hình như đề bài sai

3/y² hay y²/3

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

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

Ta có: \(\frac{x-3}{6}=\frac{2}{x-4}\Leftrightarrow\left(x-3\right)\left(x-4\right)=6.2=12\)( 1 )

Do \(x\in Z\Rightarrow\hept{\begin{cases}x-3\in Z\\x-4\in Z\end{cases}}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\left(x-3\right).\left(x-4\right)=3.4\Leftrightarrow x-3=4\Leftrightarrow x=7.\)

Một cách khác.

x - 3/6 = 2/x - 4

<=> (x - 3).(x - 4) = 2.6

<=> x^2 - 7x + 12 = 12

<=> x^2 - 7x = 12 - 12

<=> x^2 - 7x = 0

<=> x(x - 7) = 0

<=> x = 0 hoặc x - 7 = 0

                         x       = 0 + 7

                         x       = 7

=> x = 0 hoặc 7

\(a,đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}.\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(b,x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow x=\sqrt{3}-1\)

\(\Rightarrow A=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}\)

\(b,A=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}\)\(=1-\frac{4}{\sqrt{x}+2}\)

\(A\in Z\Leftrightarrow1-\frac{4}{\sqrt{x}+2}\in Z\Rightarrow\frac{4}{\sqrt{x}+2}\in Z\)

\(\Rightarrow\sqrt{x}+2\inƯ_4\)

Mà \(Ư_4=\left\{\pm1;\pm2;\pm4\right\}\)Nhưng \(\sqrt{x}+2\ge2\)\(\Rightarrow\sqrt{x}+2\in\left\{2;4\right\}\)

\(Th1:\sqrt{x}+2=2\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

\(Th2:\sqrt{x}+2=4\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

\(KL:x\in\left\{0;4\right\}\)

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

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 }

Xét đẳng thức , ta thấy :

\(\left|x+\frac{3}{4}\right|\ge0\)

\(\left|y-\frac{1}{5}\right|\ge0\)

\(\left|x+y+z\right|\ge0\)

=> \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\)

Mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\) (đề bài)

=> \(\hept{\begin{cases}\left|x+\frac{3}{4}\right|=0\\\left|y-\frac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{1}{5}\\z=-\left(-\frac{3}{4}+\frac{1}{5}\right)=\frac{11}{20}\end{cases}}\)

Ta thấy một điều phê phê thế này :v  : |a| >= 0 
=> x+3/4=0 
y-1/5=0
x+y+z=0 
=> x=-3/4 =>y=1/5 => z= 3/4 - 1/5 = 11/20 
còn Trường hợp >0 Loại vì lúc ấy phương trình vô nghiệm rồi :v

Ta có: \(x+y+y+z+z+x=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}\)

          \(x+y+z=\frac{13}{12}:2=\frac{13}{24}\)

          \(x=\frac{13}{24}-\frac{1}{3}=\frac{5}{24}\)

         \(y=\frac{13}{24}-\frac{1}{4}=\frac{7}{24}\)

         \(z=\frac{13}{24}-\frac{1}{2}=\frac{1}{24}\)

 Vậy x = ....; y = .....; z = .......

k cho mik nha