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Bài 1:
\(a,\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^2\left(x^2-1\right)\)
\(b,\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)^3\)
\(=x^6+27-27-27x^2-9x^4-x^6\)
\(=-9x^2\left(3-x^2\right)\)
Bài 5:
\(A=x^2-2x+1\)
\(=\left(x^2-2x+1\right)-2\)
\(=\left(x-1\right)^2-2\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)
Vậy Min A = -2
Để A = -2 thì \(x-1=0\Rightarrow x=1\)
b, \(B=4x^2+4x+5\)
\(=\left(4x^2+4x+1\right)+4\)
\(=\left(2x+1\right)^2+4\)
Với mọi giá trị của x ta có:
\(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+4\ge4\)
Vậy Min B = 4
Để B = 4 thì \(2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)
c, \(C=2x-x^2-4\)
\(=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Với mọi giá trị của x ta có:
\(\left(x-1\right)^2\ge0\Rightarrow-\left(x-1\right)^2\le0\Rightarrow-\left(x-1\right)^2-3\le-3\)Vậy Max C = -3
để C = -3 thì \(x-1=0\Rightarrow x=1\)
mk chỉ làm bài 1 và 1 câu bài 2 vi no tuong duong
1. x+x +2 = 86
x = số thứ nhất = 42
x+2 = số t2 = 44
2.a) x2-6x +10 = (x-3)2 +1 >0 với mọi x
(vì (x-3)2 >= 0)
a, \(A=5x-x^2=-x^2+5x=-x^2+2x\cdot2,5-\dfrac{25}{4}+\dfrac{25}{4}\)
\(=-\left(x-2,5\right)^2+\dfrac{25}{4}\)
Có: \(-\left(x-2,5\right)^2\le0\forall x\)
=> \(-\left(x-2,5\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
''='' xảy ra khi \(x-2,5=0\Rightarrow x=2,5\)
Vậy \(A_{MAX}=\dfrac{25}{4}\Leftrightarrow x=2,5\)
b, \(B=x-x^2=x^2-x=x^2-2\cdot x\cdot\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Lập luận như câu a
c, \(C=4x-x^2+3=-x^2+2\cdot x\cdot2-4+7\)
\(=-\left(x-2\right)^2+7\)
Vì \(-\left(x-2\right)^2\le0\forall x\)
=> \(-\left(x-2\right)^2+7\le7\)
Dấu ''='' xảy ra khi và chỉ khi x = 2
Vậy \(C_{MAX}=7\Leftrightarrow x=2\)
d, \(D=-x^2+6x-11=-x^2+2\cdot x\cdot3-9-2\)
\(=-\left(x-3\right)^2-2\)
Vì \(-\left(x-3\right)^2\le0\forall x\)
=> \(-\left(x-3\right)^2-2\le-2\)
Dấu ''='' xảy ra khi và chỉ khi x - 3 = 0 => x = 3
Vậy \(D_{MAX}=-2\Leftrightarrow x=3\)
e, \(E=5-8x-x^2=-x^2-8x+5=-x^2-2\cdot x\cdot4-16+21\)
\(=-\left(x+4\right)^2+21\)
Lập luận như trên
f, \(F=4x-x^2+1=-x^2+4x+1=-x^2+2\cdot x\cdot2-4+5\)
\(=-\left(x-2\right)^2+5\)
Tượng tự mấy ý trc
a) \(x^2-6x+3\)
\(=x^2-2.x.3+9-6\)
\(=\left(x-3\right)^2-\left(\sqrt{6}\right)^2\)
\(=\left(x-3-\sqrt{6}\right)\left(x-3+\sqrt{6}\right)\)
b) \(9x^2+6x-8\)
\(=\left(3x\right)^2+2.3x+1-9\)
\(=\left(3x+1\right)^2-3^2\)
\(=\left(3x+1-3\right)\left(3x+1+3\right)\)
\(=\left(3x-2\right)\left(3x+4\right)\)
d) \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
e) \(x^3+4x^2-29x+24\)
\(=x^3+8x^2-4x^2-32x+3x+24\)
\(=x^2\left(x+8\right)-4x\left(x+8\right)+3\left(x+8\right)\)
\(=\left(x+8\right)\left(x^2-4x+3\right)\)
\(=\left(x+8\right)\left(x^2-3x-x+3\right)\)
\(=\left(x+8\right)\left[x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+8\right)\left(x-3\right)\left(x-1\right)\)
a: \(9x^2-6x+3\)
\(=\left(9x^2-6x+1\right)+2\)
\(=\left(3x-1\right)^2+2\ge2\)
b: \(6x-x^2+1\)
\(=-\left(x^2-6x-1\right)\)
\(=-\left(x^2-6x+9-10\right)\)
\(=-\left(x-3\right)^2+10\le10\)
Bài 2:
a: \(A=-3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)\)
\(=-3\left(x-\dfrac{2}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Dấu '=' xảy ra khi x=2/3
b: \(B=-x^2+5x+3\)
\(=-\left(x^2-5x-3\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{37}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)
Dấu '=' xảy ra khi x=5/2
a) \(x^3-\dfrac{1}{9}x=0\)
\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(x\left(x-3\right)+x-3=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)
\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Rightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\Rightarrow x=y\\2-x+y=0\Rightarrow x-y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\left(1\right)\\\left(1\right)\Rightarrow x-x=2\left(loại\right)\end{matrix}\right.\)
d) \(x^2\left(x-3\right)+27-9x=0\)
\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)
\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)
\(\Rightarrow x-3=0\Rightarrow x=3.\)
\(x^2< 2x\)
\(\Leftrightarrow-2< x< 2\)
\(\Leftrightarrow x\in\left\{-1;0;1\right\}\)
k viết đề nữa nhé bn ,mấy bài này cũng dẽ mà
A=\(\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)
Vậy....................................
\(B=\left(2x\right)^2+2.2x+1+2016=\left(2x+1\right)^2+2016\ge2016\)
Vậy...............................
C=\(x^2-2.3x+3^2=\left(x-3\right)^2\ge0\)
Vậy.......................
Thank you !