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\(A=2x^2-20x+7=2\left(x^2-10x+25\right)-43=2\left(x-5\right)^2-43\ge-43\left(\forall x\right)\)
=> Chưa thể khẳng định A dương
\(B=9x^2-6xy+2y^2+1\)
\(B=\left(9x^2-6xy+y^2\right)+y^2+1\)
\(B=\left(3x-y\right)^2+y^2+1\ge1>0\left(\forall x\right)\)
=> đpcm
\(C=x^2-2x+y^2+4y+6\)
\(C=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\)
\(C=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1>0\left(\forall x\right)\)
=> đpcm
\(D=x^2-2x+2=\left(x^2-2x+1\right)+1=\left(x-1\right)^2+1\ge1>0\left(\forall x\right)\)
=> đpcm
1/
\(M=3x^2-4x+3=3\left(x^2-\frac{4}{3}x+1\right)=3\left(x^2-2x\cdot\frac{2}{3}+\frac{4}{9}\right)+\frac{5}{3}=3\left(x-\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)
\(N=5x^2-10x+2018=5\left(x^2-2x+1\right)+2013=5\left(x-1\right)^2+2013\ge2013>0\)
\(P=x^2+2y^2-2xy+4y+7=\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)+3=\left(x-y\right)^2+\left(y+2\right)^2+3\ge3>0\)
2/
\(A=10x-6x^2+7=-6x^2+10x+7=-6\left(x^2-\frac{10}{6}x+\frac{25}{36}\right)-\frac{11}{6}=-6\left(x-\frac{5}{6}\right)^2-\frac{11}{6}\le-\frac{11}{6}< 0\)
\(B=-3x^2+7x+10=-3\left(x^2-\frac{7}{3}x+\frac{49}{36}\right)-\frac{311}{12}=-3\left(x-\frac{7}{6}\right)^2-\frac{311}{12}\le-\frac{311}{12}< 0\)
\(C=2x-2x^2-y^2+2xy-5=\left(2x-x^2-1\right)-\left(x^2-2xy+y^2\right)-4=-\left(x^2-2x+1\right)-\left(x-y\right)^2-4=-\left(x-1\right)^2-\left(x-y\right)^2-4\)\(\le-4< 0\)
Bài 1 :
a, \(\left(x^2-2x+3\right)\left(x-4\right)=0\)
TH1 : \(x^2-2x+3=0\)
\(\left(-2\right)^2-4.3=4-12< 0\)vô nghiệm
TH2 : \(x-4=0\Leftrightarrow x=4\)
b, \(\left(2x^2-3x-1\right)\left(5x+2\right)=0\)
TH1 : \(\left(-3\right)^2-4.\left(-1\right).2=9+8=17>0\)
\(\Rightarrow x_1=\frac{3-\sqrt{17}}{4};x_2=\frac{3+\sqrt{17}}{4}\)
TH2 ; \(5x+2=0\Leftrightarrow x=-\frac{2}{5}\)
c, đưa về hệ đc ko ?
d, \(\left(5x^3-x^2+2x-3\right)\left(4x^2-x+2\right)=0\)
TH1 : \(x=0,74...\) ( bấm máy cx ra )
TH2 : \(\left(-1\right)^2-4.2.4< 0\)vô nghiệm
KL : vô nghiệm
Bài 2 :
a, \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)-\left(18x-12\right)\)
\(=6x^2+21x-2x-7-6x^2+5x-6x+5-18x+12=10\)
Vậy biểu thức ko phụ thuộc vào biến
b, \(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4y^4\)
\(=x^4+x^3y+x^2y^2+xy^3-yx^3-y^2x^2-y^3x-y^4-x^4y^4\)
\(=x^4-y^4-x^4y^4\)Vậy biểu thức phụ thuộc vào biến
a) A = 4x2 + 4x +11
=> (2x)2+2.2x+1+11-1
=> (2x+1)2+10
do (2x+1)2 \(\dfrac{>}{ }\) 0 vs mọi x
(2x+1)2 +10 \(\dfrac{>}{ }\)10 vs mọi x
GTNNA=10 khi
2x+1=0
=>x=\(\dfrac{-1}{2}\)
a)\(A=4x^2+4x+11\)
\(\Leftrightarrow A=4x^2+4x+1+10\)
\(\Leftrightarrow A=\left(2x+1\right)^2+10\)
Vì \(\left(2x+1\right)^2\ge0\)
Nên \(\left(2x+1\right)^2+10\ge10\)
Vậy GTNN của A=10 khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)
b) \(B=2x-2x^2-5\)
\(\Leftrightarrow B=-2x^2+2x-5\)
\(\Leftrightarrow B=-2x^2+2x-\dfrac{1}{2}-\dfrac{9}{2}\)
\(\Leftrightarrow B=-\left(2x^2-2x+\dfrac{1}{2}\right)-\dfrac{9}{2}\)
\(\Leftrightarrow B=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}\)
\(\Leftrightarrow B=-2\left(x^2-2.x\dfrac{1}{2}+\dfrac{1}{4}\right)-\dfrac{9}{2}\)
\(\Leftrightarrow B=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\)
Do đó \(-\left(x-\dfrac{1}{2}\right)^2\le0\)
Nên \(-\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le\dfrac{-9}{2}\)
Vậy GTLN của \(B=\dfrac{-9}{2}\) khi \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
c) \(C=4x^2-12x\)
\(\Leftrightarrow C=4x^2-12x+9-9\)
\(\Leftrightarrow C=\left(4x^2-12x+9\right)-9\)
\(\Leftrightarrow C=\left(2x-3\right)^2-9\)
Vì \(\left(2x-3\right)^2\ge0\)
Nên \(\left(2x-3\right)^2-9\ge-9\)
Vậy GTNN của \(C=-9\) khi \(2x-3=0\Leftrightarrow x=\dfrac{3}{2}\)
d) \(D=5-x^2+2x-4y^2-4y\)
\(\Leftrightarrow D=7-1-1-x^2+2x-4y^2-4y\)
\(\Leftrightarrow D=-x^2+2x-1-4y^2-4y-1+7\)
\(\Leftrightarrow D=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+7\)
\(\Leftrightarrow D=-\left(x-1\right)^2-\left(2y+1\right)^2+7\)
Vậy GTLN của \(D=7\) khi \(\left\{{}\begin{matrix}x-1=0\Leftrightarrow x=1\\2y+1=0\Leftrightarrow y=\dfrac{-1}{2}\end{matrix}\right.\)
a) x2- 2x - 4y2 - 4y = (x2 - 2x + 1) - (4y2 + 4y + 1) = (x - 1)2 - (2y + 1)2 = (x - 1 - 2y - 1)(x - 1 + 2y + 1) = (x - 2y - 2)(x + 2y)
b) x3 - 4x2 + 12x - 27 = (x3 - 3x2) - (x2 - 3x) + (9x - 27) = x2(x - 3) - x(x - 3) + 9(x - 3) = (x2 - x + 9)(x - 3)
d) x4 - 2x3 + 2x - 1 = (x4 - 2x3 + x2) - (x2 - 2x + 1) = (x2 - x)2 - (x - 1)2 = (x2 - x - x + 1)(x2 - x + x - 1)
= (x2 - 2x + 1)(x2 - 1) = (x - 1)2(x - 1)(x + 1) = (x - 1)3(x + 1)
e) x4 + 2x3 - 4x - 4 = (x4 + 2x4 + x2) - (x2 + 4x + 4) = (x2 + x)2 - (x + 2)2 = (x2 + x - x - 2)(x2 + x + x + 2) = (x2 - 2)(x2 + 2x + 2)
Ai help gấp với
a )
=-(x2-2x+3)
=-(x2-2x+1+2)
=-[(x-1)2+2]
vì (x-1)2+2 lớn hơn hoặc = 2 với mọi x nên -[(x-1)2+2] bé hơn hoặc = 0 hay -x2+2x-3 luôn âm