a. x² - 6x = -9

<=> x² - 6x + 9 = 0

<=> (x - 3)² = 0

<=> x - 3 = 0

<=> x = 3

b. 2(x + 3) - x² + 3x = 0

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

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

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

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

a, <=> x² -2x +1 + 5x -x² =8

<=> 3x +1 =8 

<=> 3x = 7

<=> x= 7/3

b, thiếu đề

c, <=> 2x³ -1 + 2x(4 -x²) = 7

<=> 2x³ + 8x -2³ = 8

<=> 8x =8

<=> x=1

b)(x-2)3-(x-3)(x2+3x+9)+6(x+1)2=49

(=) x³- 6x² +12 x -8 - ( x3 - 27 ) + 6( x² + 2x +1)

(=) x³ - 6x² +12x -8 - x³ +27 + 6x² +12x +6

(=) 24x + 25 = 49

(=) 24x = 49 - 25 = 24

(=) x = 24/24 =1

a: Ta có: \(4x\left(x-7\right)-4x^2=56\)

\(\Leftrightarrow4x^2-7x-4x^2=56\)

hay x=-8

b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)

\(\Leftrightarrow36x^2-24x-4+6x=0\)

\(\Leftrightarrow36x^2-18x-4=0\)

\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)

c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)

a) x³ - 3x² + 3x - 1 = 0

<=>(x-1)³=0

<=>x-1=0

<=>x=1

b) x³+6x² + 12x+8 =0

<=>(x+2)³=0

<=>x+2=0

<=>x=-2

c) (x-2)³+6(x+1)²-x³+9=0

<=>x³-6x²+12x-8+6x²+12x+6-x³+9=0

<=>24x+7=0

<=>24x=-7

<=>x=-7/24

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

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

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)

\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

a. \(\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\)

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

\(\Leftrightarrow2\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b: Ta có: \(4x^2-9=0\)

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

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

c: Ta có: \(x^2+6x+9=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

\(a,x^3+3x^2+3x=0\)

\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)

\(\Leftrightarrow x=0\) Vì \(x^2+3x+3>0\forall x\)

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

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

\(\Leftrightarrow x=0\)

\(c,\) bạn làm tương tự nha

c, x^3 + 6x^2 + 12x = 0

=> x(x^2 + 6x + 12) = 0

=> x(x^2 + 6x + 9 + 3) = 0

=> x[(x + 3)^2 + 3) = 0

=> x = 0 hoặc (x + 3)^2 + 3 = 0

=> x = 0 hoặc (x + 3)^2 = -3 (loại vì (x+3)^2 > 0)

vậy x = 0

a, x^3 + 3x^2 + 3x = 0

=> x(x^2 + 3x + 3) = 0

=>x(x^2 + 3x + 2,25 + 0,75) = 0

=> x[(x + 1,5)^2 + 0,75)] = 0

=> x = 0 hoặc (x + 1,5)^2 + 0,75 = 0

=> x = 0 hoặc (x + 1,5)^2 = -0,75 (loại)

vậy x = 0

b, x^3 - 3x^2 + 3x = 0

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

=> x(x^2 - 3x + 2,25 + 0,75) = 0

=> x[(x - 1,5)^2 + 0,75] = 0

=> x = 0 hoặc (x-1,5)^2 + 0,75 = 0 

=> x = 0 hoặc (x - 1,5)^2 = -0,75 (loại) 

vậy x = 0

a. 2x\(^2\)-8=0

2x\(^2\)=8

x\(^2\)=4

x=2

b.3x\(^3\)-5x=0

x(3x\(^2\)-5)=0

\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)

c.x+3x-4=0\(^{\left(\cdot\right)}\)

đặt t=x\(^2\) (t>0)

ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)

thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm

t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4

khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1

khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2

vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2

d)3x\(^2\)+6x-9=0

thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm

x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)

e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)  (ĐK: x#5; x#2 )

⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)

⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0

⇔-7x\(^2\) - 6x + 46=0

Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0

\(\sqrt{\Delta'}=\sqrt{62}\)

vậy pt có 2 nghiệm phân biệt

x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)

x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)

vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......

câu g làm tương tự câu c