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a ) MTC : \(2x\left(x+3\right)\left(x-3\right)\)
\(\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{3-2x}{x^2-9}=\frac{3-2x}{\left(x-3\right)\left(x+3\right)}=\frac{2x\left(3-2x\right)}{2x\left(x+3\right)\left(x-3\right)}\)
b ) MTC : \(2\left(-x\right)\left(x-1\right)^2\)
\(\frac{2x-1}{x-x^2}=\frac{2x-1}{-x\left(x-1\right)}=\frac{2\left(2x-1\right)\left(x-1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
\(\frac{x+1}{2-4x+2x^2}=\frac{x+1}{2\left(x^2-2x+1\right)}=\frac{-x\left(x+1\right)}{2\left(-x\right)\left(x-1\right)^2}\)
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 : \(\left(x+2\right)^3\)
\(\frac{6x^2}{x^3+6x^2+12x+8}=\frac{6x^2}{\left(x+2\right)^3}\)
\(\frac{3x}{x^2+4x+4}=\frac{3x}{\left(x+2\right)^2}=\frac{3x\left(x+2\right)}{\left(x+2\right)^3}\)
\(\frac{2}{2x+4}=\frac{1}{x+2}=\frac{\left(x+2\right)^2}{\left(x+2\right)^3}\)
\(MTC:\left(x-3\right)^2\left(x^2+3x+9\right)\)
\(\frac{x}{x^3-27}=\frac{x}{\left(x-3\right)\left(x^2+3x+9\right)}=\frac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{2x}{x^2-6x+9}=\frac{2x}{\left(x-3\right)^2}=\frac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\frac{1}{x^2+3x+9}=\frac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(MTC:2\left(x-1\right)\left(x+1\right)\)
\(\frac{x-1}{2x+2}=\frac{x-1}{2\left(x+1\right)}=\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{x+1}{2x-2}=\frac{x+1}{2\left(x-1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)
\(\frac{1}{1-x^2}=-\frac{1}{\left(x-1\right)\left(x+1\right)}=-\frac{2}{2\left(x-1\right)\left(x+1\right)}\)
\(MTC:2\left(x+1\right)\left(x^2-x+1\right)\)
\(\frac{1}{x^3+1}=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{2}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{3}{2x+2}=\frac{3}{2\left(x+1\right)}=\frac{3\left(x^2-x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
\(\frac{2}{x^2-x+1}=\frac{4\left(x+1\right)}{2\left(x+1\right)\left(x^2-x+1\right)}\)
a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)
<=> \(6x^2-5x+3-2x+9x-6x^2=0\)
<=> \(2x+3=0\)
<=> \(x=\frac{-3}{2}\)
b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)
<=> \(10x-40-6-4x=20x+4-4x\)
<=> \(6x-46-16x-4=0\)
<=> \(-10x-50=0\)
<=> \(-10\left(x+5\right)=0\)
<=> \(x+5=0\)
<=> \(x=-5\)
c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)
<=> \(8x+9x-15=36x-18-14\)
<=> \(8x+9x-36x=+15-18-14\)
<=> \(-19x=-14\)
<=> \(x=\frac{14}{19}\)
d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
<=> \(12x+10-10x-3=8x+4x+2\)
<=> \(2x-7=12x+2\)
<=> \(2x-12x=7+2\)
<=> \(-10x=9\)
<=> \(x=\frac{-9}{10}\)
e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)
<=> \(x^2-6x-12-\left(x-4^2\right)=0\)
<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)
<=> \(x^2-6x-12-x^2+8x-16=0\)
<=> \(2x-28=0\)
<=> \(2\left(x-14\right)=0\)
<=> x-14=0
<=> x=14
ko bt có sai ko nữa mà mình tìm ra câu a hai nghiệm:\(\frac{-11+\sqrt{69}}{26}\)
và \(\frac{-11-\sqrt{69}}{29}\)
d) \(\frac{1}{2x-3}-\frac{3}{x.\left(2x-3\right)}=\frac{5}{x}\)
\(\Leftrightarrow\frac{x}{x.\left(2x-3\right)}-\frac{3}{x.\left(2x-3\right)}=\frac{5.\left(2x-3\right)}{x.\left(2x-3\right)}\)
\(\Leftrightarrow x-3=5.\left(2x-3\right)\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=\left(-15\right)+3\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow9x=12\)
\(\Leftrightarrow x=12:9\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{\frac{4}{3}\right\}.\)
Chúc bạn học tốt!