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) \(3x^2+12x-66=0\)

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

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

tải photomath về bn

thank you nha 

Tui chưa nháp nhưng câu 1 thử nhân hết ra coi triệt tiêu bớt đc ko, mà chắc chắn là nhân ra sẽ mất cái 27x^3 rồi nên thành pt bậc 2 giải vô tư nhé, câu 2 tách hết ra cx lm đc vì nó là pt bậc 2 

câu 3 tách thành (x+3)(x^2-7x+9)=0 có pt bậc 2 nên ok r

(3x - 2)(9x² + 6x + 4) - (3x - 1)(9x² - 3x + 1) = x - 4

<=> 27x³ - 8 - 27x³ + 1 =  x - 4

<=> x - 4 = -7

<=> x = -3

Vậy S = {-3}

9(2x + 1) = 4(x - 5)²

<=> 4(x² - 10x + 25) - 18x - 9 = 0

<=>4x² - 40x + 100 - 18x - 9 = 0

<=> 4x² - 58x + 91 = 0

<=> (4x² - 58x + 210,25) - 119,25 = 0

<=> (2x - 14,5)² = 119,25

<=> \(\orbr{\begin{cases}2x-14,5=\sqrt{119,25}\\2x-14,5=-\sqrt{119,25}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

Vậy S = {...}

x³ - 4x²  - 12x + 27 = 0

<=> (x³ + 3x²) - (7x² + 21x) + (9x + 27) = 0

<=> x²(x + 3) - 7x(x + 3) + 9(x + 3) = 0

<=> (x² - 7x + 9)(x + 3) = 0

<=> \(\orbr{\begin{cases}x-7x+9=0\\x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x^2-7x+12,25\right)-3,25=0\\x=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-3,5\right)^2=3,25\\x=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}x-3,5=\sqrt{3,25}\\x-3,5=-\sqrt{3,25}\end{cases}}\)

hoặc x = -3

<=>  \(\orbr{\begin{cases}x=\frac{7+\sqrt{13}}{2}\\x=\frac{7-\sqrt{13}}{2}\end{cases}}\)

hoặc x = -3

Vậy S = {...}