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

a) x² - 5x - 6 = 0

=> x² - 2x - 3x - 6 = 0

=> (x² - 2x) + (-3x - 6) = 0

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

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

=> x - 2 = 0 => x = 2

     x - 3 = 0 => x = 3

còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546

Bạn giải giúp mình mấy câu kia được k=)))

a) \(x^3=x^5\)

=> \(x^3-x^5=0\)

=> \(x^3\left(1-x^2\right)=0\)

=> \(\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

b) \(4x\left(x+1\right)=x+1\)

=> \(4x^2+4x-x-1=0\)

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

=> \(\left(x+1\right)\left(4x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)

c) \(x\left(x-1\right)-2\left(1-x\right)=0\)

=> \(x\left(x-1\right)-\left[-2\left(x+1\right)\right]=0\)

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

=> \(\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

d) Kết quả ?

e) \(\left(x-3\right)^2+3-x=0\)

=> \(x^2-6x+9+3-x=0\)

=> \(x^2-7x+12=0\)

=> \(x^2-3x-4x+12=0\)

=> \(x\left(x-3\right)-4\left(x-3\right)=0\)

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

=> \(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)

f) Tương tự

\(\left(2x-1\right)^3-2x\left(2x+1\right)^2=5x-20x^2\)

\(\Rightarrow8x^3-12x^2+6x-1-2x\left(4x^2+4x+1\right)-5x+20x^2=0\)

\(\Rightarrow8x^3-12x^2+6x-1-8x^3-8x^2-2x-5x+20x^2=0\)

\(\Rightarrow-x-1=0\)

\(\Rightarrow x=-1\)

a, 5x(x - 1) - (1 - x) = 0

=> 5x(x - 1) + (x - 1) = 0

=> (x - 1)(5x + 1) = 0

=> x - 1 = 0 hoặc 5x - 1 = 0

=> x = 1 hoặc x = \(\dfrac{1}{5}\)

b, (x - 3)² - (x + 3)² = 24

=> (x - 3 + x + 3)(x - 3 - x - 3) = 24

=> 2x. (-6) = 24

=> -12x = 24

=> x = -2

c, 2x(x² - 4) = 0

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

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

d, 2(x + 5)² - x² - 5x = 0

=> 2(x + 5)² - x(x + 5) = 0

=> (x + 5) [2(x + 5) - x] = 0

=> (x + 5) (2x - 10 - x) = 0

=> (x + 5) ( x - 10) = 0

\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-10=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=10\end{matrix}\right.\)

e, (2x - 3)² - (x +5)² = 0

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

=> (3x + 2)(x - 8) = 0

\(\Rightarrow\left[{}\begin{matrix}3x+2=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=8\end{matrix}\right.\)

f, 3x² - 48x = 0

=> 3x(x - 16) = 0

\(\Rightarrow\left[{}\begin{matrix}3x=0\\x-16=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)

chúc bạn học tốt!

bạn qua trả lời câu hỏi hộ mình với

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

$\Leftrightarrow -3-6x=0$

$\Leftrightarrow 6x=-3$

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

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

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

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

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

TH1: x=0

TH2: -13x-49=0

        -13x    = 49

            x    = \(-\frac{49}{13}\)

Vậy x=0 và   x    = \(-\frac{49}{13}\)

5-10x=3x+54

   10x=3x+49

     7x=49

Vậy x=7

Vì 5x(1-2x)-3x(x+18)=0

Nên 5x(1-2x)=0=>1-2x=0=>x=0,5

 Hoặc 3x(x+18)=0=>x+18=0=>x=-18

Vậy x=-18 hoặc x=0,5

 a) 2 (x-1) -3 (x+2) = 5x-9

2x -2 - 3x -6 = 5x -9

2x - 3x - 5x = -9 +2 +6

-6x = -1

x= 1/6

b) 4x (x+1) -2x (x+2) =0

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

2x^2 =0

x^2 =0

x=0