Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 :
a, \(A=x^2-4x+6=x^2-4x+4+2=\left(x-2\right)^2+2\ge2\)
Dấu ''='' xảy ra khi x = 2
Vậy GTNN A là 2 khi x = 2
b, \(B=y^2-y+1=y^2-2.\frac{1}{2}y+\frac{1}{4}+\frac{3}{4}=\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu ''='' xảy ra khi y = 1/2
Vậy GTNN B là 3/4 khi y = 1/2
c, \(C=x^2-4x+y^2-y+5=x^2-4x+4+y^2-y+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-2\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu ''='' xảy ra khi \(x=2;y=\frac{1}{2}\)
Vậy GTNN C là 3/4 khi x = 2 ; y = 1/2
Bài 3 :
a, \(x^2-6x+10=x^2-2.3.x+9+1=\left(x-3\right)^2+1\ge1>0\)( đpcm )
b, \(-y^2+4y-5=-\left(y^2-4y+5\right)=-\left(y^2-4y+4+1\right)=-\left(y-2\right)^2-1< 0\)( đpcm )
Bài 4 :
\(B=\left(x^2+y^2\right)=\left(x+y\right)^2-2xy\)
Thay (*) ta được : \(225-2\left(-100\right)=225+200=425\)
Bài 5 :
\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)\)
\(=2y.2x=4xy=VP\)( đpcm )
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Bài 1
A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2
Bài 1:
a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)
\(A=2x^2-x-4x+2-2x^2-6x\)
\(A=-11x+2\)
b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)
\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)
\(B=-12x\)
c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)
\(C=12x^2+18x-12x^2+8x+3x-2\)
\(C=29x-2\)
d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)
\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)
\(D=36x-10\)
2: a) \(x^2-6x+2018\)
\(=\left(x^2-6x+9\right)+2009\)
\(=\left(x-3\right)^2+2009\)
Vì \(\left(x-3\right)^2\ge0\forall x\) ; \(2009>0\) nên \(\left(x-3\right)^2+2009>0\forall x\)
Hay \(x^2-6x+2018>0\forall x\) \(\left(dpcm\right)\)
b) \(4x-x^2-5\)
\(=-\left(x^2-4x+5\right)\)
\(=-\left[\left(x^2-4x+4\right)+1\right]\)
\(=-\left(x-2\right)^2-1\)
Vì \(\left(x-2\right)^2\ge0\forall x\) nên \(-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2-1< 0\)
Hay \(4x-x^2-5< 0\forall x\) \(\left(dpcm\right)\)
Bài 3:
\(A=2\left(x^2-3x\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)\)
\(=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Đẳng thức xảy ra khi x = 3/2
\(B=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+1\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+1\ge1\)
Đẳng thức xảy ra khi x = 1/2 y = -3