tìm x,y thuộc Z thỏa mãn: \(3x^2+y^2+4xy-8x-2y=0\)
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1/ \(=3^n.3^2+3^n=3^n\left(3^2+1\right)=10.3^n⋮10\)
2/ \(100.x+\left(1+2+3+...+100\right)=7450\)
Đến đây bạn tự làm nốt nhé
1. Ta có: \(3^{n+2}+3^n=3^n.\left(3^2+1\right)=3^n.\left(9+1\right)=3^n.10⋮10\)( đpcm )
2. \(\left(x+1\right)+\left(x+2\right)+.......+\left(x+100\right)=7450\)
\(\Leftrightarrow x+1+x+2+........+x+100=7450\)
\(\Leftrightarrow100x+\frac{100.101}{2}=7450\)
\(\Leftrightarrow100x+5050=7450\)
\(\Leftrightarrow100x=2400\)\(\Leftrightarrow x=24\)
Vậy \(x=24\)
1000+3,64×x+6,36×x=2015
3,64×x+6,36×x=2015-1000
3,64×x+6,36×x=1015
x×(3,64+6,36)=1015
x×10=1015
x=1015:10
x=101,5
vậy x=101,5
\(10^{35}:\left(5^{30}.4^{15}\right)=10^{35}:\left[5^{30}.\left(2^2\right)^{15}\right]=10^{35}:\left(5^{30}.2^{30}\right)\)
\(=10^{35}:10^{30}=10^5\)
a, \(\Leftrightarrow x\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(x=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
b, \(\Leftrightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
x - 1 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
x | 2 | 0 | 3 | -1 | 4 | -2 | 7 | -5 |
a) Ta có : \(8⋮x\)
\(\Rightarrow x\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Rightarrow x=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
b) Ta có : \(6⋮x-1\)
\(\Rightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Rightarrow x=\left\{0;2;-1;3;-2;4;-5;7\right\}\)
\(=4^{15}.5^{15}.5^{15}.3^{19}:3^{12}\)
\(=4^{15}.5^{15}.5^{15}.3^7\)
\(=\left(4.5.5\right)^{15}.3^7\)
\(=100^{15}.3^7\)
có thể mình sai
Sửa lại đề: \(\left(125^3.7^4-5^9.49^2\right):2005^{2006}\)
Ta có : \(125^3.7^4=\left(5^3\right)^3.7^4=5^{3.3}.7^4=5^9.7^4\)
\(5^9.49^2=5^9.\left(7^2\right)^2=5^9.7^{2.2}=5^9.7^4\)
\(\Rightarrow125^3.7^4-5^9.49^2=5^9.7^4-5^9.7^4=0\)
mà \(2005^{2006}>0\)\(\Rightarrow\left(125^3.7^4-5^9.49^2\right):2005^{2006}=0\)
\(3x^2+y^2+4xy-8x-2y=0\)
\(\Leftrightarrow4x^2+4xy+y^2-4x-2y+1-x^2-4x-4=-3\)
\(\Leftrightarrow\left(2x+y-1\right)^2-\left(x+2\right)^2=-3\)
\(\Leftrightarrow\left(2x+y-1-x-2\right)\left(2x+y-1+x+2\right)=-3\)
\(\Leftrightarrow\left(x+y-3\right)\left(3x+y+1\right)=-3\)
Do \(x,y\in Z\Rightarrow x+y-3;3x+y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Bạn lập bảng xét ước rồi tìm ra x,y thỏa mãn
Vậy \(\left(x,y\right)=\left(0,2\right);\left(-4,8\right);\left(-4;10\right);\left(0,0\right)\)