1. \(3-|2x+1|=-5\)

\(\Rightarrow|2x+1|=8\)

\(\Rightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{9}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{7}{2};-\frac{9}{2}\right\}\)

2.\(12+|3-x|=9\)

\(\Rightarrow|3-x|=-3\)

Mà \(|3-x|\ge0\forall x\)

\(\Rightarrow\)Vô lí

Vậy không có x

3.\(|x+9|=12+\left(-9\right)+2\)

\(\Rightarrow|x+9|=5\)

\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)

Vậy \(x\in\left\{-4;-14\right\}\)

4.\(5x-16=40+x\)

\(\Rightarrow5x-x=40+16\)

\(\Rightarrow4x=56\)

\(\Rightarrow x=14\)

Vậy \(x=14\)

5.\(5x-7=-21-2x\)

\(\Rightarrow5x+2x=-21+7\)

\(\Rightarrow7x=-14\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

6.\(\left(2x-1\right)\left(y-2\right)=12\)

Vì \(x,y\inℤ\)nên \(2x-1;y-2\inℤ\)

\(\Rightarrow2x-1;y-2\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Ta có bảng : (em tự xét bảng nhé)

(x-40)(x-7)=0

=>x-40=0 hoặc x-7=0

=>x=40   hoặc x=7

tíc mình nha

1.\(\left(x-40\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-40=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=40\\x=7\end{cases}}\)

2. \(\left(x-12\right)\left(x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-12=0\\x+13=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)

3. \(\left(2-x-4\right)\left(3x-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2-x-4=0\\3x-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

e: =>-40+3+33+40-x=47

=>36-x=47

=>x=-11

f: =>x(x-3)(11-x)(11+x)=0

hay \(x\in\left\{0;3;11;-11\right\}\)

g: =>-62-38-x+2x=-100

=>x-100=-100

hay x=0

i: =>x-12-2x-31=6

=>-x-43=6

=>x+43=-6

hay x=-49

h: =>(x+1)=0

=>x=-1

f: =>x(x-3)(x+11)(x-11)=0

hay \(x\in\left\{0;3;-11;11\right\}\)

a) 3/7 + 4/9 + 4/7 + 5/9

= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )

= 7/7 + 9/9

= 1  + 1 

= 2

b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45

= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5

= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5  ) + 5/5

= 2 + 2 + 2 + 2 + 1

= 2  x 4 + 1

= 8  +1 

= 9

c)  1/8 + 1/12 + 3/8 + 5/12

= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)

= 4/8 + 6/12

= 1/2 + 1/2

= 2/4 = 1/2

mỏi tay rồi

d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))

 = \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\)  x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)

= \(\dfrac{1}{100}\)

e; \(\dfrac{9}{5}\) : \(\dfrac{17}{15}\) + \(\dfrac{8}{5}\) : \(\dfrac{17}{15}\)

=  \(\dfrac{9}{5}\) x \(\dfrac{15}{17}\) + \(\dfrac{8}{5}\) x \(\dfrac{15}{17}\)

=  \(\dfrac{27}{17}\) + \(\dfrac{24}{17}\)

=  \(\dfrac{51}{17}\)

=   3

9: =3/2*5/12+3/16

=15/24+3/16

=5/8+3/16

=13/16

x^40+2.x^20+9 = [x^20 +3]^2 - 4x^20 = [x^20+3]^2 -[2x^10]^2 = [x^20-2x^10+3].[x^20+2x^10+3]

x^12+x^6+1 = x^12 + 2x^6 +1 - x^6 = [x^6 +1]^2 -[x^3]^2 = [x^6 -x^3 +1].[x^6+x^3+1]

x^16+x^8+1 =[x^8+1]^2 - [x^4]^2 = [x^8-x^4+1].[x^8+x^4+1]

x^4+x^2+1 = x^4+2x^2+1 - x^2 = [x^2+1]^2-x^2 = [x^2-x+1].[x^2+x+1]

Bài 1: 

ta có: \(x^2+2⋮x+2\)

\(\Leftrightarrow x^2-4+6⋮x+2\)

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

hay \(x\in\left\{-1;-3;0;-4;1;-5;4;-8\right\}\)

\(\dfrac{1}{6}x+\dfrac{1}{12}x+\dfrac{1}{20}x+...+\dfrac{1}{2450}x=1\)

\(x\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2450}\right)\)=1

\(x\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{49\times50}\right)\)=1

\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)

\(x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)

\(x\times\)\(\dfrac{12}{25}=1\)

\(\Rightarrow x=1\div\dfrac{12}{25}\)

\(x=1\times\dfrac{25}{12}=\dfrac{25}{12}\)

vậy \(x=\dfrac{25}{12}\)

vậy \(x=2\)\(x=2\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)\(\left(2\dfrac{2}{9}-x\right)\)=\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)

\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

nè bn!!!!

máy hết chỗ đánh làm ơn chuyển 3 dòng đầu này xuống cuối

\(\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)

\(\left(\dfrac{20}{9}-x\right)=\dfrac{2}{9}\)

\(x=\dfrac{20}{9}-\dfrac{2}{9}\)

\(x=2\)

vậy x=2

giúp mk nha !!!THANKS!!!