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\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
\(A=x^2+10x-37\)
\(=\left(x+5\right)^2-62\)
Có \(\left(x+5\right)^2\ge0\forall x\in R\)
\(\Rightarrow\left(x+5\right)^2-62\ge-62\forall x\in R\)
Dấu = xảy ra \(\Leftrightarrow x+5=0\Leftrightarrow x=-5\)
Vậy A đạt GTNN là -62 tại x=-5
1) a) Đặt biểu thức là A
\(A=2x^2+4y^2-4xy-4x-4y+2017\)
\(A=\left(x-2y\right)^2+x^2-4x-4y+2017\)
\(A=\left(x-2y\right)^2+2\left(x-2y\right)+x^2-6x+2017\)
\(A=\left(x-2y-1\right)^2+\left(x+3\right)^2+2008\)
Vậy: MinA=2008 khi x=-3; y=-2
3) a) \(A=\dfrac{1}{x^2+x+1}\)
\(B=x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(\Rightarrow B\ge\dfrac{3}{4}\Rightarrow A\ge\dfrac{4}{3}\)
Vậy MinA là \(\dfrac{4}{3}\) khi x=-0,5
a) MTC : \(\left(x+1\right)\left(x^2-x+1\right)\)
Quy đồng :
\(\frac{x-1}{x^3+1}=\frac{x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2x}{x^2-x+1}=\frac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x+1}=\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
b ) MTC : \(10x\left(2y-x\right)\left(2y+x\right)\)
\(\frac{7}{5x}=\frac{7.2.\left(2y-x\right)\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=\frac{-4.10x.\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}=\frac{-40x\left(2y+x\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{10x\left(2y-x\right)\left(2y+x\right)}\)
c ) MTC : \(\left(x+2\right)^3\)
\(\frac{6x^2}{x^3+6x^2+12x+8}=\frac{6x^2}{\left(x+2\right)^3}\)
\(\frac{3x}{x^2+4x+4}=\frac{3x}{\left(x+2\right)^2}=\frac{3x\left(x+2\right)}{\left(x+2\right)^3}\)
\(\frac{2}{2x+4}=\frac{1}{x+2}=\frac{\left(x+2\right)^2}{\left(x+2\right)^3}\)
1 a) Tìm giá trị nhỏ nhất của biểu thức:
\(A=x^2-6x+2\)
\(=\left(x-3\right)^2-7\ge-7\)
\(B=4x^2-x+2\)
\(=4\left(x-\frac{1}{8}\right)^2+\frac{31}{16}\ge\frac{31}{16}\)
\(C=4x^2+2y^2+4xy-4x-6y+2019\)
\(=4x^2+4x\left(y-1\right)+\left(y-1\right)^2+y^2-4y+4+2014\)
\(=\left(2x+y-1\right)^2+\left(y-2\right)^2+2014\ge2014\)
\(D=\frac{-3}{x^2-6x+10}\)
\(=\frac{-3}{\left(x-3\right)^2+1}\ge3\)