a,x.y=3=1x3=3x1=-1x(-3)=-3x(-1).

Vậy (x,y)=(1,3)=(3,1)=(-1,-3)=(-3,-1)

b,x.(y+1)=5=1x5=5x1=-1x(-5)=-5x(-1)

=>

Vậy (x,y)=(1,4)=(5,0)=(-1,-6)=(-1,-2).

c,(x-2)(y+3)=7=1x7=7x1=-1x(-7)=-7(-1)

=>

Vậy (x,y)=(3,4)=(9,-2)=(1,-10)=(-5,-4).

`a, x/7 =-4/14`

`=> 14x=7.(-4)`

`=>14x=-28`

`=>x=-28:14`

`=>x=-2`

`b,x/2=-2/-x`

`=>x/2=2/x`

`=>x.x=2.2`

`=>x^2=4`

`=>x= +-2`

`c,(x-1)/5=5/(x-1)`

`=>(x-1)^2 = 5.5`

`=>(x-1)^2=25`

`=>(x-1)^2=5^2`

\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

`d,x+3/2=-12/16`

`=>x=-12/16 -3/2`

`=>x= -12/16 - 24/16`

`=>x= -36/16`

`=>x=-9/4`

Mk cần gấp để nộp ạ

B2

( a³ + a²b + ab² + b³ ).( a - b ) = a⁴ - b⁴

[( a³ + b³ + ab.( a + b )].( a - b ) = a⁴ - b⁴

[( a + b ).( a² - ab + b² ) + ab.( a + b )].( a - b ) = a⁴ - b⁴

 ( a + b ).( a² - ab + b² + ab ).( a - b ) = a⁴ - b⁴

( a + b ).( a² + b² ).( a  -  b ) = a⁴ - b⁴

 ( a² - b² ).( a² + b² ) = a⁴ - b⁴

 a⁴ - b⁴ = a⁴ - b⁴  ( đpcm )

a) Ta có: \(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)

nên \(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{2x}{3}=12\\\dfrac{3y}{4}=12\\\dfrac{4z}{5}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=36\\3y=48\\4z=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=16\\z=20\end{matrix}\right.\)

Vậy: (x,y,z)=(18;16;20)

b) Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k\\y=3k\end{matrix}\right.\)

Ta có: \(x^2-y^2=4\)

\(\Leftrightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Leftrightarrow16k^2=4\)

\(\Leftrightarrow k\in\left\{\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

Trường hợp 1: \(k=\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=5\cdot\dfrac{1}{2}=\dfrac{5}{2}\\y=3k=3\cdot\dfrac{1}{2}=\dfrac{3}{2}\end{matrix}\right.\)

Trường hợp 2: \(k=-\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5k=5\cdot\dfrac{-1}{2}=\dfrac{-5}{2}\\y=3k=3\cdot\dfrac{-1}{2}=\dfrac{-3}{2}\end{matrix}\right.\)

Vậy: \(\left(x,y\right)\in\left\{\left(\dfrac{5}{2};\dfrac{3}{2}\right);\left(-\dfrac{5}{2};-\dfrac{3}{2}\right)\right\}\)

a)

Theo tính chất của dãy tỉ số bằng nhau, ta có : 

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Suy ra : 

\(x=\dfrac{12.3}{2}=18\\ y=\dfrac{12.4}{3}=16\\ z=\dfrac{12.5}{4}=15\)

b)

\(x=\dfrac{y}{3}.5=\dfrac{5y}{3}\\ x^2-y^2=4\\ \Leftrightarrow\left(\dfrac{5y}{3}\right)^2-y^2=4\\ \Leftrightarrow\dfrac{16y^2}{9}=4\Leftrightarrow y=\pm\dfrac{3}{2} \)

Với $y = \dfrac{3}{2}$ thì $x = \dfrac{5}{2}$

Với $y = \dfrac{-3}{2}$ thì $x = \dfrac{-5}{2}$

c)

\(\dfrac{x}{y+z+1}=\dfrac{y}{z+x+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2x+2y+2z}=\dfrac{1}{2}\)

Suy ra : 

\(2x=y+z+1\Leftrightarrow y+z=2x-1\)

Mặt khác : 

\(x+y+z=\dfrac{1}{2}\Leftrightarrow x+2x-1=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(2y=x+z+1=z+\dfrac{3}{2}\)

Mà \(y+z=0\Leftrightarrow z=-y\)

nên suy ra:  \(y=\dfrac{1}{2};z=-\dfrac{1}{2}\)

a) \(\frac{-9}{x}=\frac{x}{-49}\Leftrightarrow x^2=\left(-9\right)\left(-49\right)=441\Leftrightarrow x=\pm21\)

b) \(\frac{x}{3}=\frac{y}{5};x+y=-16\). Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{3}=\frac{y}{5}=\frac{x+y}{3+5}=\frac{-16}{8}=-2\Rightarrow x=-6;y=-10\)

c) \(\frac{x}{2}=\frac{y}{-5};y-x=-14\). Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{2}=\frac{y}{-5}=\frac{y-x}{-5-2}=\frac{-14}{-7}=2\Leftrightarrow x=4;y=-10\)