a) \(|x+7|+|2y-12|=0\)

Vì \(\hept{\begin{cases}|x+7|\ge0;\forall x,y\\|2y-12|\ge0;\forall x,y\end{cases}}\)\(\Rightarrow|x+7|+|2y-12|\ge0;\forall x,y\)

Do đó \(|x+7|+|2y-12|=0\)

\(\Leftrightarrow\hept{\begin{cases}|x+7|=0\\|2y-12|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-7\\y=6\end{cases}}\)

Vậy ...

các phần sau tương tự

a) Ta có :

\(\left|x+7\right|\ge0\)

\(\left|2y-12\right|\ge0\)

Để |x+7| + | 2y - 12| = 0

=> x +7 = 0      và      2y - 12= 0

     x  = 7                    2y = 12

                                    y = 12 : 2

                                   y = 6

Vậy x = 7 ; y = 6

1,(x-1)(y+5)=101

th1:x-1=101

<=>x=102

th2:y+5=101

<=>y=96

2,(x-2)(-y+5)=12

th1:x-2=12

<=>x=14

th2:-y+5=12

<=>-y=7

<=>y=-7

Trả lời :

Mk giúp bn câu a ) thôi mà sai thì thôi nhé :)))

a, \(\left|x\right|+\left|y\right|=0\)

\(\Leftrightarrow x=0;y=0\) \(\Rightarrow\left|x\right|+\left|y\right|=0\)

Vậy x = 0 ; y = 0

_Học tốt

câu a,b,c dạng tương tự nhau nha nên mình làm câu a

a)\(\left|x\right|+\left|y\right|=0\left(1\right)\)

Ta có: \(\hept{\begin{cases}\left|x\right|\ge0;\forall x,y\\\left|y\right|\ge0;\forall x,y\end{cases}\Rightarrow}\left|x\right|+\left|y\right|\ge0;\forall x,y\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\end{cases}}}\)

Vậy \(\left(x,y\right)=\left(0;0\right)\)

d) \(\left|x^2+1\right|=12\left(1\right)\)

Ta thấy \(x^2\ge0;\forall x\)

\(\Rightarrow x^2+1\ge1>0;\forall x\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x^2+1=12\)

                      \(\Leftrightarrow x^2=11\)

                      \(\Leftrightarrow x=\pm\sqrt{11}\)

Vậy \(x=\pm\sqrt{11}\)

( x+1 )(y-2)=0

x+1=0 hoặc y-2=0

x=(-1) hoặc y=2

(x-5)(y-7)=1

x-5=1 và y-7=1

x=6 và y=8

đợi mình 1 tí mình làm tiếp

(x-4)(y+3)=-3

x-4=-1 và y+3=3

x=3 và y=0

(x-12)(y-10)=-8

x-12=2 và y-10=-4

x=14 và y=6

(x+1)(x-3)=0

x+1=0 hoặc x-3=0

x=-1 hoặc x=3

a) \(\left(x-7\right)\left(x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-12\end{matrix}\right.\)

Vậy: x∈{7;-12}

b) \(\left(3x-15\right)\left(6-2x\right)=0\)

⇔\(3\left(x-5\right)\cdot2\cdot\left(3-x\right)=0\)

hay \(6\left(x-5\right)\left(3-x\right)=0\)

Vì 6≠0

nên \(\left[{}\begin{matrix}x-5=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)

Vậy: x∈{3;5}

c) \(\left(3x+9\right)\left(4y-8\right)=0\)

⇔\(3\left(x+3\right)\cdot4\left(y-2\right)=0\)

hay \(12\left(x+3\right)\left(y-2\right)=0\)

Vì 12≠0

nên \(\left\{{}\begin{matrix}x+3=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\)

Vậy: x=-3 và y=2

d) \(\left(2y-16\right)\left(8x-24\right)=0\)

⇔\(2\left(y-8\right)\cdot8\left(x-3\right)=0\)

hay 16(y-8)(x-3)=0

Vì 16≠0

nên \(\left\{{}\begin{matrix}y-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=8\\x=3\end{matrix}\right.\)

Vậy: y=8 và x=3

e) \(\left(22-11y\right)\left(9x-18\right)=0\)

⇔\(11\left(2-y\right)9\left(x-2\right)=0\)

hay 99(2-y)(x-2)=0

Vì 99≠0

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

Vậy: x=2 và y=2

g) \(\left(7y+14\right)\cdot\left(9x-18\right)=0\)

⇔7(y+2)*9(x-2)=0

hay 63(y+2)(x-2)=0

Vì 63≠0

nên \(\left\{{}\begin{matrix}y+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\)

Vậy: y=-2 và x=2

h) xy=3

⇒x,y∈Ư(3)

⇒x,y∈{1;-1;3;-3}

*Trường hợp 1:

\(\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

*Trường hợp 2:

\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

*Trường hợp 3:

\(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)

*Trường hợp 4:

\(\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)

Vậy: x∈{1;-1;3;-3} và y∈{1;-1;3;-3}

i) x*y=-5

⇔x,y∈Ư(-5)

⇔x,y∈{1;-1;5;-5}

*Trường hợp 1:

\(\left\{{}\begin{matrix}x=1\\y=-5\end{matrix}\right.\)

*Trường hợp 2:

\(\left\{{}\begin{matrix}x=-1\\y=5\end{matrix}\right.\)

*Trường hợp 3:

\(\left\{{}\begin{matrix}x=-5\\y=1\end{matrix}\right.\)

*Trường hợp 4:

\(\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)

Vậy: x∈{1;5;-1;-5} và y∈{1;5;-1;-5}

k) \(\left(x+4\right)\left(y-5\right)=-3\)

⇔x+4; y-5∈Ư(-3)

⇔x+4; y-5∈{1;3;-3;-1}

*Trường hợp 1:

\(\left\{{}\begin{matrix}x+4=-1\\y-5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=8\end{matrix}\right.\)

*Trường hợp 2:

\(\left\{{}\begin{matrix}x+4=1\\y-5=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\)

*Trường hợp 3:

\(\left\{{}\begin{matrix}x+4=3\\y-5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=4\end{matrix}\right.\)

*Trường hợp 4:

\(\left\{{}\begin{matrix}x+4=-3\\y-5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=6\end{matrix}\right.\)

Vậy: x∈{-5;-3;-1;-7} và y∈{8;2;4;6}

m) (x-9)(y-5)=-1

⇔x-9; y-5∈Ư(-1)

⇔x-9; y-5∈{1;-1}

*Trường hợp 1:

\(\left\{{}\begin{matrix}x-9=1\\y-5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=4\end{matrix}\right.\)

*Trường hợp 2:

\(\left\{{}\begin{matrix}x-9=-1\\y-5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=6\end{matrix}\right.\)

Vậy: x∈{10;8} và y∈{4;6}

n) x+3⋮x+4

⇔x+4-1⋮x+4

⇔-1⋮x+4

hay x+4∈Ư(-1)

⇔x+4∈{1;-1}

⇔x∈{-3;-5}

Vậy: x∈{-3;-5}

p)(x-5)⋮x+2

⇔x+2-7⋮x+2

hay -7⋮x+2

⇔x+2∈Ư(-7)

⇔x+2∈{1;-1;7;-7}

hay x∈{-1;-3;5;-9}

Vậy: x∈{-1;-3;5;-9}

a)9.x + 1=73

 9x=73-1

9x=72

  x=72:9

  x=8

b)2.x - 5 = -17 - 12

2x-5=-29

2x=-29+5

2x=-24

  x=-24:2

  x=-12

c)10 - x - 5 = - 5 - 7 -11

10-x-5=-12-11

10-x-5=-23

 10-x=-23+5 

  10-x=18

       x=10-18

       x=-8

d)(-9) . x + 3 = (-2) . (-7) +16

-9x+3=14+16

-9x+3=30

-9x=30-3

-9x=27

   x=27:(-9)

   x=-3

(-12) . x - 34 =2

-12x=2+34

-12x=36

     x=36:(-12)

     x=-3

(-11).x + 9 =130

-11x=130-9

-11x=121

     x=121:(-11)

     x=11 

(-5) .x + 5 = (-15) .(-4) -12

-5x+5=60-12

-5x+5=48

-5x=48-5

-5x=43

   x=43:(-5)

  x=-8,6

IxI -3=0

|x|=3

=>x=+3

(7 - IxI).(2.x - 4) =0

*7-|x|=0            * 2x-4=0

|x|=7                   2x=4

=>x=+7                 x=4:2

                             x=2

280-(x-140):35 =270

(x-140):35=280-270

(x-140):35=10

x-140=10.35

x-140=350

       x=350+140

       x=490

(1900 - 2.x ) : 35- 32 =16

1900-2x=(16+32).35

1900-2x=1680

         2x=1900-1680

         2x=220

           x=220:2

           x=110

720 :[41-(2x -5 )] =2³ .5

720:[41-(2x-5)]=40

41-(2x-5)=720:40

41-(2x-5)=18

2x-5=41-18

2x-5=23

2x=23+5

2x=28

  x=28:2

  x=14

(x - 5).(x² - 4 ) =0

* x-5=0           * x²-4=0

      x=0+5         x²=4

      x=5             x²=2²

                    => x=+2

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~