Question

Answer 1

Bài 2:

a: 5x(x-2000)-x+2000=0

=>5x(x-2000)-(x-2000)=0

=>(x-2000)(5x-1)=0

=>\(\left[\begin{array}{l}x-2000=0\\ 5x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2000\\ x=\frac15\end{array}\right.\)

b: \(5x^2=13x\)

=>\(5x^2-13x=0\)

=>x(5x-13)=0

=>\(\left[\begin{array}{l}x=0\\ 5x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac{13}{5}\end{array}\right.\)

c: \(4x^2-49=0\)

=>\(4x^2=49\)

=>\(x^2=\frac{49}{4}\)

=>\(\left[\begin{array}{l}x=\frac72\\ x=-\frac72\end{array}\right.\)

d: \(x^2-36=0\)

=>(x-6)(x+6)=0

=>\(\left[\begin{array}{l}x-6=0\\ x+6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6\\ x=-6\end{array}\right.\)

e: \(x^2-6x+8=0\)

=>\(x^2-2x-4x+8=0\)

=>x(x-2)-4(x-2)=0

=>(x-2)(x-4)=0

=>\(\left[\begin{array}{l}x-2=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=4\end{array}\right.\)

f: \(9x^2+6x-8=0\)

=>\(9x^2+12x-6x-8=0\)

=>3x(3x+4)-2(3x+4)=0

=>(3x+4)(3x-2)=0

=>\(\left[\begin{array}{l}3x+4=0\\ 3x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-4\\ 3x=2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac43\\ x=\frac23\end{array}\right.\)

Bài 8:

a: \(x^2-10x+16\)

\(=x^2-10x+25-9\)

\(=\left(x-5\right)^2-9\ge-9\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(10x-23-x^2\)

\(=-x^2+10x-25+2\)

\(=-\left(x-5\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

c: \(x^2+5x-6\)

\(=x^2+5x+\frac{25}{4}-\frac{49}{4}\)

\(=\left(x+\frac52\right)^2-\frac{49}{4}\ge-\frac{49}{4}\forall x\)

Dấu '=' xảy ra khi \(x+\frac52=0\)

=>\(x=-\frac52\)

d: \(-x^2-x+1\)

\(=-\left(x^2+x-1\right)\)

\(=-\left(x^2+x+\frac14-\frac54\right)\)

\(=-\left(x+\frac12\right)^2+\frac54\le\frac54\forall x\)

Dấu '=' xảy ra khi \(x+\frac12=0\)

=>\(x=-\frac12\)

e: \(2x^2-x-3\)

\(=2\left(x^2-\frac12x-\frac32\right)\)

\(=2\left(x^2-2\cdot x\cdot\frac14+\frac{1}{16}-\frac{25}{16}\right)\)

\(=2\left(x-\frac14\right)^2-\frac{25}{8}\ge-\frac{25}{8}\forall x\)

Dấu '=' xảy ra khi \(x-\frac14=0\)

=>\(x=\frac14\)

g: \(-3x^2+x+2\)

\(=-3\left(x^2-\frac13x-\frac23\right)\)

\(=-3\left(x^2-2\cdot x\cdot\frac16+\frac{1}{36}-\frac{25}{36}\right)=-3\left(x-\frac16\right)^2+\frac{25}{12}\le\frac{25}{12}\forall x\)

Dấu '=' xảy ra khi \(x-\frac16=0\)

=>\(x=\frac16\)