\(\Leftrightarrow\dfrac{1}{2}x+\dfrac{2}{3}x-x=-4\Leftrightarrow\dfrac{3x+4x-6x}{6}=-\dfrac{24}{6}\)

\(\Rightarrow x=-24\)

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

\(x.\dfrac{-1}{6}=-4\)

\(x=-4:\dfrac{-1}{6}\)

\(x=-24\)

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow-x^2+6x-3=-x^2+3x+1\)

\(\Leftrightarrow3x=4\)

hay \(x=\dfrac{4}{3}\)

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

\(\Leftrightarrow2x^2+6x-3x^2-3=x+1-x^2+2x\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)

Bài 2:

Với x,y,z,t là số tự nhiên khác 0

Có \(\dfrac{x}{x+y+z+t}< \dfrac{x}{x+y+z}< \dfrac{x}{x+y}\)

\(\dfrac{y}{x+y+z+t}< \dfrac{y}{x+y+t}< \dfrac{y}{x+y}\)

\(\dfrac{z}{x+y+z+t}< \dfrac{z}{y+z+t}< \dfrac{z}{z+t}\)

\(\dfrac{t}{x+y+z+t}< \dfrac{t}{x+z+t}< \dfrac{t}{z+t}\)

Cộng vế với vế \(\Rightarrow1< M< \dfrac{x+y}{x+y}+\dfrac{z+t}{z+t}=2\)

=> M không là số tự nhiên.

Bài 1:

Ta có:

\(B=\dfrac{2008}{1}+\dfrac{2007}{2}+\dfrac{2006}{3}+...+\dfrac{2}{2007}+\dfrac{1}{2008}\) 

\(B=\left(1+\dfrac{2007}{2}\right)+\left(1+\dfrac{2006}{3}\right)+...+\left(1+\dfrac{2}{2007}\right)+\left(1+\dfrac{1}{2008}\right)+1\) 

\(B=\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}\) 

\(B=2009.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)\) 

\(\Rightarrow\dfrac{A}{B}=\dfrac{2009.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}+\dfrac{1}{2008}+\dfrac{1}{2009}}=2009\)

b) 8

c) 3

b) 50-3(x+4)=14

3(x+4)=36

x+4=13

x=9

c)2⁸‐ⁿ+75=107

2⁸-ⁿ=32

2⁸-ⁿ=2⁵

8-x=5

x=3

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\)

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

Bạn xem lại đề

a)Ta có :

25-\(y^2\)=8(8-2009)

⇔ 0 ≤ \(y^2\)≤ 25

⇒y∈{1;2;3;4;5}

Mà 25-\(y^2\)⋮8(Vì x ∈ Z)

⇒y∈{1;3;5}(t/mãn y ∈ Z)

TH1:Với y =1 ,ta có:

25-\(y^2\)=\(8\left(x-2009\right)^2\)

⇔25-\(1^2\)=\(8\left(x-2009\right)^2\)

⇔\(8\left(x-2009\right)^2\) =24

⇔\(\left(x-2009\right)^2\)= 3(vô lí)

⇒TH1 loại

TH2Với y =3,ta có:

25-\(y^2\) =8(x-2009)

⇔25-\(3^2\)=\(8\left(x-2009\right)^2\) 

⇔\(8\left(x-2009\right)^2\)=16

⇔\(\left(x-2009\right)^2\)=2(vô lí)

⇒TH2 loại

TH3Với y=5,ta có:

25-\(y^2\) =\(8\left(x-2009\right)^2\)

⇔25-\(5^2\)=\(8\left(x-2009\right)^2\) 

⇔\(8\left(x-2009\right)^2\)=0

⇒x-2009=0

⇒x=2009(t/mãn x∈Z)

Vậy y=5 x=2009

\(35-5\left(x-1\right)=10\\ \Leftrightarrow35-5x+5=10\\ \Rightarrow40-5x=10\)

\(\Rightarrow-5x=10-40\\ \Rightarrow-5x=-30\\ \Rightarrow x=\dfrac{-30}{-5}=6\)

c) 

\(24\left(x-16\right)=12^2\)

\(\Rightarrow24x-384=144\\ \Rightarrow24x=144+384\\ \Rightarrow24x=528\\ \Rightarrow x=\dfrac{528}{24}=22\)

d) 

\(\left(x^2-10\right)\div5=3\\ \Rightarrow\left(x^2-10\right)=3\times5\\ \Rightarrow x^2-10=15\)

\(\Rightarrow x^2=15+10\\ \Rightarrow x^2=25\\ \Rightarrow x^2=5^2\Rightarrow x=5\)

Ơ, bn ơi @Kenny sai sai ở đâu thì phải.

`4x^2=13`

`<=>x^2=13/4`

`<=>x=[+-\sqrt{13}]/2`

Vậy `S={[+-\sqrt{13}]/2}`

`4x^2 = 13`.

`=> x^2 = 13/4`.

`=> x = (sqrt 13)/(sqrt 4)`

`=> x = (+-sqrt 13)/2`.

Vậy `S = (+-sqrt 13)/2`.

b ) - 25 + ( - 16 + x ) = 0

               ( - 16 + x ) = 0 - ( - 25 )

                 - 16 + x   = 25

                          x    = 25 - ( - 16 )

                          x    = 41

Vậy x = 41

x=7,y=1

_HT_