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mk gợi ý, phần còn lại tự làm
a) \(A=x^2+2x+5=\left(x+1\right)^2+4\ge4\)
b) \(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
c) \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
d) \(D=x^2-2x+y^2-4y+7=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)
e) \(E=x^2-4xy+5y^2+10x-22y+28=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
a) A = x2 + 2x + 5
= x2 + 2x + 1 + 4
= ( x + 1 )2 + 4
Nhận xét :
( x + 1 )2 > 0 với mọi x
=> ( x + 1 )2 + 4 > 4
=> A > 4
=> A min = 4
Dấu " = " xảy ra khi : ( x + 1 )2 = 0
=> x + 1 = 0
=> x = - 1
Vậy A min = 4 khi x = - 1
b) B = 4x2 + 4x + 11
= ( 2x )2 + 4x + 1 + 10
= ( 2x + 1 )2 + 10
Nhận xét :
( 2x + 1 )2 > 0 với mọi x
=> ( 2x + 1 )2 + 10 > 10
=> B > 10
=> B min = 10
Dấu " = " xảy ra khi : ( 2x + 1 )2 = 0
=> 2x + 1 = 0
=> x = \(\frac{-1}{2}\)
Vậy Bmin = 10 khi x = \(\frac{-1}{2}\)
c) C = ( x - 1 ) ( x + 2 ) ( x + 3 ) ( x + 6 )
= [ ( x - 1 ) ( x + 6 ) ] [ ( x + 2 ) ( x + 3 ) ]
= ( x2 + 5x - 6 ) ( x2 + 5x + 6 )
= ( x2 + 5x ) 2 - 62
= ( x2 + 5x )2 - 36
Nhận xét :
( x2 + 5x )2 > 0 với mọi x
=> ( x2 + 5x )2 - 36 > - 36
=> C > - 36
=> C min = - 36
Dấu " = " xảy ra khi : ( x2 + 5x )2 = 0
=> x2 + 5x = 0
=> x ( x + 5 ) = 0
=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy C min = - 36 khi x = 0 hoặc x = - 5
d) D = x2 - 2x + y2 - 4y + 7
= ( x2 - 2x + 1 ) + ( y2 - 4x + 4 ) + 2
= ( x - 1 )2 + ( y - 2 )2 + 2
Nhận xét :
( x - 1 )2 > 0 với mọi x
( y - 2 )2 > 0 với mọi y
=> ( x - 1 )2 + ( y - 2 )2 > 0
=> ( x - 1 )2 + ( y - 2 )2 + 2 > 2
=> D > 2
=> D min = 2
Dấu " = " xảy ra khi : \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)
=> \(\hept{\begin{cases}x-1=0\\y-2=0\end{cases}}\)
=> \(\hept{\begin{cases}x=1\\y=2\end{cases}}\)
Vậy D min = 2 khi x = 1 và y = 2
\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)
a) \(A=x^2-3x+5\)
\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\("="\Leftrightarrow x=5\Rightarrow x=0;5\)
c) \(C=4x-x^2+3\)
\("="\Leftrightarrow x=7\Rightarrow x=2;7\)
d) \(D=x^4+x^2+2\)
\("="\Leftrightarrow x=2\Rightarrow x=0;2\)
\(A=x^2-4x-x\left(x-4\right)-15\)
\(=x^2-4x-x^2+4x-15=-15\) => đpcm
\(B=5x\left(x^2-x\right)-x^2\left(5x-5\right)-13\)
\(=5x^3-5x^2-5x^3+5x^2-13=-13\) => đpcm
\(C=-3x\left(x-5\right)+3\left(x^2-4x\right)-3x+7\)
\(=-3x^2+15x+3x^2-12x-3x+7=7\) => đpcm
\(D=7\left(x^2-5x+3\right)-x\left(7x-35\right)-14\)
\(=7x^2-35x+21-7x^2+35x-14=7\) => đpcm
\(E=4x\left(x^2-7+2\right)-4\left(x^3-7x+2x-5\right)\)
\(=4x^3-20x-4x^3+20x+20=20\) => đpcm
\(H=x\left(5x-3\right)-x^2\left(x-1\right)+x\left(x^2-6x\right)-10+3x\)
\(=5x^2-3x-x^3+x^2+x^3-6x^2-10x+3x=-10\) => đpcm
b/ \(3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow-101x=-303\)
\(\Rightarrow x=3\)
c/ \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-79x=-158\)
\(\Rightarrow x=2\)
d/ \(3\left(3x+2\right)-\left(3x+1\right)=12x+10\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow-6x=5\)
\(\Rightarrow x=-\frac{5}{6}\)
e/ \(30x-6\left(2x-5\right)+5\left(x+8\right)=210+10\left(x-1\right)\)
\(\Leftrightarrow30x-12x+30+5x+40=210+10x-10\)
\(\Leftrightarrow13x=130\)
\(\Rightarrow x=10\)
\(A=x^2-4x+1=\left(x-2\right)^2-3\ge-3\)
\(\Rightarrow A_{min}=-3\) khi \(x=2\)
\(B=4x^2+4x+11=\left(2x+1\right)^2+10\ge10\)
\(\Rightarrow B_{min}=10\) khi \(x=-\frac{1}{2}\)
\(C=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow C_{min}=-36\) khi \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=-x^2-8x-16+21=21-\left(x+4\right)^2\le21\)
\(\Rightarrow C_{max}=21\) khi \(x=-4\)
\(E=-x^2+4x-4+5=5-\left(x-2\right)^2\le5\)
\(\Rightarrow E_{max}=5\) khi \(x=2\)
a, A = (x-1)(x+6) (x+2)(x+3)
= (x^2 + 5x -6 ) (x^2 + 5x + 6)
Đặt t = x^2 +5x
A= (t-6)(t+6)
= t^2 - 36
GTNN của A là -36 khi và ck t= 0
<=> x^2 +5x = 0
<=> x=0 hoặc x=-5
Vậy...
\(N=\left|x-4\right|\left(2-\left|x-4\right|\right)\)
\(=-\left(\left|x-4\right|\right)^2+2\left|x-4\right|\)
\(=-\left[\left(\left|x-4\right|\right)^2-2\left|x-4\right|+1\right]+1\)
\(=-\left(\left|x-4\right|-1\right)^2+1\) \(\le1\)
Dấu = xảy ra \(\Leftrightarrow\left(\left|x-4\right|-1\right)^2=0\Leftrightarrow\left|x-4\right|-1=0\)
\(\Leftrightarrow\left|x-4\right|=1\Leftrightarrow\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Vậy \(Max_N=1\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
\(G=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x\right)^2-25\ge-25\)
Dấu = xảy ra \(\Leftrightarrow x^2+4x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy \(Min_G=-25\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
2. \(Q=\left(x-3\right)\left(4x+5\right)+2019\)
\(Q=4x^2+5x-12x-15+2019\)
\(Q=4x^2-7x+2004\)
\(Q=\left(2x\right)^2-2.2x.\frac{7}{4}+\frac{49}{16}+2019-\frac{49}{16}\)
\(Q=\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\)
\(Do\) \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\) \(Nên\) \(\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\ge\frac{32255}{16}\)
\(\Rightarrow Q\ge\frac{32255}{16}\)
\(Vậy\) \(MinQ=\frac{32255}{16}\Leftrightarrow x=\frac{7}{8}\)
3. \(T=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)
\(T=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\)
\(T=4\left(a^2-ab+b^2\right)-6a^2-6b^2\) (do a+b=1)
\(T=4a^2-4ab+4a^2-6a^2-6b^2\)
\(T=-2a^2-4ab-2b^2\)
\(T=-2\left(a^2+2ab+b^2\right)\)
\(T=-2\left(a+b\right)^2\)
\(T=-2.1^2=-2.1=-2\) (do a+b=1)
\(C=4x^2+3+4x\)
\(C=\left[\left(2x\right)^2+2.2x+1\right]+2\)
\(C=\left(2x+1\right)^2+2\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2+2\ge2\forall x\)
\(C=2\Leftrightarrow\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)
Vậy \(C=2\Leftrightarrow x=-\frac{1}{2}\)