bài 2 tìm x

a,106- ( x+ 7) =9

x+7 = 106 - 9

x+7 = 107

x= 107 - 7

x=100

b, 2 x ( x+ 4) + 5 =65

2 x (x+4) = 65 - 5

2 x (x+4) = 60

x+4 = 60:2

x+4= 30

x= 30 - 4

x=26

c, (16x x -32) x 45=0

16 x X - 32 = 0: 45

16 x X - 32 =0

16 x X = 0 + 32

16 x X = 32

X= 32:16

X=2

d, x+4 x x = 100 : 5

X + 4 x X = 20

(1+4) x X = 20

5 x X  = 20

X= 20:5

X=4

Bài 1:

a, 24 x 66 - 24 + 36 x 24

= 24 x (66-1+36)

= 24 x 101

= 24 x 100 + 24

= 2400+24

=2424

b, 29 x 173 - 29 x 72 - 29

= 29 x (173 - 72 -1)

= 29 x 100

= 2900

c, 1+5+10+15+...+55+60

= (5+60) x 6 + 1 (có 6 cặp)

= 65 x 6 + 1

= 390 +1 =391

d, D= 1 x 2 + 2 x 3 + 3 x 4 +...+18 x 19 + 19 x 20

3D= 1 x 2 x 3+ 2 x 3 x (4-1) + 3 x 4 x (5-2) +...+ 18 x 19 x (20 - 17) + 19 x 20 x (21-18)

3D= 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 +.... + 19 x 20 x 21 - 18 x 19 x 20

3D= 19 x 20 x 21 =7980

=> D= 7980:3=2660

a: x=-5/11+2/11=-3/11

b: =>x=-3/24+20/24+1/24=18/24=3/4

c: =>5/8-x=1/9+5/4=4/36+45/36=49/36

=>x=5/8-49/36=-53/72

d: =>2/3-x=1/3

=>x=1/3

e: =>1/5:x=12/35

=>x=7/12

a) 2(2x - 8) = 24

2x - 8 = 12

2x = 20

x = 10

b)3.(2^x - 26) = 18

2^x - 26 = 6

2^x = 32

x = 5

Bài 2:

a: \(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)

b: \(\Leftrightarrow x+3\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{-2;-4;0;-6;2;-8;12;-18\right\}\)

c: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)

d: =>x+1+15 chia hết cho x+1

=>\(x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

` 8/23 . 46/24 =1/3 .x`

`=>8/23 . 23/12 =1/3 . x`

`=> 1/3 . x=2/3`

`=>x=2/3 : 1/3`

`=>x=2/3 . 3`

`=> x= 6/3`

`=>x=2`

`----`

`1/5 : x= 1/5-1/7`

`=>1/5 : x=  7/35 - 5/35`

`=> 1/5 :x= 2/35`

`=>x= 1/5 : 2/35`

`=>x=1/5 . 35/2`

`=>x=7/2`

`----`

`4/9 - (x-1/2)^2 =1/3`

`=> (x-1/2)^2 =4/9-1/3`

`=> (x-1/2)^2 =4/9- 3/9`

`=> (x-1/2)^2 =1/9`

`=> (x-1/2)^2 = (+- 1/3)^2`

`@ TH1`

`x-1/2=1/3`

`=>x=1/3+1/2`

`=>x= 2/6 + 3/6`

``=>x= 5/6`

`@ TH2`

`x-1/2=-1/3`

`=>x=-1/3 +1/2`

`=>x= -2/6 + 3/6`

`=>x=1/6`

`----`

`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`

`=> 3,2 . x-22/15 : 11/3 = 7/10`

`=>  3,2 . x-22/15 = 7/10 . 11/3`

`=>  3,2 . x-22/15 =77/30`

`=> 3,2 .x= 77/30 + 22/15`

`=> 3,2 .x=121/30`

`=>x= 121/30. 5/16`

`=>x= 121/96`

Câu b,

= 50x + (1 + 3 + 5 + ..... + 99) = 2550

= 50x + 2500 = 2550

= 50x = 2550 - 2500

= 50x = 50

= x = 50 : 50

= x = 1

refer

A) 2448:[119-[x-6]]=24

<=> 199 - ( x-6 ) = 2448 : 24 = 102

<=> x - 6 = 199 - 102 = 97

<=> x = 97 +6 = 103

B) (x+1)+(x+3)+(x+5)+......+(x+99) = 2550

(x+x+x+....+x)+(1+3+5+....+99) = 2550

50x + 2500 = 2550

50x = 2550 - 2500

50x = 50

x = 50 : 50

x = 1

a) 6/-x=x/-24

=> -x.x=6.-24

=>-x.x=-144

=>x=12 hay x=-12

b)9/x=-35/105

=>9/x=-1/3

=>x=9.3/-1=-27

=>x=-27

c)x-1/8=5/8

=>x=5/8+1/8

=>x=3/4

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

=>-x+4/2=-2

=>-x/2=0

=>x=0

e)x+1/3=-12/5.10/6

x+1/3=-4

x=-4-1/3

x= -13/3

#YM

a: 3(x+7)-2x+5>0

=>3x+21-2x+5>0

=>x+26>0

=>x>-26

Sửa đề: \(\dfrac{x+2}{18}-\dfrac{x+3}{8}< \dfrac{x-1}{9}-\dfrac{x-4}{24}\)

=>\(\dfrac{4\left(x+2\right)}{72}-\dfrac{9\left(x+3\right)}{72}< \dfrac{8\left(x-1\right)}{72}< \dfrac{3\left(x-4\right)}{72}\)

=>\(4\left(x+2\right)-9\left(x+3\right)< 8\left(x-1\right)-3\left(x-4\right)\)

=>\(4x+8-9x-27< 8x-8-3x+12\)

=>-5x-19<5x+4

=>-10x<23

=>\(x>-\dfrac{23}{10}\)

b: \(3x+2+\left|x+5\right|=0\left(1\right)\)

TH1: x>=-5

(1) trở thành: 3x+2+x+5=0

=>4x+7=0

=>\(x=-\dfrac{7}{4}\left(nhận\right)\)

TH2: x<-5

=>x+5<0

=>|x+5|=-x-5

Phương trình (1) sẽ trở thành:

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

=>2x-3=0

=>2x=3

=>\(x=\dfrac{3}{2}\)

a) Ta có: \(\left|4-5x\right|=24\)

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

Vậy: \(x\in\left\{-4;\dfrac{28}{5}\right\}\)

b) Ta có: \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

Vậy: \(x\in\left\{-8;6\right\}\)

a) \(\left|4-5x\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)

b) \(\left(8+x\right)\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

a.|4-5.x|=24

TH1: 4-5.x=24

        ⇔5x=-20

        ⇔x=-4

TH1: 4-5.x=-24

        ⇔5x=28

        ⇔x=28/5

b.(8+x).(6-x)=0

\(\left\{{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)

a: \(\dfrac{-8}{3}\cdot\dfrac{15}{24}=\dfrac{-8}{24}\cdot\dfrac{15}{3}=\dfrac{-1}{3}\cdot5=-\dfrac{5}{3}\)

b: \(=\dfrac{3}{4}\cdot\dfrac{1}{-9}=\dfrac{-1}{12}\)

c: \(x=-\dfrac{7}{6}+\dfrac{5}{8}=-\dfrac{13}{24}\)

d: \(x=-\dfrac{14}{25}-\dfrac{3}{4}=\dfrac{-56-75}{100}=\dfrac{-131}{100}\)