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a) H=x2 - 4x +16
<=> H=x2 -4x + 4 + 12
<=> H=(x-2)2 +12 \(\ge12\)
Vậy Min H = 12
Dấu "=" xảy ra khi x=2
\(K=x^2-6xy+9y^2+4\left(x-3y\right)+4+x^2-12x+36+1978\)
\(K=\left(x-3y\right)^2+4\left(x-3y\right)+2^2+\left(x-6\right)^2+1978\)
\(K=\left(x-3y+2\right)^2+\left(x-6\right)^2+1978\ge1978\)
Vậy Min K =1978
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-3y+2=0\\x-6=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=\dfrac{8}{3}\\x=6\end{matrix}\right.\)
a)\(P=-x^2-4x+16\)
\(=-x^2-4x-4-12\)
\(=-\left(x^2+4x+4\right)-12\)
\(=-\left(x+2\right)^2-12\le-12\)
Đẳng thức xảy ra khi \(x=-2\)
b)\(-x^2+2xy-4y^2+2x+10y-2017\)
\(=\left(-x^2+2xy-y^2+2x-2y-1\right)+\left(-3y^2+12y-12\right)-2004\)
\(=-\left(x^2-2xy+y^2-2x+2y+1\right)-3\left(y^2-4y+4\right)-2004\)
\(=-\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]-3\left(y-2\right)^2-2004\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2-2004\le-2004\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
1/ a/ Ta có:
\(P\left(2\right)=m.2^2+\left(2m+1\right).2-10=16\)
\(\Leftrightarrow m-3=0\)
\(\Leftrightarrow m=3\)
b/ Theo câu a thì
\(P\left(x\right)=3x^2+7x-10=0\)
\(\Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{10}{3}\end{cases}}\)
2/ Tương tự a phân tích nhân tử hộ thôi nha
a/ \(1-5x=0\)
b/ \(x^2\left(x+2\right)=0\)
c/ \(\left(x-1\right)\left(2x-3\right)=0\)
d/ \(\left(x-2\right)^2+4x^{2018}\ge0\) vì dấu = không xảy ra nên đa thức vô nghiệm
a) P(x)=8x6-4x2+5x5-12x+7x2-2x5
=8x6+(-4x2+7x2)+(5x5-2x5)-12x
=8x6+3x2+3x5-12x
b) P(x)=8x6+3x2+3x5-12x
=8x6+3x5+3x2-12x
P(x)-Q(x)=(8x6+3x5+3x2-12x)-(2x5-6x2+8x-2x6)
=8x6+3x5+3x2-12x-2x5+6x2-8x+2x6
=(8x6+2x6)+(3x5-2x5)+(3x2+6x2)+(-12x-8x)
=10x6+x5+9x2-20x
R(x)-Q(x)=4x6-8x2
R(x) =(4x6-8x2)+Q(x)
R(x) =(4x6-8x2)+(2x5-6x2+8x-2x6)
R(x) =4x6-8x2+2x5-6x2+8x-2x6
R(x) =(4x6-2x6)+(-8x2-6x2)+2x5+8x
R(x) =2x6-14x2+2x5+8x
\(2x^2+9y^2-6xy+4x+5\)
\(=\left(x^2-6xy+9y^2\right)+\left(x^2+4x+4\right)+1\)
\(=\left(x-3y\right)^2+\left(x+2\right)^2+1>0\) ;\(\forall x;y\)
\(10x^2+10xy+25y^2-8x+20\)
\(=x^2+10xy+25y^2+9x^2-8x+\frac{16}{9}+\frac{164}{9}\)
\(=\left(x+5y\right)^2+\left(3x-\frac{4}{3}\right)^2+\frac{164}{9}>0\); \(\forall x;y\)
a) x2 - 2x + y2 - 4y + 5 = 0
<=>x^2-2x+1 + y^2-4y+4=0
<=>(x-1)^2 + (y-1)^2 =0
<=>x=1 và y=2
a) \(x^2-2x+y^2-4y+5=0\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-4y+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2++\left(y-2\right)^2=0\)
Mà \(\left(x-1\right)^2\ge0\)và \(\left(y-2\right)^2\ge0\)
Dấu "=" xảy ra khi và chỉ khi x-1=0 và y-2=0
=> x=1 và y=2
Nỗi hứng lm cho vui!
Bài 1:
a) H = \(x^2-4x+16=\left(x^2-4x+4\right)+12=\left(x-2\right)^2+12\)
Vì \(\left(x-2\right)^2\ge0\) => H \(\ge\) 12
=> Dấu = xảy ra <=> \(x=2\)
b) K = \(2x^2+9y^2-6xy-8x-12y+2018\)
= \(\left(x^2-6xy+9y^2\right)+4\left(x-3y\right)+\left(x^2-12x+36\right)+1982\)
= \(\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-6\right)^2+1978\)
= \(\left(x-3y+2\right)^2+\left(x-2\right)^2+1978\)
Vì \(\left\{{}\begin{matrix}\left(x-3y+2\right)^2\ge0\\\left(x-6\right)^2\ge0\end{matrix}\right.\) => K \(\ge\) 1978
=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=\dfrac{2+x}{3}\\x=6\end{matrix}\right.\) => \(x=6;y=\dfrac{8}{3}\)
Bài 2:
a) P = \(-x^2-4x+16=-\left(x^2+4x+4\right)+20\)
= \(-\left(x+2\right)^2+20\le20\)
=> Dấu = xảy ra <=> \(x=-2\)
b) \(Q=-x^2+2xy-4y^2+2x+10y-2017\)
= \(-\left[\left(x^2-2xy+y^2\right)+3\left(y^2-4y+4\right)-2\left(x-y\right)+2005\right]\)
= \(-\left[\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-2\right)^2+2004\right]\)
= \(-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2\right]-2004\)
Vì \(\left\{{}\begin{matrix}-\left(x-y-1\right)^2\le0\\3\left(y-2\right)^2\le0\end{matrix}\right.\) => Q \(\le-2004\)
=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x=y+1\\y=2\end{matrix}\right.\) <=> \(x=3;y=2\)