Tìm GTNN của 3x²+6x+8
\(A=\left(3x^2+6x+3\right)+5=3\left(x^2+2x+1\right)+5=3\left(x+1\right)^2+5\ge5\)
\(A_{min}=5\) khi \(x=-1\)
5: \(=\dfrac{x^3-x+2x-x-1+x^2-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{x^3+x^2-2}{\left(x+1\right)\left(x-1\right)}\)
1-3x/2x + 3x-2/2x-1 + 2-3x/4x2-2x
\(2x;2x-1;4x^2-2x=2x\left(2x-1\right)\)
\(MTC=2x\left(2x-1\right)\)
\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{2-3x}{4x^2-2x}\)
\(=\dfrac{\left(1-3x\right).2x\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{\left(3x-2\right).2x}{\left(2x-1\right).2x}+\dfrac{2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{2x\left(1-3x\right)\left(2x-1\right)+2x\left(2x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{-8x^2+4x+4x^2-4x+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2-3x+2}{2x\left(2x-1\right)}\)
#AEZn8
\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{2-3x}{4x^2-2x}=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}+\dfrac{2-3x}{2x\left(2x-1\right)}=\dfrac{-6x^2+5x-1}{2x\left(2x-1\right)}+\dfrac{6x^2-4x}{2x\left(2x-1\right)}+\dfrac{2-3x}{2x\left(2x-1\right)}=\dfrac{\left(-6x^2+6x^2\right)+\left(5x-4x-3x\right)+\left(-1+2\right)}{2x\left(2x-1\right)}=\dfrac{-2x}{2x\left(2x-1\right)}=\dfrac{-1}{2x-1}\)
\(=\dfrac{-6x^2+5x-1+6x^2-4x-3x+2}{2x\left(2x-1\right)}=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}=\dfrac{-1}{2x}\\ x=\dfrac{1}{534}\Leftrightarrow BT=\dfrac{-1}{2\cdot\dfrac{1}{534}}=-267\)
Mn ơi, giúp em với ạ, em ko có vt lộn đề
hình như là sai rồi hay sao ý, phải là 4 chứ sao lại 4x^2
Thực hiện phép tính sau:
\(\dfrac{3}{2x^2+y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(\dfrac{3}{2x^2+y}+\dfrac{5}{xy^2+}+\dfrac{x}{y^3}\)
=\(\dfrac{3xy^5}{xy^2.y^3\left(2x^2+y\right)+}+\dfrac{10y^3x^2+5y^4}{xy^2.y^3\left(2x^2+y\right)}+\dfrac{2x^4y^2+x^2y^3}{xy^2.y^3\left(2x^2+y\right)}\)
=\(\dfrac{3xy^5+10y^3x^2+5y^4+2x^4y^2+x^2y^3}{xy^5\left(2x^2+y\right)}\)
=\(\dfrac{3xy^5+11y^3x^2+5y^4+2x^4y^2}{xy^5\left(2x^2+y\right)}\)
ủa đáp án cứ sao sao:<