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A) \(A=-3x^2+x+1\)
\(A=-3\left(x^2-\dfrac{1}{3}x-\dfrac{1}{3}\right)\)
\(A=-3\left(x^2-2\cdot\dfrac{1}{6}\cdot x+\dfrac{1}{36}-\dfrac{13}{36}\right)\)
\(A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\)
Mà: \(-3\left(x-\dfrac{1}{6}\right)^2\le0\forall x\)
\(\Rightarrow A=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\le\dfrac{13}{12}\forall x\)
Dấu "=" xảy ra khi:
\(x-\dfrac{1}{6}=0\Rightarrow x=\dfrac{1}{6}\)
Vậy: \(A_{max}=\dfrac{13}{12}.khi.x=\dfrac{1}{6}\)
B) \(B=2x^2-8x+1\)
\(B=2\left(x^2-4x+\dfrac{1}{2}\right)\)
\(B=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)
\(B=2\left(x-2\right)^2-7\)
Mà: \(2\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow B=2\left(x-2\right)^2-7\ge-7\forall x\)
Dấu "=" xảy ra khi:
\(x-2=0\Rightarrow x=2\)
Vậy: \(B_{min}=2.khi.x=2\)
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x^2 - 4x + 4 = 5 ( x - 2 )
x^2 - 4x + 4 - 5 ( x - 2 ) = 0
( x - 2 ) ^2 - 5 ( x - 2 ) = 0
( x - 2 ) ( x - 2 - 5 ) = 0
( x - 2 ) ( x - 7 ) = 0
x - 2 = 0 hoặc x - 7 = 0
x = 2 hoặc x = 7
c) (x-2)^2=5(x-2)
=> x-2=5 hoặc x-2 =0
=> x=7 hoặc x=2
d) (2x-3)^2=(5-x)^2
=> 2x-3=5-x
=> x=8/3
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\(A=\frac{x^2}{x^4+x^2+1}\)
\(\Rightarrow\)\(3A=\frac{3x^2}{x^4+x^2+1}=\frac{x^4+x^2+1-x^4+2x^2-1}{x^4+x^2+1}\)
\(=\frac{\left(x^4+x^2+1\right)-\left(x^2-1\right)^2}{x^4+x^2+1}=1-\frac{\left(x^2-1\right)^2}{x^4+x^2+1}\le1\)
\(\Rightarrow\)\(A\le\frac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=\pm1\)
Vậy Max A = 1/3 <=> \(x=\pm1\)
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\(\left(x-3\right)\left(1-x\right)-2=-x^2+4x-3-2=-x^2+4x-5=-\left(x^2-4x+4\right)-1=-\left(x-2\right)^2-1\le-1\)
\(ĐTXR\Leftrightarrow x=2\)
\(\left(x-3\right)\left(1-x\right)-2=4x-x^2-1=-\left(x^2-4x+4\right)+3=-\left(x-2\right)^2+3\le3\)
Dấu \("="\Leftrightarrow x=2\)
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\(B=\frac{x^2}{\left(x+2000\right)^2}\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
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a) \(\left(x+2\right)^2=4\left(2x-1\right)^2\)
\(\left(x+2\right)^2-4\left(2x-1\right)^2=0\)
\(\left(x+2\right)^2-\left[2\left(2x-1\right)\right]^2=0\)
\(\left(x+2\right)^2-\left(4x-2\right)^2=0\)
\(\left(x+2-4x+2\right)\left(x+2+4x-2\right)=0\)
\(6x\left(-3x+4\right)=0\)
\(\Rightarrow6x=0\) hoặc \(-3x+4=0\)
*) \(6x=0\)
\(x=0\)
*) \(-3x+4=0\)
\(3x=4\)
\(x=\dfrac{4}{3}\)
Vậy \(x=0;x=\dfrac{4}{3}\)
b) \(4x\left(x-2019\right)-x+2019=0\)
\(4x\left(x-2019\right)-\left(x-2019\right)=0\)
\(\left(x-2019\right)\left(4x-1\right)=0\)
\(\Rightarrow x-2019=0\) hoặc \(4x-1=0\)
*) \(x-2019=0\)
\(x=2019\)
*) \(4x-1=0\)
\(4x=1\)
\(x=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{4};x=2019\)
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Bài 1 :
a, \(A=x\left(x-6\right)+10\)
=x^2 - 6x + 10
=x^2 - 2.3x+9+1
=(x-3)^2 +1 >0 Với mọi x dương