\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{4}{3}.\frac{-1}{3}=\frac{-4}{9}\)

 k nha

\(\frac{-4}{9}\)k mk nha

\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\)

\(=\frac{4}{3}.\left(\frac{1}{6}-\frac{3}{6}\right)\)

\(=\frac{4}{3}.\frac{-2}{6}\)

\(=\frac{4}{3}.\frac{-1}{3}\)

\(=\frac{-4}{9}\)

-xx=-2x4

-xx=-8

xx=8

x²=8

x= căn bâc của 8

a; \(\dfrac{-x}{4}\) = \(\dfrac{-2}{x}\)

    -\(x.x\) = -2.4

    -\(x^2\) = -8

      \(x^2\) = 8

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

Vậy \(x\in\) {-\(\sqrt{8}\); \(\sqrt{8}\)}

b; \(\dfrac{-4}{8}\) = \(\dfrac{x}{-10}\) = \(\dfrac{-7}{y}\) = \(\dfrac{7}{21}\)

     \(-\dfrac{4}{8}\) = - \(\dfrac{1}{2}\) ≠  \(\dfrac{1}{3}\) = \(\dfrac{7}{21}\)

Vậy pt vô nghiệm

a)ta có xy=7*9=7*3*3

vậy x =9;21 , y=7;3

b) xy=-2*5

mà x<0<y

nên x=-2 ,y=5

c)x-y=5 hay x=y+5

\(\frac{y+5+4}{y-5}=\frac{4}{3}\Rightarrow3y+27=4y-20\Rightarrow y=47\Rightarrow x=52\)

câu c mk nhầm đề sr bạn nha

\(\frac{y+5-4}{y-5}=\frac{4}{3}\Rightarrow3y+3=4y-5\Rightarrow y=8\Rightarrow x=13\)

\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

\(x-\frac{5}{6}+x-x=-\frac{2}{3}\)

\(x=\frac{-2}{3}+\frac{5}{6}\)

\(x=\frac{-4}{6}+\frac{5}{6}\)

\(x=\frac{1}{6}\)

\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

\(x-\frac{5}{6}+x=x-\frac{2}{3}\)

\(\Rightarrow x+x-\frac{5}{6}=x-\frac{2}{3}\Rightarrow x+x-x=-\frac{2}{3}+\frac{5}{6}\)

\(\Rightarrow x=\frac{1}{6}\Rightarrow\)x ko tồn tại

\(x-\left(\frac{5}{6}-x\right)=x-\frac{2}{3}\)

\(x-\frac{5}{6}+x=x-\frac{2}{3}\)

\(2x-x=\frac{-2}{3}+\frac{5}{6}\)

\(x=\frac{1}{6}\)

\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

 \(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right)\div\left(\frac{x\left(x-2\right)}{x\left(x^2-x+1\right)}\right)\)

\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\div\frac{x-2}{x^2-x+1}\)

\(=\left(\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\times\frac{x^2-x+1}{x-2}\)

\(=\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{x^2-x+1}{x-2}\)

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

b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)

<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)

Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)

c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z

=> -2x ⋮ x + 1

=> -2x - 2 + 2 ⋮ x + 1

=> -2( x + 1 ) + 2 ⋮ x + 1

Vì -2( x + 1 ) ⋮ ( x + 1 )

=> 2 ⋮ x + 1

=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }

Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

Vậy x ∈ { -3 ; -2 ; 0 ; 1 }

Bài 2  :x+1/3=x-3/4                                  <=>4.(x+1)=3.(x-3)                             4x+4=3x-9                                                   4x-3x=-9-4                                                    x=-13

Bài 1: 

ta có: \(\frac{17}{x+1}.\frac{x}{6}=\frac{17x}{6x+6}\)

Để 17x/6x+6 thuộc Z

=> 17x chia hết cho 6x + 6

=> 102x chia hết cho 6x + 6

102x + 102 - 102 chia hết cho 6x + 6

17.(6x+6) - 102 chia hết cho 6x+6

mà 17.(6x+6) chia hết cho 6x + 6

=> 102 chia hết cho 6x + 6

=> ...

bn tự lm típ nha!

Bài 2:

ta có: \(\frac{x+1}{3}=\frac{x-3}{4}\)

\(\Rightarrow4x+4=3x-9\)

\(\Rightarrow4x-3x=-9-4\)

\(x=-13\)

Bài 3: ( đề phần a;b bn chép nhầm r, vì mk ko thấy y)

a) ta có: \(\frac{x}{3}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=3k\\y=5k\end{cases}}\)

mà x.y = 60 => 3k.5k = 60 => 15.k² = 60 => k² = 4 => k = 2 hoặc k = -2

nếu k = 2

=> x = 3.2 => x = 6

y = 5.2 => y = 10

nếu k = -2

=> x = 3.(-2) => x = -6

y = 5.(-2) => y = -10

KL:...

b) \(\frac{x}{7}=\frac{y}{10}=k\Rightarrow\hept{\begin{cases}x=7k\\y=10k\end{cases}\Rightarrow y^2=100.k^2}\)

mà x.y² = 5600 => 7k.100k² = 5600 => 700.k³ = 5600 => k³ = 8 => k = 2

=> x = 7.2 => x = 14

y = 10.2 => y = 20

KL:...

a) Điều kiện : \(x\ne2;x\ne3\)

 \(B=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+4}{3-x}=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}-\frac{x+3}{x-2}+\frac{2x+4}{x-3}\)

\(=\frac{2x-9-\left(x-3\right)\left(x+3\right)+2\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{2x-9-x^2+9+2x^2-8}{\left(x-2\right)\left(x-3\right)}=\frac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+4}{x-3}\)

b) Điều kiện \(x\in Z;x\ne2;x\ne3\)

Có \(B=\frac{x+4}{x-3}\in Z\), mà x+4 và x-3 nguyên do x nguyên, nên

\(x+4⋮x-3\Leftrightarrow7⋮x-3\), do đó \(x-3\inƯ\left(7\right)=\left\{1;7;-1;-7\right\}\Rightarrow x\in\left\{4;10;2;-4\right\}\)

mà do x khác 2 (điều kiện) nên ta kết luận \(x\in\left\{4;10;-4\right\}\)