{xyz}=-48\)
\(a^3+8b^3+1=6ab\)
\(\Rightarrow\left(a+2b\right)^3-6a^2b-12ab^2+1-6ab=0\)
\(\Rightarrow\left(a+2b\right)^3+1-6ab\left(a+2b+1\right)=0\)
\(\Rightarrow\left(a+2b+1\right)\left[\left(a+2b\right)^2-\left(a+2b\right)+1\right]-6ab\left(a+2b+1\right)=0\)
\(\Rightarrow\left(a+2b+1\right)\left(a^2+4ab+4b^2-a-2b+1-6ab\right)=0\)
\(\Rightarrow\left(a+2b+1\right)\left(a^2-2ab+4b^2-a-2b+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\a^2-2ab+4b^2-a-2b+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a^2-2a\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b\right)+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a^2-2a+1-1\right)+\dfrac{1}{2}\left(a^2-4ab+4b^2\right)+2\left(b^2-b+\dfrac{1}{4}-\dfrac{1}{4}\right)+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a-1\right)^2-\dfrac{1}{2}+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2-\dfrac{1}{2}+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\\dfrac{1}{2}\left(a-1\right)^2+\dfrac{1}{2}\left(a-2b\right)^2+2\left(b-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a+2b+1=0\\a=1;b=\dfrac{1}{2}\end{matrix}\right.\)
*\(a+2b+1=0\Rightarrow a+2b=-1\)
*\(a=1;b=\dfrac{1}{2}\Rightarrow a+2b=1+2.\dfrac{1}{2}=2\)