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\(x^2-10x+16\)
\(=\left(x^2-2x\right)-\left(8x-16\right)\)
\(=x.\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x-8\right)\)
Tham khảo nhé~
\(x^2+4x-y^2+4\)
\(=\left(x^2+2.x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right).\left(x+2+y\right)\)
Tham khảo nhé~
\(x^2+4x-y^2+4\)
\(=x^2+4x+4-y^2\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x^2+2x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left[\left(x+2\right)+y^2\right].\left[\left(x+2\right)-y^2\right]\)
\(=\left(x+2+y^2\right)\left(x+3-y^2\right)\)
a: \(3x^2+y^2+10x-2xy+26=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(2x^2+10x+\dfrac{5}{2}\right)+\dfrac{47}{2}=0\)
\(\Leftrightarrow\left(x-y\right)^2+2\cdot\left(x+\dfrac{5}{2}\right)^2+\dfrac{47}{2}=0\)(vô lý)
b: \(\Leftrightarrow3x^2-12x+12+6y^2-20y+\dfrac{50}{3}+\dfrac{34}{3}=0\)
\(\Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\)(vô lý)
\(x^2-2x-4y^2-4y=\left(x^2-4y\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x-2y-2\right)\left(x+2y\right)\)
\(b,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
\(c,x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
f) \(x^2-6x+5=\left(x^2-x\right)+\left(-5x+5\right)=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
g) \(x^4+64=\left(x^2+4x+8\right)\left(x^2-4x+8\right)\)
\(x^2-6x+5\)
\(=\left(x^2-2.3x+3^2\right)-4\)
\(=\left(x-3\right)^2-2^2\)
\(=\left(x-3-2\right)\left(x-3+2\right)\)
\(=\left(x-5\right)\left(x-1\right)\)
c/ Ta có:
\(x^2-3xy+x-3y\)
\(=x^2+x-3xy-3y\)
\(=x\left(x+1\right)-3y\left(x+1\right)\)
\(=\left(x+1\right)\left(x-3y\right)\)
d/ Ta có:
\(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^2-3xy+x-3y\)
\(=x\left(x-3y\right)+\left(x-3y\right)\)
\(=\left(x+1\right)\left(x-3y\right)\)
\(x^3-x^2-5x+125\) k có nghiệm
\(x^3-4x^2+4x-1\)
\(=x^3-x^2-3x^2+3x+x-1\)
\(=x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-3x+1\right)\)
Ta co:
\(x^2+x-2011.2012=x^2+x-\left(2012-1\right).2012\)
\(=x^2+x-2012^2+2012=\left(x^2-2012^2\right)+\left(x+2012\right)\) \(=\left(x+2012\right)\left(x-2012\right)+\left(x+2012\right)=\left(x+2012\right)\left(x-2012+1\right)\)
\(=\left(x+2012\right)\left(x-2011\right)\)
Đề ghi sai tùm lum sao giải được em?