Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41

\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)

\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)

Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)

b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

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

vậy ...

d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)

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

a: Ta có: \(x^4-2x^3+5x^2-10x=0\)

\(\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0\)

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

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

b:Ta có: \(\left(3x+5\right)^2=\left(2x-2\right)^2\)

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

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

\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

a)(x+2).(x+3)-(x-2).(x+5)=10

  ( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10

 x^2 +3x+2x+6-x^2 -5x+2x+10-10=0

 2x+6=0

2x=-6

x=-3

a)    (x + 2)(x + 3) - (x - 2)(x + 5) = 0
<=> x² + 3x + 2x + 6 - (x² + 5x - 2x - 10) = 0
<=> x² + 3x + 2x + 6 - x² - 5x + 2x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8

b)    (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> (2x + 3)(x - 4) + (x - 5)(x - 2) - (3x - 5)(x - 4) = 0
<=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 - (3x² - 12x - 5x + 20) = 0
<=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 - 3x² + 12x + 5x - 20 = 0
<=> 5x = 12 - 10 + 20
<=> 5x = 22
<=>   x = 22/5

c)    (8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
<=> 8x + 16 - 5x² - 10x + (4x - 8)(x + 1) + 2(x² - 4) = 0
<=> 8x + 16 - 5x² - 10x + 4x² + 4x - 8x - 8 + 2x² - 8 = 0
<=> x² - 6x = 0
<=> x(x - 6) = 0
<=> x = 0 hay     x - 6 = 0
                  I<=> x      = 6

d)    (8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
<=> 24x² + 16x - 9x - 6 - (4x² + 16x + 7x + 28) = 10x² - 2x + 5x - 1 - 33
<=> 24x² + 16x - 9x - 6 - 4x² - 16x - 7x - 28 - 10x² + 2x - 5x + 1 + 33 = 0
<=> 10x² - 19x = 0
<=> x(10x - 19) = 0
<=> x = 0 hay      10x - 19 = 0
                  I <=> 10x       = 19
                  I <=>    x       = 19/10

a)  \(\left(x+6\right)^2-x\left(x+9\right)=0\)

\(\Leftrightarrow\)\(x^2+12x+36-x^2-9x=0\)

\(\Leftrightarrow\)\(3x+36=0\)

\(\Leftrightarrow\)\(x=-12\)

Vậy...

b) \(6x\left(2x+5\right)-\left(3x+4\right)\left(4x-3\right)=9\)

\(\Leftrightarrow\)\(12x^2+30x-12x^2-7x+12=9\)

\(\Leftrightarrow\)\(23x+12=9\)

\(\Leftrightarrow\)\(x=-\frac{3}{23}\)

Vậy

c) \(2x\left(8x+3\right)-\left(4x+1\right)=13\)

\(\Leftrightarrow\)\(16x^2+6x-4x-1=13\)

\(\Leftrightarrow\)\(16x^2+2x-14=0\)

\(\Leftrightarrow\)\(8x^2+x-7=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(8x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=\frac{7}{8}\end{cases}}\)

Vậy

d) \(\left(x-4\right)^2-x\left(x+4\right)=0\)

\(\Leftrightarrow\)\(x^2-8x+16-x^2-4x=0\)

\(\Leftrightarrow\)\(-12x+16=0\)

\(\Leftrightarrow\)\(x=\frac{4}{3}\)

Vậy

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

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

\(\Leftrightarrow\)\(-x^2-3x+10=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

Vậy

\(a.3\left(x^2-2x+1\right)-3x^2+15x-2=0\)

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

\(9x+1=0\)

\(x=-\frac{1}{9}\)

\(b.4x^2-12x+9=0\)

\(4x^2-6x-6x+9=0\)

\(2x\left(x-3\right)-3\left(x-3\right)=0\)

\(\left(2x-3\right)\left(x-3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)

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

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

\(\left(x-8\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

a) 3(x - 1)² - 3x(x - 5) = 2

=> 3(x² - 2x + 1) - 3x² + 15x = 2

=> 3x² - 6x + 3 - 3x² + 15x = 2

=> 9x = 2 - 3

=> 9x = -1

=> x = -1/9

b) 4x² - 12x = -9

=> 4x² - 12x + 9 = 0

=> (2x - 3)² = 0

=> 2x - 3 = 0

=> 2x = 3

=>  x = 3/2

c) (2x - 3)² = (x  + 5)²

=> (2x - 3)² - (x + 5)² = 0

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

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

=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)

d) \(\left(x^4-2x^3+4x^2-8x\right):\left(x^2+4\right)-2x=-4\)

=> \(\left[x^3\left(x-2\right)+4x\left(x-2\right)\right]:\left(x^2+4\right)-2x=-4\)

=> \(x\left(x-2\right)\left(x^2+4\right):\left(x^2+4\right)-2x=-4\)

=> \(x^2-2x-2x+4=0\)

=> \(\left(x-2\right)^2=0\)

=> x - 2 = 0

=> x = 2

e) khđ

Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.

f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)

=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0

=>6x-24=0

=>x=4

e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2

=>-5x^2-2x+16+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

Chúng ta sẽ giải từng phương trình một:

a. Đặt �=�2−2y=x2−2, ta có: (−2+�2)(�2−2)4=1(−2+x2)(

a) x³ - 16x = 0

x(x² - 16) = 0

=> x = 0 hoặc x² - 16 = 0

x = 4

Vậy x = 0 hoặc x = 4

b) x⁴ -2x³ + 10x² - 20x = 0

x³ (x - 2) + 10x(x - 2) = 0

(x - 2)(x³ + 10x) = 0

=> x - 2 = 0 hoặc x³ + 10x = 0

x = 2 x(x² + 10) = 0

+ TH1: x = 0

+ TH2: x² + 10 = 0

x² = -10 (vô lí)

Vậy x = 2 hoặc x = 0

c) (2x - 3)² = (x + 5)²

(2x)² + 2 . 2x . 3 + 3² = x² + 2.x.5 + 5²

4x² + 12x + 9 = x² + 10x + 25

4x² + 12x - x² - 10x = 25 - 9

3x² + 2x = 16

x(3x + 2) = 16

Đến đây bạn làm nốt câu c nhé!