a) \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}+\dfrac{5\left(1+2x\right)}{15}=\dfrac{90}{15}\)

\(\Leftrightarrow3\left(x-3\right)+5\left(1+2x\right)=90\)

\(\Leftrightarrow3x-9+5+10x=90\)

\(\Leftrightarrow13x-4=90\)

\(\Leftrightarrow13x=90+4\)

\(\Leftrightarrow13x=94\)

\(\Leftrightarrow x=\dfrac{94}{13}\)

Vậy \(S=\left\{\dfrac{94}{13}\right\}\)

b) \(\left(2x-3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(tm\right)\\x^2=1\left(ktm\right)\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)

c) \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\)

\(\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow x-5=3x-11\)

\(\Leftrightarrow x-3x=5-11\)

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

\(\Leftrightarrow x=3\)

Vậy \(S=\left\{3\right\}\)

\(\dfrac{x+1}{x+2}-\dfrac{5}{x-2}=\dfrac{20}{4-x^2}\) (\(ĐK:x\)≠\(2;-2\))

⇔ \(\dfrac{\left(x+1\right)\left(x-2\right)-5\left(x+2\right)}{x^2-4}=\dfrac{20}{4-x^2}\)

⇔ \(-\left(x+1\right)\left(x-2\right)+5\left(x+2\right)=20\)

⇔ \(-\left(x^2-2x+x-2\right)+5x+10=20\)

⇔ \(-x^2+x+2+5x+10-20=0\)

⇔ \(-x^2+6x-8=0\)

⇔ \(-\left(x^2-6x+9\right)=-1\)

⇔ \(\left(x-3\right)^2=1\)

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

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

Vậy ...

b: \(\Leftrightarrow20-5\left(3x+2\right)>4\left(x+7\right)\)

=>20-15x-10>4x+28

=>-15x+10-4x-28>0

=>-19x-18>0

=>-19x>18

hay x<-18/19

b) Đặt \(t=x^2\left(t\ge0\right)\). PT trở thành: \(t^2+5t-6=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+6\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\left(TM\right)\\t=-6\left(L\right)\end{matrix}\right.\)

\(\Rightarrow x^2=1\Leftrightarrow x=\pm1\).

Vậy tập nghiệm của PT là \(S=\left\{\pm1\right\}\).

Các câu d, f em làm tương tự nhé! 

3.(2X+3)=-X.(X-2)-1 <=>6X+9=-\(x^2\)+2X-1 <=> \(x^2\) +4x+10=0 (\(\Delta\)' =4-10=-6 nhỏ hơn 0)

pt vô nghiệm

`x^2 -x=12`

`<=>x^2 -x-12=0`

`<=> x^2+3x-4x-12=0`

`<=> x(x+3)-4(x+3)=0`

`<=>(x+3)(x-4)=0`

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

`---`

`2x^2-3x=15-4x`

`<=> 2x^2-3x+4x=15`

`<=>2x^2 +x-15=0`

`<=>2x^2+6x-5x-15=0`

`<=> 2x(x+3)-5(x+3)=0`

`<=>(x+3)(2x-5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

`---`

`x(x-5)=24`

`<=> x^2 -5x-24=0`

`<=>x^2+3x-8x-24=0`

`<=>x(x+3) -8(x+3)=0`

`<=>(x+3)(x-8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)

`----`

`x(x-3)=10(x-4)`

`<=> x^2 -3x =10x -40`

`<=>x^2 -3x-10x +40=0`

`<=> x^2 -13x+40=0`

`<=>x^2-5x-8x+40=0`

`<=> x (x-5) - 8(x-5)=0`

`<=>(x-5)(x-8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=8\end{matrix}\right.\)

5. \(x^2-x=12\Leftrightarrow x^2-x-12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

6. \(2x^2-3x=15-4x\Leftrightarrow2x^2+x-15=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-3\end{matrix}\right.\)

7. \(x\left(x-5\right)=24\Leftrightarrow x^2-5x-24=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

8. \(x\left(x-3\right)=10\left(x-4\right)\Leftrightarrow x^2-3x=10x-40\)

\(\Leftrightarrow x^2-13x+40=0\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

b: \(\Leftrightarrow\left(x^2+3x-1\right)^2+4\left(x^2+3x-1\right)-2\left(x^2+3x-1\right)-8=0\)

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

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

\(\Delta=3^2-4\cdot1\cdot\left(-3\right)=9+12=21>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{21}}{2}\\x_2=\dfrac{-3+\sqrt{21}}{2}\end{matrix}\right.\)

d: \(\Leftrightarrow\left(x^2-3x\right)^2+6\left(x^2-3x\right)+8=3\)

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

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

\(\Delta=\left(-3\right)^2-4\cdot1\cdot1=5>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{5}}{2}\\x_2=\dfrac{3+\sqrt{5}}{2}\end{matrix}\right.\)

a: 3x^2-4x+1=0

a=3; b=-4; c=1

Vì a+b+c=0 nên phương trình có hai nghiệm là:

x1=1 và x2=c/a=1/3

b: -x^2+6x-5=0

=>x^2-6x+5=0

a=1; b=-6; c=5

Vì a+b+c=0 nên phương trình có hai nghiệm là;
x1=1; x2=5/1=5

TRẢ LỜI:

Phương trình bậc hai 5x2 – 6x + 1 = 0

Có: a = 5; b’ = -3; c = 1.; Δ’ = (b’)2 – ac = (-3)2 – 5.1 = 4 > 0

Phương trình có hai nghiệm phân biệt:

Ta có: \(5x^2-6x+1=0\)

a=5; b=-6; c=1

Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=1;x_2=\dfrac{c}{a}=\dfrac{1}{5}\)

Bài 1: 

c) |2x - 1| = x + 2

<=> 2x - 1 = +(x + 2) hoặc -(x + 2)

* 2x - 1 = x + 2      

<=> 2x - x = 2 + 1

<=> x = 3

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

<=> 2x - 1 = x - 2

<=> 2x - x = -2 + 1

<=> x = -1

Vậy.....