Bài 3: 

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

hay \(x\in\left\{0;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)

=>x-1=0

hay x=1

d: \(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

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

a. 9x² - 6x - 3 = 0

<=> 3(3x² - 2x - 1) = 0

<=> 3(3x² - 3x + x - 1) = 0

<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)

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

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

b. (2x + 1)² - 4(x + 2)² = 9

<=> (2x + 1)² - \(\left[2\left(x+2\right)\right]^2=9\)

<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9

<=> -3(4x + 5) = 9

<=> 4x + 5 = -3

<=> 5 + 3 = -4x

<=> -4x = 8

<=> -x = 2

<=> x = -2

a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)

\(\Leftrightarrow\left(3x-1\right)^2-4=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

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

b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)

c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)

*vn:vô nghiệm.

a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

d) \(2x^2+5x-7=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)

b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1

=>-3/4x+1/6x=2+5-4/5=24/5

=>x=-288/35

c: =>6x^2+3x-30x-15=6x^2+10x-21x-35

=>-27x-15=-11x-35

=>-16x=-20

=>x=5/4

a) \(6x^2-72x=0\)

\(6x\left(x-12\right)=0\)

\(6x=0\) hoặc \(x-72=0\)

*) \(6x=0\)

\(x=0\)

*) \(x-12=0\)

\(x=12\)

Vậy \(x=0;x=12\)

b) \(-2x^4+16x=0\)

\(-2x\left(x^3-8\right)=0\)

\(-2x=0\) hoặc \(x^3-8=0\)

*) \(-2x=0\)

\(x=0\)

*) \(x^3-8=0\)

\(x^3=8\)

\(x=2\)

Vậy \(x=0;x=2\)

c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)

\(x^2-5x-x^2+6x-9=0\)

\(x-9=0\)

\(x=9\)

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

\(x^3-6x^2+12x-8-x^3+8=0\)

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

\(-6x\left(x-2\right)=0\)

\(-6x=0\) hoặc \(x-2=0\)

*) \(-6x=0\)

\(x=0\)

*) \(x-2=0\)

\(x=2\)

Vậy \(x=0;x=2\)

2:

a: =>x-1=0 hoặc 3x+1=0

=>x=1 hoặc x=-1/3

b: =>x-5=0 hoặc 7-x=0

=>x=5 hoặc x=7

c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)

d: =>x=0 hoặc x^2-1=0

=>\(x\in\left\{0;1;-1\right\}\)

Bạn tách ra từng câu thoi nhe .

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)