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a) \(\frac{3x+6}{x^2-4}=\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{3}{x-2}\)( ĐKXĐ : x ≠ ±2 )
\(\frac{2x+6}{x^3+3x^2-9x-27}=\frac{2\left(x+3\right)}{x^2\left(x+3\right)-9\left(x+3\right)}=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x^2-9\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)( ĐKXĐ : x ≠ ±3 )
MTC : ( x - 2 )( x - 3 )( x + 3 )
=> \(\hept{\begin{cases}\frac{3}{x-2}=\frac{3\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3\left(x^2-9\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{3x-27}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\\\frac{2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{4x-4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\end{cases}}\)
b) \(\frac{x^2-4x+4}{2x^2-3x+1}=\frac{\left(x-2\right)^2}{2x^2-2x-x+1}=\frac{\left(x-2\right)^2}{2x\left(x-1\right)-\left(x-1\right)}=\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}\)( ĐKXĐ : \(\hept{\begin{cases}x\ne1\\x\ne\frac{1}{2}\end{cases}}\))
\(\frac{x+4}{2x-2}=\frac{x+4}{2\left(x-1\right)}\)( ĐKXĐ : x ≠ 1 )
MTC : \(2\left(x-1\right)\left(2x-1\right)\)
=> \(\hept{\begin{cases}\frac{\left(x-2\right)^2}{\left(x-1\right)\left(2x-1\right)}=\frac{2\left(x^2-4x+4\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2-8x+8}{2\left(x-1\right)\left(2x-1\right)}\\\frac{x+4}{2\left(x-1\right)}=\frac{\left(x+4\right)\left(2x-1\right)}{2\left(x-1\right)\left(2x-1\right)}=\frac{2x^2+7x-4}{2\left(x-1\right)\left(2x-1\right)}\end{cases}}\)
c) \(\frac{6a}{a-b}\)( ĐKXĐ : a ≠ b ) ; \(\frac{2b}{b-a}=\frac{-2b}{a-b}\)( ĐKXĐ : a ≠ b) ; \(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)( ĐKXĐ : a ≠ ±b )
MTC : \(\left(a-b\right)\left(a+b\right)\)
=> \(\frac{6a}{a-b}=\frac{6a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{6a^2+6ab}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{-2b}{a-b}=\frac{-2b\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}=\frac{-2ab-2b^2}{\left(a-b\right)\left(a+b\right)}\)
\(\frac{5}{a^2-b^2}=\frac{5}{\left(a-b\right)\left(a+b\right)}\)
d) \(\frac{x}{x^2+11x+30}=\frac{x}{x^2+5x+6x+30}=\frac{x}{x\left(x+5\right)+6\left(x+5\right)}=\frac{x}{\left(x+5\right)\left(x+6\right)}\)( ĐKXĐ : x ≠ -5 ; x ≠ -6 )
\(\frac{5}{x^2+9x+20}=\frac{5}{x^2+4x+5x+20}=\frac{5}{x\left(x+4\right)+5\left(x+4\right)}=\frac{5}{\left(x+4\right)\left(x+5\right)}\)( ĐKXĐ : x ≠ -4 ; x ≠ -5 )
MTC : \(\left(x+4\right)\left(x+5\right)\left(x+6\right)\)
=> \(\hept{\begin{cases}\frac{x}{\left(x+5\right)\left(x+6\right)}=\frac{x\left(x+4\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{x^2+4x}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\\\frac{5}{\left(x+4\right)\left(x+5\right)}=\frac{5\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}=\frac{5x+30}{\left(x+4\right)\left(x+5\right)\left(x+6\right)}\end{cases}}\)
Sai chỗ nào bạn bỏ qua nhé
a) \(5x-20\le0\\ \Leftrightarrow5x\le20\\ \Leftrightarrow x\le\frac{20}{5}\\ \Leftrightarrow x\le4\)
b)\(3x+7>-7x+2\\ \Leftrightarrow3x+7x>2-7\\ \Leftrightarrow10x>-5\\ \Leftrightarrow x>-\frac{5}{10}\\ \Leftrightarrow x>-\frac{1}{2}\)
c)\(-9x-5< 4x+21\\ \Leftrightarrow-9x-4x< 21+5\\ \Leftrightarrow-13x< 26\\ \Leftrightarrow x>\frac{-26}{13}\\ \Leftrightarrow x>-2\)
d) 3(2x-5) >8x-13
<=> 6x -15> 8x-13
<=> 6x-8x>-13+15
<=>-2x>2
<=> x< -2/2
<=>x<-1
e) \( 2(3x-5) ≥ 5(2x+6)\\ \Leftrightarrow6x-10\ge10x+30\\ \Leftrightarrow6x-10x\ge30+10\\ \Leftrightarrow-4x\ge40\\ \Leftrightarrow x\le-\frac{40}{4}\\ \Leftrightarrow x\le-10\)
f) \(\frac{2x-3}{4}\le\frac{3x+1}{6}\\ \Leftrightarrow\frac{3.\left(2x-3\right)}{12}\le\frac{2.\left(3x+1\right)}{12}\\ \Leftrightarrow6x-9\le6x+2\\ \Leftrightarrow6x-6x\le2+9\\ \Leftrightarrow0x\le11\)
=>Luôn đúng => Bpt vô số nghiệm
h. Ta có : \(2x+3\left(x-3\right)\ge10x-\left(3x+2\right)\)
=> \(2x+3x-9\ge10x-3x-2\)
=> \(2x+3x-9-10x+3x+2\ge0\)
=> \(-2x-7\ge0\)
=> \(x\ge-\frac{7}{2}\)
\(a,2x-6< 0\Leftrightarrow2x>6\Leftrightarrow x>3\)
\(b,5x+2x< 4+25\Leftrightarrow7x< 29\Leftrightarrow x< \frac{29}{7}\)
\(c,-5x+6>8-10+8x\Leftrightarrow-5x-8x>8-10-6\)
\(-13x>-8\Leftrightarrow x< \frac{8}{13}\)
\(d,3x-12\le2-4x\Leftrightarrow3x+4x\le2+12\)
\(\Leftrightarrow7x\le14\Leftrightarrow x\le2\)
\(e,\frac{3\left(x-3\right)}{6}>\frac{2\left(2x-5\right)}{6}+\frac{6}{6}\Rightarrow3x-9>4x-10+6\)
\(\Leftrightarrow3x-4x>-4+9\Leftrightarrow x>-5\)
\(f,3\left(2x-3\right)>1+2\left(2+2x\right)\Leftrightarrow6x-9>1+4+4x\)
\(6x-4x>14\Leftrightarrow2x>14\Leftrightarrow x>7\)
Tự biểu diễn nha!
a) \(6x^2+6\)
\(=6\left(x^2+1\right)\)
b) \(2x^2-18\)
\(=2\left(x^2-9\right)\)
\(=2\left(x-3\right)\left(x+3\right)\)
c) \(3x^2-3xy+4x-4y\)
\(=\left(3x^2-3xy\right)+\left(4x-4y\right)\)
\(=3x\left(x-y\right)+4\left(x-y\right)\)
\(=\left(3x-4\right)\left(x-y\right)\)
a) \(\left(x^3-9x^2+27x-27\right)\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^3\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^2\)
c) \(\frac{x^2-4}{2x}:\frac{3x-6}{6}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{2x}.\frac{6}{3\left(x-2\right)}\)
\(=\frac{\left(x+2\right)}{x}\)