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`A=x^2+6x+y^2+4y+15`
`=(x^2+6x+9)+(y^2+4y+4)+2`
`=(x+3)^2+(y+2)^2+2`
Vì `(x+3)^2+(y+2)^2 >=0 forall x,y`
`=>A_(min)=2 <=> x=-3; y=-2`.
Ta có: \(A=x^2+6x+y^2+4y+15\)
\(=x^2+6x+9+y^2+4y+4+2\)
\(=\left(x+3\right)^2+\left(y+2\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi (x,y)=(-3;-2)
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Ta có: \(15=x+y+xy\le x+y+\frac{\left(x+y\right)^2}{4}\Rightarrow\frac{t^2}{4}+t\ge15\)(\(t=x+y\))
\(\Leftrightarrow\left(t-6\right)\left(t+10\right)\ge0\Leftrightarrow\orbr{\begin{cases}t\ge6\\t\le-10\end{cases}}\)
\(P=x^2+y^2=\frac{1}{2}.2\left(x^2+y^2\right)\ge\frac{1}{2}\left(x+y\right)^2\ge\frac{1}{2}.6^2=18\)
Dấu \(=\)xảy ra khi \(x=y=3\).
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Bài 1:
a)x2-10x+9
=x2-x-9x+9
=x(x-1)-9(x-1)
=(x-9)(x-1)
b)x2-2x-15
=x2+3x-5x-15
=x(x+3)-5(x+3)
=(x-5)(x+3)
c)3x2-7x+2
=3x2-x-6x+2
=x(3x-1)-2(3x-1)
=(x-2)(3x-1)x^3-12+x^2
d)x3-12+x2
=x3+3x2+6x-2x2-6x-12
=x(x2+3x+6)-2(x2+3x+6)
=(x-2)(x2+3x+6)
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1/ \(\left(x^2+1\right)\left(x-2\right)+2x=4.\)
\(\left(x^2+1\right)\left(x-2\right)+2x-4=0\)
\(\left(x^2+1\right)\left(x-2\right)+\left(2x-4\right)=0\)
\(\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^2+1+2\right)=0\)
\(\left(x-2\right)\left(x^2+3\right)=0\)
TH1:\(x-2=0\Rightarrow x=2\)
TH2: \(x^2+3=0\)
\(\Rightarrow x^2=-3\)(vô lí)
\(\Rightarrow x\in\left\{2\right\}\)
2/ \(A=a\left(b-3\right)-b\left(b-1\right)\)
đề sai f ko ạ, do mik đâu thấy C mà bạn lại cho đề c=2???
\(B=xy\left(x+y\right)-2x-2y\)
\(B=xy\left(x+y\right)-\left(2x+2y\right)\)
\(B=xy\left(x+y\right)-2\left(x+y\right)\)
\(B=\left(x+y\right)\left(xy-2\right)\)
có xy=8 ; x+y=7
\(\Rightarrow B=\left(x+y\right)\left(xy-2\right)\)
\(\Rightarrow B=8\cdot\left(8-2\right)=8\cdot6=48\)
Ta có: x.y = 15
=> x = \(\frac{15}{y}\)
Ta có x + y = -8
\(\frac{15}{y}\)+ y= 8
=> 15 + \(y^2\)= 8y => \(y^2-8y+15=0\)
=> y = 3 hoặc y = 5
=> y = 3 => x=5
y=5 => x=3
\(x^2+y^2=3^2+5^2=34\)
\(x^2+y^2=x^2+2xy+y^2=\left(x+y\right)^2-2xy\)
Vì x+y=-8,xy=15 nên:
\(\left(x+y\right)^2+2xy=\left(-8\right)^2+2.15=34\)