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\(\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(y-1\right)\)
\(=\left(y-1\right)\left[\left(\frac{2}{5}x-\frac{2}{5}y\right)\right]\)
\(=\left(y-1\right)\frac{2}{5}\left(x-y\right)\)
a, \(\frac{6x+1}{x^2+7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(11x^3-31x^2-72x-240=3\left(x+2\right)\left(x+5\right)\left(x-2\right)\)
\(11x^3-31x^2-72x-240-3\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)
\(8x^3-46x^2-60x-180=0\)
=> vô nghiệm
b) \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\left(x\ne0;x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2x}{\left(x-2\right)\left(x+2\right)x}-\frac{\left(x+2\right)\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{x^2+x-2}{x\left(x-2\right)\left(x+2\right)}+\frac{x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x-x^2-x+2+x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3x-6}{x\left(x-2\right)\left(x+2\right)}=0\)
=> 3x-6=0
<=> x=2 (ktm)
Vậy pt vô nghiệm
a) \(\frac{4x-8}{2x^2+1}=0\)
\(\Rightarrow4x-8=0\left(2x^2+1\ne0\right)\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
b)
\(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+2\right)}{x-3}=0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
a) \(9x^2+6x+1=\left(3x+1\right)^2\)
b)\(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
c)\(x^2y^4-2xy^2+1=\left(xy^2-1\right)^2\)
d) \(x^2+\frac{2}{3}x+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2\)
a) 9x2 + 6x + 1 = ( 3x + 1 )2
b) x2 - x + 1/4 = ( x - 1/2)2
c) x2 . y4 - 2xy2 + 1 = ( xy2 - 1 ) 2
d) x2 + 2/3x + 1/9 = (x+1/3)2
\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2-5\right)\left(a^2+5\right)\)
\(\left(a+b\right)^2-1=\left(a+b\right)^2-1^2=\left(a+b-1\right)\left(a+b-1\right)\)
\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b-m+n\right)\left(a+b+m-n\right)\)
\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+3^2\right)\)
\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2+\frac{4}{3}x+\frac{1}{9}\right)\)
Tham khảo~
\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x+3\right)\left(2x-3\right)\)
\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2+5\right)\left(a^2-5\right)\)
\(\left(a+b\right)^2-1=\left(a+b+1\right)\left(a+b-1\right)\)
\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b+m-n\right)\left(a+b-m+n\right)\)
\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)
\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2-\frac{4}{3}x+\frac{1}{9}\right)\)
(x+3)2 =x2+6x+9
(x-1/2)(x+1/2)=x2-1/2
Câu 2 phải là thế này nhé: (x-1/2)(x+1/2)=x2-1/4