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Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
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...
\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
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/ \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left[\left(x+1\right)\left(x-6\right)\right].\left[\left(x-2\right)\left(x-3\right)\right]\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
Suy ra Min A = -36 <=> \(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b/ \(B=19-6x-9x^2=-9\left(x-\frac{1}{3}\right)^2+20\le20\)
Suy ra Min B = 20 <=> x = 1/3
a) \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)
\(\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\)
Vì \(\left(x^2-5x\right)^2\ge0\)
=> \(\left(x^2-5x\right)^2-36\ge-36\)
Vậy GTNN của A là -36 khi \(x^2-5x=0\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b) \(B=19-6x-9x^2=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\)
Vì \(-\left(3x+1\right)^2\le0\)
=> \(-\left(3x+1\right)+20\le20\)
Vậy GTLN của B là 20 khi \(x=-\frac{1}{3}\)
a) \(A=\left(x-3\right)\left(x+5\right)+20\)
\(\Leftrightarrow A=x^2+5x-3x-15+20\)
\(\Leftrightarrow A=x^2+2x+5\)
\(\Leftrightarrow A=x^2+2x+1+4\)
\(\Leftrightarrow A=\left(x+1\right)^2+4\ge4\)
GTNN của A = 4
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy ..........................
c, đề : \(C=x^2+2x+1\) đước ko chị ?
\(A=\left(x-3\right)\left(x+5\right)+20\)
\(=x^2+5x-3x-15+20\)
\(=x^2+2x+5=x^2+2x+1+4\)
\(=\left(x+1\right)^2+4\ge4\)
Dấu = xảy ra \(< =>x=-1\)
Vậy \(A_{mịn}=4\)khi \(x=-1\)
b) \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(\Leftrightarrow B=\left(x^2+2x-x-2\right)\left(x^2+6x+3x+18\right)\)
\(\Leftrightarrow B=\left(x^2+x-2\right)\left(x^2+9x+18\right)\)
\(\Leftrightarrow B=x^4+9x^3+18x^2+x^3+9x^2+18x-2x^2-18x-36\)
\(\Leftrightarrow B=x^4+10x^3+25x^2-36\)
\(\Leftrightarrow B=x^2\left(x^2+10x+25\right)-36\)
\(\Leftrightarrow B=x^2\left(x+5\right)^2-36\ge-36\)
GTNN của B = -36
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}}\)
Vậy .............
c) \(C=x^2+x+1\)
\(\Leftrightarrow C=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(\Leftrightarrow C=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
GTNN của \(C=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy .................
\(C=x^2+x+1=x^2+2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra khi và chỉ khi \(x=-\frac{1}{2}\)
Vậy \(C_{min}=-\frac{1}{2}\)đạt được khi \(x=-\frac{1}{2}\)
Bài giải
\(a,\text{ }A=\left(x-3\right)\left(x+5\right)+20\)
\(A=x^2-3x+5x-15+20\)
\(A=x^2+2x+5=x^2+2x+1+4\)
\(A=\left(x+1\right)^2+4\ge4\)
Dấu " = " xảy ra khi :
\(\left(x+1\right)^2=0\text{ }\Rightarrow\text{ }x+1=0\text{ }\Rightarrow\text{ }x=-1\)
\(\Rightarrow\text{ }Min_A=4\text{ khi }x=-1\)
\(b,\text{ }B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(B=\left(x^2-x+2x-2\right)\left(x^2+3x+6x+18\right)\)
\(B=\left(x^2+x-2\right)\left(x^2+9x+18\right)\)
\(B=x^4+x^3-2x^2+9x^3+9x^2-18x+18x^2+18x-36\)
\(B=x^4+10x^3+25x^2-36\)
\(B=x^2\left(x^2+10x+25\right)-36=x^2\left(x+5\right)^2-36\ge-36\)
Dấu " = " xảy ra khi \(x^2\left(x+5\right)^2=0\)\(\Rightarrow\orbr{\begin{cases}x^2=0\\\left(x+5\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Vậy \(Min_B=-36\text{ khi }x=0\text{ hoặc }x=-5\)
\(c,\text{ }C=x^2+x+1\)
\(=x^2+2\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu " = " xảy ra khi \(\left(x+\frac{1}{2}\right)^2=0\text{ }\Rightarrow\text{ }x+\frac{1}{2}=0\text{ }\Rightarrow\text{ }x=-\frac{1}{2}\)
Vậy \(Min_C=\frac{3}{4}\text{ khi }x=-\frac{1}{2}\)
A = ( x - 3 )( x + 5 ) + 20
A = x2 + 2x - 15 + 20
A = x2 + 2x + 1 + 4
A = ( x + 1 )2 + 4
\(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+4\ge4\)
Dấu " = " xảy ra <=> x + 1 = 0 => x = -1
Vậy MinA = 4 , đạt được khi x = -1
B = ( x - 1 )( x + 2 )( x + 3 )( x + 6 )
B = [( x - 1 )( x + 6 )][( x + 2 )( x + 3 )]
B =( x2 + 5x - 6 )( x2 + 5x + 6 )
B = ( x2 + 5x )2 - 36
\(\left(x^2+5x\right)^2\ge0\forall x\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
Dấu " = " xảy ra <=> x2 + 5x = 0
<=> x( x + 5 ) = 0
<=> x = 0 hoặc x = -5
Vậy MinB = -36, đạt được khi x = 0 hoặc x = -5
C = x2 + x + 1
C = x2 + 2.1/2.x + 1/4 + 3/4
C = ( x + 1/2 )2 + 3/4
\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu " = " xảy ra <=> x + 1/2 = 0 => x = -1/2
Vậy MinC = 3/4 , đạt được khi x = -1/2