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) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

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

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

a: 3x-15=0

nên 3x=15

hay x=5

b: 4x+20=0

nên 4x=-20

hay x=-5

c: -5x-20=0

nên -5x=20

hay x=-4

bạn tự kết luận nhé ! 

a, \(4x-3=2\left(x-3\right)\Leftrightarrow4x-3=2x-6\)

\(\Leftrightarrow2x=-3\Leftrightarrow x=-\frac{3}{2}\)

b, \(5x^2+x=0\Leftrightarrow x\left(5x+1\right)=0\Leftrightarrow x=-\frac{1}{5};x=0\)

c, \(\left(3x-5\right)\left(x+7\right)=0\Leftrightarrow x=-7;x=\frac{5}{3}\)

d, \(\frac{2}{x-3}-\frac{3}{x+3}=\frac{7x-1}{x^2-9}\)ĐK : \(x\ne\pm3\)

\(\Leftrightarrow\frac{2\left(x+3\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{7x-1}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x+6-3x+9=7x-1\Leftrightarrow-x+15=7x-1\)

\(\Leftrightarrow-8x=-16\Leftrightarrow x=2\)( tmđk )

e, \(\left(12x-1\right)\left(6x-1\right)\left(4x-1\right)\left(3x-1\right)=330\)

\(\Leftrightarrow\left(12x-1\right)\left(12x-2\right)\left(12x-3\right)\left(12x-4\right)=330.24=7920\)

\(\Leftrightarrow\left(12x-1\right)\left(12x-4\right)\left(12x-2\right)\left(12x-3\right)=7920\)

\(\Leftrightarrow\left(144x^2-60x+4\right)\left(144x^2-60x+6\right)=7920\)

Đặt \(144x^2-60x+4=t\)

\(t\left(t+2\right)=7920\Leftrightarrow t^2+2t-7920=0\)

\(\Leftrightarrow\left(t-88\right)\left(t+90\right)=0\Leftrightarrow t=88;t=-90\)

suy ra :TH1 :  \(144x^2-60x+4=88\Leftrightarrow12\left(12x+7\right)\left(x-1\right)=0\Leftrightarrow x=-\frac{7}{12};x=1\)

TH2 : \(144x^2-60x+4=-90\Leftrightarrow144x^2-60x+94=0\)

\(\Leftrightarrow x=\frac{5\pm3\sqrt{39}i}{24}\)

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)

(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0

(3x+2)(x+1)(1-2x)=0

b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

a) ( 3.x + 1 ) . ( 7.x + 3 ) = (5.x-7 ) . ( 3.x + 1 )  

<=> ( 3.x + 1 ) . ( 7.x + 3 ) - ( 5.x - 7) . ( 3.x + 1 ) = 0

<=> ( 3.x + 1 ) . ( 7.x + 3 - 5.x + 7 ) = 0

<=> ( 3.x + 1 ) . ( 2.x + 10 ) = 0

<=> \(\orbr{\begin{cases}3.x+1=0\\2.x+10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-5\end{cases}}}\)

Vậy x = { \(\frac{-1}{3};-5\)} 

b) x² + 10.x + 25 - 4.x . ( x + 5 ) = 0 

<=> ( x + 5 )² -4.x . (x + 5 ) = 0

<=> ( x+ 5 ) . ( x + 5 - 4.x ) = 0

<=> ( x + 5 ) . ( 5 - 3.x )  = 0

<=> \(\orbr{\begin{cases}x+5=0\\5-3.x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{3};-5\right\}\)

c) (4.x - 5 )² - 2. ( 16.x² -25 ) = 0 

<=> ( 4.x-5)² -2 .( 4.x-5) .( 4.x + 5 ) = 0

<=> (  4.x -5 )² - ( 8.x+ 10 ) . ( 4.x -5 ) = 0

<=> ( 4.x -5 ) . ( 4.x-5 - 8.x - 10 ) = 0

<=> ( 4.x - 5 ) . ( -4.x - 15 ) = 0

<=> \(\orbr{\begin{cases}4.x-5=0\\-4.x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{-15}{4}\end{cases}}}\)

Vậy x = \(\left\{\frac{5}{4};\frac{-15}{4}\right\}\)

d) ( 4.x + 3 )² = 4. ( x² - 2.x + 1 ) 

<=> 16.x² + 24.x + 9 - 4.x²  + 8.x - 4 = 0

<=> 12.x² + 32.x + 5 =0 

<=> 12. ( x +\(\frac{1}{8}\) ) . ( x + \(\frac{5}{2}\)) = 0 

<=> \(\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)

Vậy x = \(\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)

e) x² -11.x + 28 = 0

<=> x² -4.x  - 7.x + 28 = 0

<=> ( x - 7 ) . ( x - 4 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)

Vậy x = { 4 ; 7 } 

f ) 3.x.3 - 3.x² - 6.x = 0

<=> 3.x. ( x² -x - 2 ) = 0 

<=> 3.x. ( x - 2 ) . ( x + 1 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

        \([x=0\)                \([x=0\)

( Lưu ý :Lưu ý này không cần ghi vào vở :  Chị nối 2 ý đó làm 1 nha cj ! ) 

Vậy x = { 2 ; -1 ; 0 }