Bài 3:
`a)(2x-1)²-4x²+1=0`
`⇔(2x-1)²-(4x²-1)=0`
`⇔(2x-1)²-[(2x)²-1²]=0`
`⇔(2x-1)²-(2x+1)(2x-1)=0`
`⇔(2x-1)[(2x-1)-(2x+1)]=0`
`⇔(2x-1)(2x-1-2x-1)=0`
`⇔(2x-1).(-2)=0`
`⇔2x-1=0`
`⇔2x=1`
`⇔x=1/2`
Vậy `x=1/2`
`b)6x³-24x=0`
`⇔6x(x²-4)=0`
`⇔6x(x²-2²)=0`
`⇔6x(x+2)(x-2)=0`
`⇔`$\left[\begin{matrix} 6x=0\\ x+2=0\\x-2=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=0\\ x=-2\\x=2\end{matrix}\right.$
Vậy `x∈{0;-2;2}`
`c)2x(x-3)-4x+12=0`
`⇔2x(x-3)-(4x-12)=0`
`⇔2x(x-3)-4(x-3)=0`
`⇔(x-3)(2x-4)=0`
`⇔2(x-3)(x-2)=0`
`⇔(x-3)(x-2)=0`
`⇔`$\left[\begin{matrix} x-3=0\\ x-2=0\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=3\\ x=2\end{matrix}\right.$
Vậy `x∈{3;2}`
`d)x³-5x²+x-5=0`
`⇔(x³-5x²)+(x-5)=0`
`⇔x²(x-5)+(x-5)=0`
`⇔(x-5)(x²+1)=0`
Ta có:`x²≥0∀x`
`⇒x²+1≥1>0∀x`
`⇒` vô nghiệm
`⇔x-5=0`
`⇔x=5`
Vậy `x=5`
Bài 4:
`a)A=x²+x+5`
`=x²+x+1/4+19/4`
`=(x²+x+1/4)+19/4`
`=[x²+2.x. 1/2+(1/2)^2]+19/4`
`=(x+1/2)^2+19/4`
Ta có:`(x+1/2)^2≥0∀x`
`⇒(x+1/2)^2+19/4≥19/4∀x`
Vậy `A_(min)=19/4` khi `x+1/2=0⇔x=-1/2`
`b)B=x²+5y²-4xy+6x-14y+15`
`=x^2+y^2+4y^2-4xy+6x-12y-2y+1+9+5`
`=(x^2-4xy+4y^2)+(6x-12y)+9+(y^2-2y+1)+5`
`=(x-2y)²+6(x-2y)+9+(y-1)²+5`
`=[(x-2y)²+6(x-2y)+9]+(y-1)²+5`
`=[(x-2y)²+2.(x-2y).3+3²]+(y-1)²+5`
`=(x-2y+3)²+(y-1)²+5`
Ta có:`(x-2y+3)²≥0∀x,y`
`(y-1)²≥0∀y`
`⇒(x-2y+3)²+(y-1)²≥0∀x,y`
`⇒(x-2y+3)²+(y-1)²+5≥5∀x,y`
Dấu `'='` xảy ra khi$\begin{cases} x-2y+3=0\\y-1=0 \end{cases}$`⇔`$\begin{cases} x=-1\\y=1\end{cases}$
Vậy `B_(min)=5` khi `x=-1` và `y=1`