a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

Lời giải:

a.

PT $\Leftrightarrow 3x^2+\frac{x}{2}-3x^2+3x+2=0$
$\Leftrightarrow \frac{7}{2}x+2=0$
$\Leftrightarrow \frac{7}{2}x=-2$
$\Leftrightarrow x=-2: \frac{7}{2}=\frac{-4}{7}$
b.

PT $\Leftrightarrow 5x^2-3-5x^2-6x=0$

$\Leftrightarrow -3-6x=0$

$\Leftrightarrow 6x=-3$

$\Leftrightarrow x=\frac{-3}{6}=\frac{-1}{2}$

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

a: a(x)=x^3+3x^2+5x-18

b(x)=-x^3-3x^2+2x-2

b: m(x)=a(x)+b(x)

=x^3+3x^2+5x-18-x^3-3x^2+2x-2

=7x-20

c: m(x)=0

=>7x-20=0

=>x=20/7

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

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

-2x^2(x-1)^2=0

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)

\(\Leftrightarrow5x-10x^2-3x^2-54x\)

\(\Leftrightarrow-13x^2-49x=0\)

\(\Leftrightarrow x\left(-13x-49\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-13x=49\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{49}{13}\end{matrix}\right.\)

Vậy ..

\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)

=>\(5x-10x^2-3x^2-54x=0\)

=> \(-49x-13x^2=0\)

=> \(x\left(-49-13x\right)=0\)

=> \(\left[{}\begin{matrix}x=0\\-49-13x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\-49=13x\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{-49}{13}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{-49}{13};0\right\}\)

<=> 5x - 10x² - 3x² -54x = 0

<=> -13x² - 49x =0

<=> -x. (13x + 49) = 0

<=> \(\begin{matrix}-x=0\\13x+49=0\end{matrix}\)

<=> \(\begin{matrix}x=0\\x=-\dfrac{49}{13}\end{matrix}\)

=> đáp số .....

\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)

=>  \(5x-10x^2-3x^2-54x=0\)

=> \(-13x^2-49x=0\)

=> \(-x\left(13x+49\right)=0\)

=> \(-x=0\) hoặc : \(13x+49=0\)

=> \(x=0\) hoặc: \(13x=-49\Rightarrow x=\frac{-49}{13}\)

\(5x\left(1-2x\right)-3x\left(x+18\right)=0\)

=> \(5x-10x^2-3x^2-54x=0\)

=> \(-49x-13x^2=0\)

=> \(-x\left(49+13x\right)=0\)

=> \(-x=0\) hoặc \(49+13x=0\)

=> \(x=0\) hoặc  \(13x=-49\Rightarrow x=-\frac{49}{13}\)

cậu nói rõ hơn đc ko tớ **** cho