\(\Leftrightarrow x\left(5x^2-7x+5x-7\right)=0\\ \Leftrightarrow x\left[5x\left(x+1\right)+7\left(x+1\right)\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\5x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=\dfrac{7}{5}\end{matrix}\right.\)

\(\Leftrightarrow5x^3+5x^2-7x^2-7x=0\)

\(\Leftrightarrow5x^2\left(x+1\right)-7x\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(5x^2-7x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\5x^2-7x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

Đặt `x^2=t (t>=0)`, có:

`2t^2+5t+2=0`

`\Delta = 5^2-4.2.2=9>0`

`=>` PT có 2 nghiệm:

`t_1=-1/2 (KTM)`

`t_2=-2 (KTM)`

Vậy PTVN.

(4x - 3)² - (2x + 1)² = 0

\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0

\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy ...

3x - 12 - 5x(x - 4) = 0

\(\Leftrightarrow\) 3x - 12 - 5x² + 20x = 0

\(\Leftrightarrow\) -5x² + 23x - 12 = 0

\(\Leftrightarrow\) 5x² - 23x + 12 = 0

\(\Leftrightarrow\) 5x² - 20x - 3x + 12 = 0

\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0

\(\Leftrightarrow\) (x - 4)(5x - 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy ...

(8x + 2)(x² + 5)(x² - 4) = 0

\(\Leftrightarrow\) (8x + 2)(x² + 5)(x - 2)(x + 2) = 0

Vì x² \(\ge\) 0 \(\forall\) x nên x² + 5 > 0 \(\forall\) x

\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)

Vậy ...

Chúc bn học tốt!

a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)

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

\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)

b) Ta có: \(3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)

c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)

mà \(2>0\)

và \(x^2+5>0\forall x\)

nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)

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

Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)

c: =>(x+2)(x+3)(x-5)(x-6)=180

=>(x^2-3x-10)(x^2-3x-18)=180

=>(x^2-3x)^2-28(x^2-3x)=0

=>x(x-3)(x-7)(x+4)=0

=>\(x\in\left\{0;3;7;-4\right\}\)

c: =>(x-3)(x+2)(2x+1)(3x-1)=0

=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

\(a.-3x^2+15x=0\)

\(\Leftrightarrow3x\left(-x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\-x+5=0\end{matrix}\right.\)

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

\(b.2x^2-32=0\)

\(\Leftrightarrow2x^2=32\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow\left|x\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(c.2x^2-5x+1=0\)

\(a=2;b=-5;c=1\)

\(\Delta=\left(-5\right)^2-4.2.1=17>0\)

Do \(\Delta>0\) nên phương trình có 2 nghiệm phân biệt:

\(x_1=\dfrac{5+\sqrt{17}}{4}\)

\(x_2=\dfrac{5-\sqrt{17}}{4}\)

\(a,-3x^2+15x=0\\ -3x\left(x-5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\end{matrix}\right.\) 

\(b,\\ 2\left(x^2-16\right)=0\\ \Leftrightarrow x^2-16=0\\ \Leftrightarrow\left(x-4\right)\left(x+4\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\) 

\(c,\\ \Delta=5^2-4.2=17\\ \Rightarrow x_1,x_2=\dfrac{\Delta\pm b}{2ac}\\ =\dfrac{5\pm\sqrt{17}}{4}\)

Phân tích nhân tử ra được \(\left(x^2-x-3\right)\left(2x^2+7x-6\right)=0\)

Giải 2 pt này ra có 4 nghiệm 

\(x\in\left\{\frac{1}{2}\pm\frac{\sqrt{13}}{2};-\frac{7}{4}\pm\frac{\sqrt{97}}{4}\right\}\)

MK làm lại câu b hồi nãy mk chép nhầm đề :))

b) / 2x + 1/ - / 5x - 2/ = 3 ( 1)

Lập bảng xét dấu , ta có :

+) Với : x < \(\dfrac{-1}{2}\) , ta có :

( 1) ⇔ - 2x - 1 + 5x - 2 = 3

⇔ 3x = 6

⇔ x = 2 ( KTM)

+) Với : \(\dfrac{-1}{2}\) ≤ x < \(\dfrac{2}{5}\) , ta có :

( 1) ⇔ 2x + 1 + 5x - 2 = 3

⇔ 7x = 4

⇔ x = \(\dfrac{4}{7}\) ( KTM)

+) Với : x ≥ \(\dfrac{2}{5}\) , ta có :

( 1) ⇔ 2x + 1 - 5x + 2 = 3

⇔ -3x = 0

⇔ x = 0 ( KTM)

Vậy , phương trình đã cho vô nghiệm

a)\(\left|1+4x\right|-\left|7x-2\right|=0\)

\(\left|1+4x\right|=\left|7x-2\right|\\\Leftrightarrow\left[{}\begin{matrix}1+4x=7x-2\\1+4x=-\left(7x-2\right)\end{matrix}\right.\)

TH1:

\(1+4x=7x-2\\ \Leftrightarrow4x-7x=-2-1\\ \Leftrightarrow-3x=-3\\ \Leftrightarrow x=1\)

TH2:

\(1+4x=-\left(7x-2\right)\\ \Leftrightarrow1+4x=-7x+2\\\Leftrightarrow4x+7x=2-1\\ \Leftrightarrow11x=1\\ \Leftrightarrow x=\dfrac{1}{11} \)

Vậy tập nghiệm của phương trình: S={1;\(\dfrac{1}{11}\)}

a) Cách khác :

/ 1 + 4x / - / 7x - 2/ = 0

⇔ / 1 + 4x/ = / 7x - 2/

Bình phương hai vế của phương trình , ta được :

( 1 + 4x)² = ( 7x - 2)²

⇔ ( 1 + 4x)² - ( 7x - 2)² = 0

⇔ ( 1 - 4x - 7x + 2)( 1 + 4x + 7x - 2) = 0

⇔ ( 3 - 3x)( 11x - 1) = 0

⇔ x = 1 hoặc x = \(\dfrac{1}{11}\)

KL.....