dài quá à :(

1) x - 43 = (35 - x) - 48

=> x + x = 35 - 48 + 43

=> x + x = 30

=> x = 30 : 2

=> x = 15

2) 305 - x + 14 = 48 + (x + 23)

=> 305 - x + 14 = 48 + x + 23

=> -x - x = 48 + 23 - 14 - 305

=> -x - x = -248

=> -x = -248 : 2

=> -x = -124

=> x = 124

3) - (x - 6 + 85) = (x + 51) - 54

=> -x + 6 - 85 = x + 51 - 54

=> -x - x = 51 - 54 + 85 - 6

=> -x - x = 76

=> -x = 76 : 2

=> -x = 38

=> x = -38

4) - (35 - x - 37 - x) = 33 - x

=> -35 + x + 37 + x = 33 - x

=> x + x + x = 33 + 35 - 37

=> x + x + x = 31

=> x = 31 : 3

=> x \(=\dfrac{31}{3}\)

Vì x \(\in\) Z nên không có giá trị x nào thỏa mãn trong câu này.

5) 13 - | x | = | -4 |

=> 13 - |x| = 4

=> |x| = 13 - 4

=> |x| = 9

=> \(\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)

6) | x | - 3 + 6 = 16

=> |x| = 16 - 6 + 3

=> |x| = 13

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

7) 35 - | 2x - 1 | = 14

=> |2x - 1| = 35 - 14

=> |2x - 1| = 21

=> \(\left[{}\begin{matrix}2x-1=21\\2x-1=-21\end{matrix}\right.=>\left[{}\begin{matrix}2x=21+1\\2x=-21+1\end{matrix}\right.=>\left[{}\begin{matrix}2x=22\\2x=-20\end{matrix}\right.=>\left[{}\begin{matrix}x=22:2\\x=-20:2\end{matrix}\right.=>\left[{}\begin{matrix}x=11\\x=-10\end{matrix}\right.\)

8) | 3x - 2 | + 5 = 9 - x

=> |3x - 2| = 9 - 5 - x

=> |3x - 2| = 4 - x

=> \(\left[{}\begin{matrix}3x-2=4-x\\3x-2=x-4\end{matrix}\right.=>\left[{}\begin{matrix}3x+x=4+2\\3x-x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}4x=6\\2x=-2\end{matrix}\right.=>\left[{}\begin{matrix}x=6:4\\x=-2:2\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{6}{4}\\x=-1\end{matrix}\right.\)

Vì x \(\in\) Z nên x = -1.

9) x - ( -25 + 7 ) > 12 - ( 15 - 14 )

=> x - (-18) > 12 - 1

=> x + 18 > 11

=> x > 11 - 18

=> x > -7

10) | 17 + ( x - 15 ) | < 4

=> \(\left[{}\begin{matrix}17+\left(x-15\right)< 4\\17+\left(x-15\right)< -4\end{matrix}\right.=>\left[{}\begin{matrix}x-15< 4-17\\x-15< -4-17\end{matrix}\right.=>\left[{}\begin{matrix}x-15< -15\\x-15< -21\end{matrix}\right.=>\left[{}\begin{matrix}x< -15+15\\x< -21+15\end{matrix}\right.=>\left[{}\begin{matrix}x< 0\\x< -6\end{matrix}\right.=>x< -6\)

11) x2 - 5x = 0

=> x . (2 - 5) = 0

=> x . (-3) = 0

=> x = 0 : (-3)

=> x = 0

12) | x-9 | . (-8) = -16

=> |x - 9| = (-16) : (-8)

=> |x - 9| = 3

=> \(\left[{}\begin{matrix}x-9=3\\x-9=-3\end{matrix}\right.=>\left[{}\begin{matrix}x=3+9\\x=-3+9\end{matrix}\right.=>\left[{}\begin{matrix}x=12\\x=6\end{matrix}\right.\)

13) | 4 - 5x | = 24 với x < hoặc = 0

=> \(\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.=>\left[{}\begin{matrix}5x=4-24\\5x=4-\left(-24\right)\end{matrix}\right.=>\left[{}\begin{matrix}5x=-20\\5x=28\end{matrix}\right.=>\left[{}\begin{matrix}x=-20:5\\x=28:5\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

Vì x \(\le\) 0 nên x = -4

14) x . ( x - 2 ) > 0

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

15) x . ( x - 2 ) < 0

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\end{matrix}\right.=>\left[{}\begin{matrix}2>x< 0\left(loại\right)\\0< x< 2\left(chọn\right)\end{matrix}\right.=>0< x< 2\)

16) (x-1) . (y+1) = 5

=> \(\left[{}\begin{matrix}x-1=5\\y+1=1\end{matrix}\right.=>\left[{}\begin{matrix}x=5+1\\y=1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=6\\y=0\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=1\\y+1=5\end{matrix}\right.=>\left[{}\begin{matrix}x=1+1\\y=5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=-1\\y+1=-5\end{matrix}\right.=>\left[{}\begin{matrix}x=-1+1\\y=-5-1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x-1=-5\\y+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-5+1\\y=-1-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)

17) x . ( y +2 ) = -8

=> \(\left[{}\begin{matrix}x=1\\y+2=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-1\\y+2=8\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=8-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-8\\y+2=1\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-8\\y=-1\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=8\\y+2=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-1-2\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-3\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=2\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=-6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-2\\y+2=4\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=4\\y+2=-4\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-4-2\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\y=-6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}x=-4\\y+2=2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=2-2\end{matrix}\right.=>\left[{}\begin{matrix}x=-4\\y=0\end{matrix}\right.\)

18) xy - 2x - 2y = 0

=> x . (y - 2) - 2y = 0

=> x . (y - 2) - 2y - 4 = -4

=> x . (y - 2) - 2 . (y - 2) = -4

=> (y - 2) . (x - 2) = -4

=> \(\left[{}\begin{matrix}y-2=1\\x-2=-4\end{matrix}\right.=>\left[{}\begin{matrix}y=1+2\\x=-4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}y-2=-1\\x-2=4\end{matrix}\right.=>\left[{}\begin{matrix}y=-1+2\\x=4+2\end{matrix}\right.=>\left[{}\begin{matrix}y=1\\x=6\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}y-2=2\\x-2=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2+2\\x=-2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=4\\x=0\end{matrix}\right.\)

hoặc

=> \(\left[{}\begin{matrix}y-2=-2\\x-2=2\end{matrix}\right.=>\left[{}\begin{matrix}y=-2+2\\x=2+2\end{matrix}\right.=>\left[{}\begin{matrix}y=0\\x=4\end{matrix}\right.\)

19) 2x - 5 \(⋮\) x - 1

=> (2x - 2) - (5 - 2) \(⋮\) x - 1

=> 2(x - 1) - 3 \(⋮\) x - 1

Vì 2(x - 1) \(⋮\) x - 1 nên 3 \(⋮\) x - 1

=> x - 1 \(\in\) Ư(3) = {-3; -1; 1; 3}

=> x \(\in\) {-2; 0; 2; 4}

P/s: Mình không bảo đảm là đúng hết nên câu nào sai thì bạn thông cảm nha~

Bạn xem lại đề nhé

a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)

\(\Leftrightarrow3x+2x-10=6-5x+1\)

\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm 

b, \(x^3-3x^2-x+3=0\)

\(\Leftrightarrow x\left(x^2-1\right)-3\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)

Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }

c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)

\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)

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

\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn 

Vậy ... 

a) 4( 18 - 5x ) - 12( 3x - 16 ) = 15( 2x - 16 ) - 6( x + 14 )

<=> 72 - 20x - 36x + 192 = 30x - 240 - 6x - 84

<=> -20x - 36x - 30x + 6x = -240 - 84 - 72 - 192

<=> -80x = -588

<=> x = -588/-80 = 147/20

b) ( x + 3 )( x + 2 ) - ( x - 2 )( x + 5 ) = 6

<=> x² + 5x + 6 - ( x² + 3x - 10 ) = 6

<=> x² + 5x + 6 - x² - 3x + 10 = 6

<=> 2x + 16 = 6

<=> 2x = -10

<=> x = -5

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

<=> -x² - 3x + 2 = 4x² - 3x - 1 + 2x

<=> -x² - 3x - 4x² + 3x - 2x = -1 - 2

<=> -5x² - 2x = -3

<=> -5x² - 2x + 3 = 0

<=> -( 5x² + 2x - 3 ) = 0

<=> -( 5x² + 5x - 3x - 3 ) = 0

<=> -[ 5x( x + 1 ) - 3( x + 1 ) ] = 0

<=> -( x + 1 )( 5x - 3 ) = 0

<=> \(\orbr{\begin{cases}x+1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}\)

d) ( 2x + 3 )( x - 3 ) - ( x - 3 )( x + 1 ) = ( 2 - x )( 3x + 1 ) + 3 

<=> 2x² - 3x - 9 - ( x² - 2x - 3 ) = -3x² + 5x + 2 + 3

<=> 2x² - 3x - 9 - x² + 2x + 3 = -3x² + 5x + 2 + 3

<=> 2x² - 3x - x² + 2x + 3x² - 5x = 2 + 3 + 9 - 3

<=> 4x² - 6x = 11

<=> 4x² - 6x - 11 = 0

=> Vô nghiệm ( Lớp 8 chưa học nghiệm vô tỉ nên để vậy ) :))

vẫn làm được nha quỳnh !

\(4x^2-6x-11=0\)

\(< =>\left(4x^2-6x+\frac{9}{4}\right)-13\frac{1}{4}=0\)

\(< =>\left(2x-\frac{3}{2}\right)^2=\frac{53}{4}\)

\(< =>\orbr{\begin{cases}2x-\frac{3}{2}=\frac{\sqrt{53}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{53}}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}2x=\frac{3+\sqrt{53}}{2}\\2x=\frac{3-\sqrt{53}}{2}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{3+\sqrt{53}}{4}\\x=\frac{3-\sqrt{53}}{4}\end{cases}}\)

:vc dcv ơi, lớp 8 chưa học số vô tỉ ớ nên cho PT đó vô nghiệm là đúng, cái j mà ko làm đc:)) haha ( pịt momx lại ko bị chửi mù chữ:)

a, \(4\left(18-5x\right)-12\left(3x-16\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Leftrightarrow72-20x-36x+192=30x-240-6x-84\)

\(\Leftrightarrow264-56x=24x-324\Leftrightarrow x=\frac{147}{20}\)

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

\(\Leftrightarrow-x^2-3x+2=4x^2-4x+x-1+2x\)

\(\Leftrightarrow-x^2-3x+2=4x^2-x-1\)

\(\Leftrightarrow5x^2+2x-3=0\Leftrightarrow\left(x+1\right)\left(5x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}\)

bai 1

1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0

<=>(2x)^2-5^2=0

<=>(2x+5)*(2x-5)=0

<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự