1 , 10

2 , 8

3 , 5

4 , 2

5 , 1

6 ,

1/ `|x|=10<=> x=\pm 10`

2/ `|x-8|=0<=>x-8=0<=>x=8`

3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`

4/ `|x+1|=3`

$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$

5/ `15-x=16-(14-42)`

`<=>15-x=16+28`

`<=>15-x=44`

`<=>x=-29`

6/ `210-(x-12)=168`

`<=>210-x+12=168`

`<=>222-x=168`

`<=>x=54`

1.

\(\left|x\right|=10\Leftrightarrow x=\pm10\)

2.

\(\left|x-8\right|=0\Leftrightarrow x-8=0\Leftrightarrow x=8\)

3.

\(7+\left|x\right|=12\Leftrightarrow\left|x\right|=5\Leftrightarrow x=\pm5\)

b) 2.| x - 3 | + 7 = 15

    2.| x - 3 |       =  15 - 7

    2.| x - 3 |       =   8

       | x - 3 |       =   8 : 2 

       | x - 3 |       =   4

Ta có 2 trường hợp :

TH 1 : x - 3 = -4                                                                       TH 2 : x - 3 = 4  

           x      = -4 +3                                                                             x      = 4 + 3   

           x      = -1                                                                                  x      = 7

 Vậy x là -1 hoặc 7

a) (x+10)(2y-5) = 143

=> (x+10);(2y-5) thuộc Ư(143)={-1,-143,1,143}

\(\orbr{\begin{cases}x+10=-143\\2y-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-153\\y=2\end{cases}}\)

\(\orbr{\begin{cases}x+10=-1\\2y-5=-143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\y=-69\end{cases}}\)

\(\orbr{\begin{cases}x+10=1\\2y-5=143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-9\\y=74\end{cases}}\)

\(\orbr{\begin{cases}x+10=143\\2y-5=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=133\\y=3\end{cases}}\)

Vậy ta có các cặp x,y thõa mãn : (-153,2);(-11,-69);(-9,74);(113,3)

b) x+(x+1)+(x+2)+..+(x+30)=1240

=> (x+x+x+...+x)+(1+2+3+...+30)=1240

=> 31x+465=1240

31x = 1240-465

31x = 775

x = 775 : 31

x= 25

c) 1+2+3+...+x=210

\(\frac{\left(x-1\right)}{1}+1=x\)

=> \(\frac{\left(x+1\right).x}{2}=210\)

(x+1)x = 210:2

(x+1)x = 105

chắc ko có x thõa mãn

d) 2+4+6+...+2x=210

=> 2(1+2+3+...+x)=210

1+2+3+..+x= 210:2 = 105

\(\frac{\left(x-1\right)}{1}+1\) = x

\(\frac{\left(x+1\right).x}{2}=105\)

(x+1)x = 105:2

(x+1)x = 52,5

ko có x thõa mãn đề bài

a, x + 10 và 2y - 5 thuộc Ư(143) = {1;-1;143;-143}

 b, x+(x+1)+(x+2)+........+(x+30) = 1240

=> x+x+1+x+2+...+x+30=1240

=> 31x+(1+2+...+30) = 1240

=> 31x + 465 = 1240

=> 31x = 775

=> x = 25

c, 1+2+...+x=210

=> \(\frac{x\left(x+1\right)}{2}=210\)

=> x(x+1) = 420

Mà 420 = 20.21

=> x = 20

d, 2+4+...+2x = 210

=> 2(1+2+...+x) = 210

=> \(\frac{2x\left(x+1\right)}{2}=210\)

=> x(x + 1) = 210

Mà 210 = 14.15

=> x  = 14

e, 1+3+5+...+(2x-1) = 225

=> \(\frac{\left[\left(2x-1\right)+1\right].x}{2}=225\)

=> \(\frac{2x^2}{2}=225\)

=> x² = \(\left(\pm15\right)^2\)

=> x = 15 hoặc x = -15

c,d ST làm đúng . Mình lận nhân thành chia :(

1. Ta có : 1 + 2 + ....+ x = 210 .

Ta có : \(\left(x-1\right):1+1=x-1+1=x\)

=> Lượng số hạng là x .

Ta có : Tổng \(=\frac{x\left(x+1\right)}{2}=210\)

=> \(x\left(x+1\right)=420\)

=> \(x^2+x-420=0\)

=> \(x^2-20x+21x-420=0\)

=> \(x\left(x-20\right)+21\left(x-20\right)=0\)

=> \(\left(x+21\right)\left(x-20\right)=0\)

=> \(\left[{}\begin{matrix}x+21=0\\x-20=0\end{matrix}\right.\)

Mà theo quy luật dãy số : \(0< x< 210\)

=> \(x-20=0\)

=> \(x=20\)

2, Ta có : \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

=> \(\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

=> \(\left(2x-15\right)^3\left(\left(2x-15\right)^2-1\right)=0\)

=> \(\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\end{matrix}\right.\)

Vậy ...

3, Ta có : \(x^{10}=x\)

=> \(x^{10}-x=0\)

=> \(x\left(x^9-1\right)=0\)

=> \(\left[{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy ...

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

=> \(\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)

=> \(\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy ...

a)    \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)

\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)

\(=\frac{1}{2}+\frac{1}{3}\)

\(=\frac{5}{6}\)

b)   \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)

\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)

\(=1\times\frac{1}{2}=\frac{1}{2}\)

c)   \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)

\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)

\(=\frac{247}{420}+\frac{72}{420}\)

\(=\frac{319}{420}\)

còn phần D đâu bn 

tích cho chị nha

 câu a)     = 5/6

 câu b = 1/2

 câu c = 319/420

x3x

= 14/15 x20/21 x 27/28 x ... x 209/210

= 28/30 x 40/42 x 54/56 x ... x 418/420

= 4x7/5x6 x 5x8/6x7 x 6x9/7x8 x ... x 19x22/20x21

= (4x7x5x8x6x9x...x19x22) / (5x6x6x7x7x8x...x20x21)

bạn gộp vào và tối giản đi, cuối cùng đc : 11/15 ( mik ko chắc chắn đâu nên bạn thử lại nhé)

thx bạn Ngô Khánh Linh

\(\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)....\left(1-\frac{1}{210}\right)\)

\(=\frac{14}{15}\cdot\frac{20}{21}\cdot\frac{27}{28}\cdot.....\cdot\frac{209}{210}\)

\(=\frac{28}{30}\cdot\frac{40}{42}\cdot\frac{54}{56}\cdot....\cdot\frac{418}{420}\)

\(=\frac{4\cdot7}{5\cdot6}\cdot\frac{5\cdot8}{6\cdot7}\cdot\frac{6\cdot9}{7\cdot8}\cdot...\cdot\frac{19\cdot22}{20\cdot21}\)

\(=\frac{4\cdot5\cdot6\cdot...\cdot19}{5\cdot6\cdot7\cdot8\cdot20}\cdot\frac{7\cdot8\cdot9\cdot....\cdot22}{6\cdot7\cdot8\cdot....\cdot21}\)

\(=\frac{4}{20}\cdot\frac{22}{6}=\frac{11}{15}\)

Bui Bao Ngoc oi! phan a,la (x+1)/10 hay x+(1/10)?

cac phan khac nua!