Ta có: \(\left(2x+1\right)\left(5x-1\right)=20x^2-16x-1\)

\(\Leftrightarrow10x^2-2x+5x-1-20x^2+16x+1=0\)

\(\Leftrightarrow-10x^2+19x=0\)

\(\Leftrightarrow x\left(-10x+19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-10x+19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-10x=-19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{10}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{19}{10}\right\}\)

a: \(\Leftrightarrow8x^3-4x+27=8x^3+8x^2+12x^2+12x+18x+18\)

\(\Leftrightarrow8x^3+20x^2+30x+18=8x^3-4x+27\)

\(\Leftrightarrow20x^2+34x-9=0\)

hay \(x\in\left\{\dfrac{-17+\sqrt{469}}{20};\dfrac{-17-\sqrt{469}}{20}\right\}\)

b: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1=10x^2+3x-1\)

\(\Leftrightarrow10x^2-19x=0\)

=>x=0 hoặc x=19/10

a: \(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x+3\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2-8x+36x-24-\left(12x^2-3x+12x-3\right)=-27\)

\(\Leftrightarrow12x^2+28x-24-12x^2-9x+3=-27\)

\(\Leftrightarrow19x-21=-27\)

=>19x=-6

hay x=-6/19

b: \(\left(x+1\right)\left(3x^2-x+1\right)+x^2\left(4-3x\right)=\dfrac{5}{2}\)

\(\Leftrightarrow3x^3-x^2+x+3x^2-x+1+4x^2-3x^3=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2+1=\dfrac{5}{2}\)

\(\Leftrightarrow6x^2=\dfrac{3}{2}\)

\(\Leftrightarrow x^2=\dfrac{3}{12}=\dfrac{1}{4}\)

=>x=1/2 hoặc x=-1/2

c: \(\Leftrightarrow2\left(x^2-4\right)-4\left(x^2-x-2\right)+\left(5x+8\right)\left(x+2\right)=0\)

\(\Leftrightarrow2x^2-8-4x^2+4x+8+5x^2+10x+8x+16=0\)

\(\Leftrightarrow3x^2+22x+16=0\)

\(\text{Δ}=22^2-4\cdot3\cdot16=292>0\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{-22-2\sqrt{73}}{6}=\dfrac{-11-\sqrt{73}}{3}\\x_2=\dfrac{-11+\sqrt{73}}{3}\end{matrix}\right.\)

d: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1\)

\(\Leftrightarrow10x^2-19x=0\)

=>x(10x-19)=0

=>x=0 hoặc x=19/10

\(A=\dfrac{3x^2+9x+17}{3x^2+9x+7}=1+\dfrac{10}{3x^2+9x+7}=1+\dfrac{10}{3\left(x^2+2.x.\dfrac{9}{2}+\dfrac{81}{4}\right)-\dfrac{215}{4}}\\ =1+\dfrac{10}{3\left(x+\dfrac{9}{2}\right)^2-\dfrac{215}{4}}\le\dfrac{35}{43}\)

Câu khác giải TT

#)Giải :

Câu 1 :

5x(1 - 2x ) - 3x ( x+18) = 0 

<=> 5x - 10x^2 - 3x^2 - 54x = 0 

<=> -13x^2 - 49x = 0 

<=> x= 0 hoặc x = - 49/13

Vậy x có hai giá trị là 0 và - 49/13

#)Giải :

Câu 2 :

( 12x - 5 )( 4x - 1 ) + ( 3x - 7 )( 1 - 16x ) = 81

<=> 48x² - 32x + 5 - 48x + 115x - 7 = 81

<=> 83x - 2 = 81

<=> x = 1

Vậy x = 1

\(\dfrac{2}{x^2-x-6}+\dfrac{x+1}{x^2+x-12}=\dfrac{x}{x^2+6x+8}\)

\(\Leftrightarrow\dfrac{2}{\left(x-3\right)\left(x+2\right)}+\dfrac{x+1}{\left(x-3\right)\left(x+4\right)}=\dfrac{x}{\left(x+2\right)\left(x+4\right)}\)

=> 2(x+4)+(x+1)(x+2)=x(x-3)

⇔2x+8+x²+2x+x+2=x²-3x

⇔x²+5x+10=x²-3x

⇔x²-x²+5x+3x=-10

⇔8x=-10

\(\Leftrightarrow\dfrac{-5}{4}\)

Vậy S={-\(\dfrac{5}{4}\)}

a/ 3x(12x-4)-9x(4x-3)

=36x²-12x-36x²+27x

=(36x²-36x²)-12x+27x

=15x

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

=5x-2x²+2x²-2x

=(5x-2x)-(-2x²+2x²)

=3x

c/ 5x(12x+7)-3x(20x-5)

=60x²+35x-60x²+15x

=(60x²-60x²)+(35x+15x)\

=50x

d/ 3x(2x-7)+2x(5-3x)

=6x²-21x+10x-6x²

=(6x²-6x²)+(10x-21x)

=-11x

e/ đề sai hay sao ý ra số to lắm @@

giải

5x-(4-2x+x^2)(x+2)+x(x-1)(x+1)=0

5x-(4x+8-2x^2-4x+x^3+2x^2)+x(x^2-1)=0

5x-4x-8+2x^2+4x-x^3-2x^2+x^3-1x=0

(5x-4x+4x-1x)+(-8)+(2x^2-2x^2)+(-x^3+x^3)=0

4x+(-8)=0

4x=0+8

4x=8

x=8:4

x=2

D)(4x+1)(16x^2-4x+1)-16x(4x^2-5)=17

64x^3-16x^2+4x+16x^2-4x+1-64x^3+80x=17

80x+1=17

80x=17-1

80x=16

x=1/5