a) (2x + 1)(1 - 2x) + (1 - 2x)² = 18

= ( 1 - 2x) \(\left[\left(2x+1+1-2x\right)\right]\) = 18

= 2(1 - 2x)  - 18 = 0

= 2 - 4x - 18 = 0

= -16 - 4x = 0

= -4x = 16

= x = \(\dfrac{16}{-4}=-4\)

b) 2(x + 1)² -(x - 3)(x + 3) - (x - 4)² = 0

= 2 (x² + 2x + 1) - (x² - 9) - (x² - 8x + 16) = 0

= 2x² + 4x + 2 - x² + 9 - x² + 8x - 16 = 0

= 12x - 5 = 0

= 12x = 5

= x = \(\dfrac{5}{12}\)

c) (x - 5)² - x(x - 4) = 9

= x² - 10x + 25 - x² + 4x - 9 = 0

= -6x + 16 = 0

= -6x = -16

= x = \(\dfrac{-16}{-6}=\dfrac{8}{3}\)

d) (x - 5)² + (x - 4)(1 - x)

= x² - 10x + 25 + 5x - x² - 4 = 0

= -5x + 21 = 0

= -5x = -21

= x = \(\dfrac{-21}{-5}=\dfrac{21}{5}\) 

 Chúc bạn học tốt

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 

Bài 2: 

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

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

Tự chép đề bài nhé bạn

a) 5x + 10x² - 3x² - 54x = 0

\(\Rightarrow\) 7x² - 49x = 0

\(\Leftrightarrow\) 7x( x - 7 ) = 0

\(\Leftrightarrow\left\{{}\begin{matrix}7x=0\\x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

Vậy ....

b) x² + 5x + 6 - x² - 3x - 10 = 0

\(\Leftrightarrow\) 2x - 4 = 0

\(\Leftrightarrow\) 2x = 4

\(\Leftrightarrow\) x = 2

Vậy .....

a) 5x(1-2x) - 3x(x+18) = 0

\(\Leftrightarrow5x-10x^2-3x^2-54x=0\)

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

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

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

Vậy phương trình có tập nghiệm \(S=\left\{0;\frac{-49}{13}\right\}\)

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

\(\Leftrightarrow x^2+5x+6-\left(x^2+3x-10\right)=0\)

\(\Leftrightarrow2x+16=0\)

\(\Leftrightarrow x=-8\)

Vậy phương trình có nghiệm x = -8

c) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow3x^2-3x^2-12x+17x-2-20=0\)

\(\Leftrightarrow5x-22=0\)

\(\Leftrightarrow x=\frac{22}{5}=4,4\)

Vậy phương trình có nghiệm x = 4,4.

1.

a,  \(x-14=3x+18\)                                                                       

\(\Rightarrow x-3x=18+14\)                                                                 

\(\Rightarrow-2x=32\Rightarrow x=\frac{32}{-2}=-16\)

b, \(\left(x+7\right).\left(x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=9\end{cases}}}\)

c, \(\left|2x-5\right|-7=22\)                                                                     

\(\Rightarrow\left|2x-5\right|=22+7\)

\(\Rightarrow\left|2x-5\right|=29\)

\(\Rightarrow\orbr{\begin{cases}2x+5=29\\2x-5=29\end{cases}}\Rightarrow\orbr{\begin{cases}2x=24\\2x=34\end{cases}\Rightarrow}\orbr{\begin{cases}x=12\\x=17\end{cases}}\)

d,  \(\left(\left|2x\right|-5\right)-7=22\)

\(\Rightarrow\left(\left|2x\right|-5\right)=29\)

\(\Rightarrow\left|2x\right|=29+5\Rightarrow\left|2x\right|=34\Rightarrow x=\pm17\)

e, \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\)

Vì \(\left|x+3\right|\ge0;\left|x+9\right|\ge0;\left|x+5\right|\ge0;4x\ge0\)

Nên \(\left|x+3\right|+\left|x+9\right|+\left|x+5\right|=4x\ge0\)

\(\Rightarrow\left|x+3\right|>0\Rightarrow\left|x+3\right|=x+3\)

     \(\left|x+9\right|>0\Rightarrow\left|x+9\right|=x+9\)

      \(\left|x+5\right|>0\Rightarrow\left|x+5\right|=x+5\)

Ta có : 

\(x+3+x+9+x+5=4x\)

\(\Rightarrow3x+\left(3+9+5\right)=4x\)

\(\Rightarrow4x-3x=17\)

\(\Rightarrow x=17\)

2. a , b sai đề bn 

c, \(\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(\text{ }Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

d, \(5xy-5x+y=5\)

\(\Rightarrow\left(5xy-5x\right)+y=5\)

\(\Rightarrow5x.\left(y-1\right)+y=5\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)=4\)

\(\Rightarrow\left(5x+1\right).\left(y-1\right)\inƯ\left(4\right)\)

\(Ư\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Ta có bảng sau : 

x - 14 = 3x + 18

x - 3x = 18 + 14

-2x= 32

x= 32 : (-2)

x=-16

Tìm x

a) Ta có: \(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)-15=0\)

\(\Leftrightarrow16x^2-16x^2+40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow40x=40\)

hay x=1

Vậy: x=1

b) Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4\left(x^2-1\right)-49=0\)

\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)

\(\Leftrightarrow12x-36=0\)

\(\Leftrightarrow12x=36\)

hay x=3

Vậy: x=3

d) Ta có: \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2\left(x^2+2x+1\right)-\left(x^2-9\right)-\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x-5=0\)

\(\Leftrightarrow12x=5\)

hay \(x=\frac{5}{12}\)

Vậy: \(x=\frac{5}{12}\)

e) Ta có: \(\left(x-5\right)^2-x\left(x-4\right)=9\)

\(\Leftrightarrow x^2-10x+25-x^2+4x-9=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow6x=16\)

hay \(x=\frac{8}{3}\)

Vậy: \(x=\frac{8}{3}\)

f) Ta có: \(\left(x-5\right)^2-\left(x-4\right)\left(1-x\right)=0\)

\(\Leftrightarrow x^2-10x+25-\left(x-x^2-4+4x\right)=0\)

\(\Leftrightarrow x^2-10x+25-x+x^2+4-4x=0\)

\(\Leftrightarrow2x^2-15x+29=0\)

\(\Leftrightarrow2\left(x^2-\frac{15}{2}x+\frac{29}{2}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{15}{4}+\frac{225}{16}+\frac{7}{16}=0\)

\(\Leftrightarrow\left(x-\frac{15}{4}\right)^2+\frac{7}{16}=0\)(vô lý)

Vậy: x∈∅

a) 16x^2 - (4x - 5)^2 = 15

<=> 16x^2 - 16x^2 + 40x - 25 = 15

<=> 40x = 40

<=> x = 1

b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49

<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49

<=> 12x + 13 = 49

<=> 12x = 36

<=> x = 3

c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18

<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18

<=> 2 - 4x = 18

<=> -4x = 16

<=> x = -4

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

<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0

<=> 12x - 5 = 0

<=> 12x = 5

<=> x = 5/12

e) (x - 5)^2 - x(x - 4) = 9

<=> x^2 - 10x + 25 - x^2 + 4x = 9

<=> -6x + 25 = 9

<=> -6x = 9 - 25

<=> -6x = -16

<=> x = -16/-6 = 8/3

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

<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0

<=> -5x + 21 = 0

<=> -5x = -21

<=> x = 21/5