Tìm 3 số nguyên a,b,c thỏa mãn:a+b=-4;b+c=-6;c+a=12
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Đặt \(\left(x;y;z\right)=\left(a-4;b-5;c-6\right)\) \(\Rightarrow x;y;z\ge0\)
\(\left(x+4\right)^2+\left(y+5\right)^2+\left(z+6\right)^2=90\)
\(\Leftrightarrow x^2+y^2+z^2+8x+10y+12z=13\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2xz+2yz+12\left(x+y+z\right)=13+2\left(xy+xz+yz\right)+4x+2y\)
\(\Leftrightarrow\left(x+y+z\right)^2+12\left(x+y+z\right)=13+2\left(xy+xz+yz\right)+2\left(2x+y\right)\ge13\)
\(\Leftrightarrow\left(x+y+z\right)^2+12\left(x+y+z\right)-13\ge0\)
\(\Leftrightarrow\left(x+y+z+13\right)\left(x+y+z-1\right)\ge0\)
\(\Leftrightarrow x+y+z\ge1\)
\(\Leftrightarrow a-4+b-5+c-6\ge1\)
\(\Leftrightarrow a+b+c\ge16\)
\(\Rightarrow P_{min}=16\) khi \(\left(x;y;z\right)=\left(0;0;1\right)\) hay \(\left(a;b;c\right)=\left(4;5;7\right)\)
Theo đề bài ta có:
a + b = -8
b + c = -6
c + a = 16
\(\Rightarrow\)(a + b) + (b + c) + (c + a) = (-8) + (-6) + 16 = 2
Mà (a + b) + (b + c) + (c + a) = a + b + b + c + c + a = 2a + 2b + 2c =2(a+b+c)
\(\Rightarrow a+b+c=2\div2=1\)
\(\Rightarrow a=\left(a+b+c\right)-\left(b+c\right)=1-\left(-6\right)=7\)
\(\Rightarrow b=\left(a+b+c\right)-\left(c+a\right)=1-16=-15\)
\(\Rightarrow c=\left(a+b+c\right)-\left(a+b\right)=1-\left(-8\right)=9\)
Vậy a = 7; b = -15; c = 9
1,https://diendantoanhoc.net/topic/157361-t%C3%ACm-c%C3%A1c-s%E1%BB%91-nguy%C3%AAn-x-y-tho%E1%BA%A3-m%C3%A3n-x3y32016/
\(a+b=-4;b+c=-6;c+a=12\\ \Rightarrow a+b+b+c+c+a=\left(-6\right)+\left(-4\right)+12=2\\ \Rightarrow2\left(a+b+c\right)=2\\ \Rightarrow a+b+c=1\)
\(\Rightarrow c=1-\left(-4\right)=5\\ \Rightarrow b=\left(-6\right)-5=-11\\ \Rightarrow a=7\)
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