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1.
\(x^2\)+\(y^2\)+2y-6x+10=0
=> \(x^2\)-6x+9 +\(y^2\)+2y+1=0
=> (x-3)\(^2\)+(y+1)\(^2\)=0
pt vô nghiệm
4.
=> \(x^2\)+8x+16+(3y)\(^2\)-2.3.2y+4=0
=> (x+4)\(^2\)+(3y-2)\(^2\)=0
pt vô nghiệm
a) \(\left(x-4\right)\left(x+4\right)-x\left(x+2\right)=10\)
<=> \(x^2-16-x^2-2x=10\)
<=> \(-16-2x-10=0\)
<=> \(x=-13\)
Vậy pt có tập nghiệm S\(\)={-13}
b) \(\frac{\left(x+3\right)}{2}-\frac{\left(x-2\right)}{3}=2-\frac{\left(x+3\right)}{2}\)
<=> \(3\left(x+3\right)-2\left(x-2\right)=2.6-3\left(x+3\right)\)
<=> \(3x+9-2x+4=12-3x-9\)
<=> \(3x+9-2x+4-12+3x+9=0\)
<=> \(4x+10=0\)
<=> \(x=\frac{-5}{2}\)
Vậy pt có tập nghiệm S={\(\frac{-5}{2}\)}
a) \(\left(x-4\right)\left(x+4\right)-x\left(x+2\right)=10\)
\(\Leftrightarrow x^2-16-x^2-2x-10=0\)
\(\Leftrightarrow-26=2x\Leftrightarrow x=\frac{-26}{2}=-13\)
b) \(\frac{\left(x+3\right)}{2}-\frac{\left(x-2\right)}{3}=2-\frac{\left(x+3\right)}{2}\)
\(\Leftrightarrow\left(\frac{3\left(x+3\right)-2\left(x-2\right)}{6}\right)=\frac{12-3\left(x+3\right)}{6}\)
\(\Leftrightarrow3x+9-2x+4=12-3x-9\)
\(\Leftrightarrow x+13=-3x+3\)
\(\Leftrightarrow x+3x=-13+3\)
\(\Leftrightarrow4x=-10\Leftrightarrow x=\frac{-10}{4}=-2,5\)
c,3a2−6ab+3b2−12c2=3(a2−2ab+b2−4c2)=3.((a−b)2−(2c)2)
=3(a−b−2c).(a−b+2c)
d,x2−25+y2−2xy=(x2−2xy+y2)−52=(x−y)2−52
=(x−y+5)(x−y−5)
e,a2+2ab+b2−ac−bc=(a+b)2−c(a+b)=(a+b)(a+b−c)
ƒ ,x2−2x−4y2−4y=(x2−4y2)−(2x+4y)=(x−2y)(x+2y)−2(x+2y)
=(x+2y)(x−2y−2)
h,x2(x−1)+16(1−x)=x2(x−1)−16(x−1)=(x−1)(x2−16)=
=(x−1)(x−4)(x+4)
\(=x^2-y^2+9x-9y\)
\(=\left(x-y\right)\left(x+y\right)+9\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+9\right)\)