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\(\left(12x-5\right)\left(3x-1\right)-36x^2=6\)
\(36x-12x-15x-5-36x^2=6\)
\(-12x-15x-5=6\)
\(-27x=5+6\)
\(x=11:\left(-27\right)\)
\(x=-\frac{11}{27}\)
chuk bn hok giỏi
a) \(x^2y+2xy+y=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)
b) \(4x^2-4xy-6y^2+6xy=4x\left(x-y\right)+6y\left(x-y\right)=\left(x-y\right)\left(4x+6y\right)\)
\(=2\left(x-y\right)\left(2x+3y\right)\)
c) \(18x^5y+18x^3y-2x^3y^5-2xy^5=18x^3y\left(x^2+1\right)-2xy^5\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(18x^3y-2xy^5\right)=2xy\left(x^2+1\right)\left(9x^2-y^4\right)=2xy\left(x^2+1\right)\left(3x-y^2\right)\left(3x+y^2\right)\)
d)
d) \(-12x^5-12x^3y-3xy^2+36x^4+36x^2y+9y^2=-3x\left(4x^4+4x^2y+y^2\right)+9y\left(4x^4+4x^2y+y^2\right)\)\(=\left(4x^4+4x^2y+y^2\right)\left(9-3x\right)\)
\(36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow-12x+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
36x2-12x-36x2+27x=30
=>(36x2-36x2)-12x+27x=30
=>0-12x+27x=30
=>-12x+27x=30
=>15x=30
=>x=30:15
=>x=2
\(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
đkxđ \(x\ne\pm\frac{1}{3}\)
\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow\frac{\left(24x+2\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{\left(36x-20\right)\left(3x-1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}=\frac{-36x^2+10x-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow72x^2+6x+24x+2-108x^2+60x+36x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\left(tmđk\right)\)
\(\left(36x^2+12x+1\right)+\left(36x^2-12x+1\right)-2\left(1+6x\right)\left(1-6x\right)\)
\(=36x^2+12x+1+36x^2-12x+1-2\left(1-36x^2\right)\)
\(=72x^2+2-2+72x^2=144x^2\)
Chúc bạn học tốt!!!
\(A=\dfrac{12x^2+36x+39}{4x^2+12x+11}\\ =\dfrac{3\left(4x^2+12x\right)+39}{4x^2+12x+11}\\ =\dfrac{3.\left(2x+3\right)^2+6+6}{\left(2x+3\right)^2+2}\\ =\dfrac{3\left(\left(2x+3\right)^2+2\right)+6}{\left(2x+3\right)^2+2}\\ =3+\dfrac{6}{\left(2x+3\right)^2+2}\le6\)
Dấu "=" xảy ra khi x=-3/2
Vậy ...
\(A=\dfrac{12x^2+36x+39}{4x^2+12x+11}=\dfrac{\left(12x^2+36x+33\right)+6}{4x^2+12x+11}\)
\(=\dfrac{3\left(4x^2+12x+11\right)+6}{4x^2+12x+11}\)
\(=3+\dfrac{6}{4x^2+12x+11}\)
Ta có:
\(4x^2+12x+11=\left(4x^2+12x+9\right)+2\)
\(=\left(2x+3\right)^2+2\)
Vì \(\left(2x+3\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+3\right)^2+2\ge2\forall x\)
\(\Rightarrow4x^2+12x+11\ge2\forall x\)
\(\Rightarrow\dfrac{6}{4x^2+12x+11}\le3\forall x\)
\(\Rightarrow3+\dfrac{6}{4x^2+12x+11}\le6\forall x\)
\(\Rightarrow A\le6\forall x\)
Dấu "="xảy ra khi \(\left(2x+3\right)^2=0\)
\(\Rightarrow2x+3=0\)
\(\Rightarrow x=-\dfrac{3}{2}\)
Vậy \(A_{max}=6\) tại \(x=-\dfrac{3}{2}\)