\(3x^2+12x-66=0\)

\(\Rightarrow3\left(x^2+4x+4\right)-78\) = 0

\(\Rightarrow3\left(x+2\right)^2=78\)

\(\Rightarrow\left(x+2\right)^2=26\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{26}-2\\x=-\sqrt{26}-2\end{matrix}\right.\)

Vậy ..........

\(3x^2+12x-66=0\)

=>\(3\left(x+2\right)^2-12-66=0\)

=>\(3\left(x+2\right)^2-78=0\)

=>\(3\left(x+2\right)^2=78\)

=>\(\left(x+2\right)^2=26\)

=>\(x=\sqrt{26}-2\)

\(\Leftrightarrow3\left(x^2+4x-22\right)=0\)

\(\Leftrightarrow x^2+4x-22=0\)

\(\Leftrightarrow\left(x+2\right)^2-26=0\)

\(\Leftrightarrow\left(x+2+\sqrt{26}\right)\left(x+2-\sqrt{26}\right)=0\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{26}\\x=-2-\sqrt{26}\end{matrix}\right.\)

1.\(3x^2+12x-66=0\)

\(\Rightarrow\)\(3\left(x^2+4x+4\right)-78=0\)

\(\Rightarrow3\left(x+2\right)^2=78\)

\(\Rightarrow\left(x+2\right)^2=26\)

\(\Rightarrow x+2=\sqrt{26}\)hoặc \(x+2=-\sqrt{26}\)

\(\Rightarrow x=\sqrt{26}-2\)hoặc \(x=-\sqrt{26}-2\)

a)

\(3x^2+12x-66=0\)

\(\Leftrightarrow x^2+4x-22=0\)

\(\Leftrightarrow x^2+4x+4=26\Leftrightarrow (x+2)^2=26\)

\(\Rightarrow x+2=\pm \sqrt{26}\Rightarrow x=-2\pm \sqrt{26}\)

b)

\(9x^2-30x+225=0\)

\(\Leftrightarrow (3x)^2-2.3x.5+25+200=0\)

\(\Leftrightarrow (3x-5)^2=-200< 0\) (vô lý nên pt vô nghiệm)

c)

\(x^2+3x-10=0\)

\(\Leftrightarrow x^2-2x+5x-10=0\)

\(\Leftrightarrow x(x-2)+5(x-2)=0\Leftrightarrow (x+5)(x-2)=0\)

\(\Rightarrow x=-5\) hoặc $x=2$

d)

$3x^2-7x+1=0$

$\Leftrightarrow 3(x^2-\frac{7}{3}x)+1=0$

$\Leftrightarrow 3(x^2-\frac{7}{3}x+\frac{7^2}{6^2})=\frac{37}{12}$

$\Leftrightarrow 3(x-\frac{7}{6})^2=\frac{37}{12}$
$\Leftrightarrow (x-\frac{7}{6})^2=\frac{37}{36}$

$\Rightarrow x-\frac{7}{6}=\frac{\pm \sqrt{37}}{6}$

$\Rightarrow x=\frac{7\pm \sqrt{37}}{6}$

e)

$3x^2+7x+2=0$

$\Leftrightarrow 3(x^2+\frac{7}{3}x+\frac{7^2}{6^2})=\frac{25}{12}$

$\Leftrightarrow 3(x+\frac{7}{6})^2=\frac{25}{12}$

$\Leftrightarrow (x+\frac{7}{6})^2=\frac{25}{36}$

$\Rightarrow x+\frac{7}{6}=\pm \frac{5}{6}$

$\Rightarrow x=\frac{-1}{3}$ hoặc $x=-2$

f)

$4x^2-12x+9=0$

$\Leftrightarrow (2x)^2-2.2x.3+3^2=0$

$\Leftrightarrow (2x-3)^2=0\Rightarrow 2x-3=0\Rightarrow x=\frac{3}{2}$

g) Trùng câu e

h)

$x^2-4x+1=0$

$\Leftrightarrow x^2-4x+4-3=0$

$\Leftrightarrow (x-2)^2=3\Rightarrow x-2=\pm \sqrt{3}$

$\Rightarrow x=2\pm \sqrt{3}$

i)

$2x^2-6x+1=0$

$\Leftrightarrow 2(x^2-3x+\frac{3^2}{2^2})=\frac{7}{2}$

$\Leftrightarrow 2(x-\frac{3}{2})^2=\frac{7}{2}$

$\Leftrightarrow (x-\frac{3}{2})^2=\frac{7}{4}$

$\Rightarrow x-\frac{3}{2}=\pm \frac{\sqrt{7}}{2}$

$\Rightarrow x=\frac{3\pm \sqrt{7}}{2}$

j)

$3x^2+4x-4=0$

$\Leftrightarrow 3x^2+6x-2x-4=0$

$\Leftrightarrow 3x(x+2)-2(x+2)=0$

$\Leftrightarrow (x+2)(3x-2)=0$

$\Rightarrow x+2=0$ hoặc $3x-2=0$

$\Rightarrow x=-2$ hoặc $x=\frac{2}{3}$

a)

a) $3x^2+12x-66=0$

=> 3(x + 2)² - 12 - 66 = 0

=> 3(x + 2)² - 78 = 0

=> 3(x + 2)² = 78

=> (x + 2)² = 26

=> x = \(\sqrt{26}-2\)

b)$9x^2-30x+225=0$

=> (3x - 5)² - 25 + 225 = 0

=> (3x - 5)² + 200 = 0

=> (3x - 5)² = -200

9x² - 30x + 225 không có ngiệmc)$x^2+3x-10=0$=> (x + 1,5)² - 2,25 - 10 = 0

=> (x + 1,5)² - 12,25 = 0

=> (x + 1,5)² = 12, 25

=> x + 1,5 = 3,5

=> x = 2

d)$3x^2-7x+1=0$=> 3(x - \(\dfrac{7}{6}\))² - \(\dfrac{49}{12}\) + 1 = 0

=> 3(x - \(\dfrac{7}{6}\))² - \(\dfrac{37}{12}\) = 0

=> 3(x - \(\dfrac{7}{6}\))² = \(\dfrac{37}{12}\)

=> (x - \(\dfrac{7}{6}\))² = \(\dfrac{37}{36}\)

=> x = \(\dfrac{\sqrt{37}}{6}+\dfrac{7}{6}=\dfrac{\sqrt{37}+7}{6}\)

e) $3x^2-7x+8=0$

=> 3(x - \(\dfrac{7}{6}\))² - \(\dfrac{49}{12}\)+ 8 = 0

=> 3(x - \(\dfrac{7}{6}\))² + \(\dfrac{47}{12}\) = 0

=> 3(x - \(\dfrac{7}{6}\))² = \(-\dfrac{47}{12}\)

KL : Không có ngiệm

a) \(3x^2+12x-66=0\)

Ta có \(\Delta=12^2+4.3.66=936,\sqrt{\Delta}=6\sqrt{26}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-12+6\sqrt{26}}{6}=-2+\sqrt{26}\\x=\frac{-12-6\sqrt{26}}{6}=-2-\sqrt{26}\end{cases}}\)

b) \(9x^2-30x+225=0\)

Ta có \(\Delta=33^2-4.9.225=-7011\)

\(\Delta< 0\)nên pt vô nghiệm

c) \(x^2+3x-10=0\)

Ta có \(\Delta=3^2+4.10=49,\sqrt{\Delta}=7\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-3+7}{2}=2\\x=\frac{-3-7}{2}=-5\end{cases}}\)

d) \(3x^2-7x+1=0\)

Ta có \(\Delta=7^2-4.3.1=37,\sqrt{\Delta}=\sqrt{37}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+\sqrt{37}}{6}\\x=\frac{7-\sqrt{37}}{6}\end{cases}}\)