Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x+2}{4}=\dfrac{2y+2}{5}=\dfrac{3x+2y+4}{4,5x}=\dfrac{3x+2+2y+2-3x-2y-4}{4+5-4,5x}=\dfrac{0}{9-4,5x}=0\)

\(\Rightarrow\left\{{}\begin{matrix}3x+2=0\\2y+2=0\\3x+2y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=-2\\2y=-2\\3x+2y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=-1\end{matrix}\right.\)

Áp dụng t/c dãy tỉ số bằng nhau :

\(\dfrac{3x+2}{4}=\dfrac{2y+2}{5}=\dfrac{3x+2+2y+2}{4+5}=\dfrac{3x+2y+4}{9}\)

Mà \(\dfrac{3x+2}{4}=\dfrac{2y+2}{5}=\dfrac{3x+2y+4}{4,5x}\)

=> \(\dfrac{3x+2y+4}{9}=\dfrac{3x+2y+4}{4,5x}\)

=> 9 = 4,5x

=> x = 9 : 4,5 = 2

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

\(\dfrac{3.2+2}{4}=\dfrac{2y+2}{5}\) ( Thay x = 2)

\(2=\dfrac{2y+2}{5}\)

=> 2y = 2.5 - 2 = 8

=> y = 8 : 2 = 4

Vậy x = 2, y = 4

a, \(\dfrac{x^3+27}{x^2-3x+9}=\dfrac{x+3}{M}\Leftrightarrow\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{x+3}{M}\)

\(\Rightarrow M=\dfrac{x+3}{x+3}=1\)

b, \(\dfrac{M}{x+4}=\dfrac{x^2-8x+16}{16-x^2}=\dfrac{\left(x-4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{4-x}{x+4}\)

\(\Rightarrow M=\dfrac{\left(4-x\right)\left(x+4\right)}{x+4}=4-x\)

c, tương tự 

a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

cho mik sửa lại câu

b) \(2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

b) \(2y-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

\(=\dfrac{2y\left(3x+2y\right)}{3x+2y}-\dfrac{6xy+2y}{3x+2y}+\dfrac{2y-9x^2}{3x+2y}\)

\(=\dfrac{2y\left(3x+2y\right)-\left(6xy+2y\right)+\left(2y-9x^2\right)}{3x+2y}\)

\(=\dfrac{6xy+4y^2-6xy-2y+2y-9x^2}{3x+2y}\)

\(=\dfrac{4y^2-9x^2}{3x+2y}\)

\(=\dfrac{-\left(9x^2-4y^2\right)}{3x+2y}\)

\(=\dfrac{-\left[\left(3x\right)^2-\left(2y\right)^2\right]}{3x+2y}\)

\(=\dfrac{-\left(3x-2y\right)\left(3x+2y\right)}{3x+2y}\)

\(=-\left(3x-2y\right)\)

\(=-3x+2y\)

a)\(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{\left(1+x\right)+\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2\left(1+x^2\right)+2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4\left(1+x^4\right)}{\left(1+x^4\right)\left(1-x^4\right)}+\dfrac{4\left(1-x^4\right)}{\left(1+x^4\right)\left(1-x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4\left(1+x^4\right)+4\left(1-x^4\right)}{\left(1+x^4\right)\left(1-x^4\right)}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)

\(=\dfrac{16\left(1+x^{16}\right)+16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)

\(=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}\)

\(=\dfrac{32}{1-x^{32}}\)

Từ tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x+2}{3}=\dfrac{2y-6}{9}=\dfrac{\left(3x+2\right)+\left(2y-6\right)}{3+9}=\dfrac{3x+2y-4}{12}=\dfrac{3x+2y-4}{6x}\)

Suy ra 6x = 12 <=> x = 12 : 6 = 2

Khi đó \(\dfrac{3x+2}{3}=\dfrac{3\cdot2+2}{3}=\dfrac{8}{3}\)

Suy ra \(\dfrac{2y-6}{9}=\dfrac{8}{3}\Leftrightarrow2y-6=\dfrac{8\cdot9}{3}=24\)

\(\Leftrightarrow2y=24+6=30\Leftrightarrow y=30:2=15\)

Vậy x = 2; y = 15

minh trang là nữ hay nam

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

=>\(\dfrac{4\left(3x-2y\right)}{4.4}=\dfrac{3\left(2z-4x\right)}{3.3}=\dfrac{2\left(4y-3z\right)}{2.2}\)

=>\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8y+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)

=>\(\dfrac{12x-8y}{16}=0\)

=>12x-8y=0

=>12x=8y

=>\(\dfrac{12x}{24}=\dfrac{8y}{24}\)

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

Lại có \(\dfrac{8y-6z}{4}=0\)

=>8y-6z=0

=>8y=6z

=>\(\dfrac{8y}{24}=\dfrac{6z}{24}\)

=>\(\dfrac{y}{3}=\dfrac{z}{4}\)(2)

từ (1) và (2)=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)

a) Để y nguyên thì \(6x-4⋮2x+3\)

\(\Leftrightarrow-13⋮2x+3\)

\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)

\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)

hay \(x\in\left\{-1;-2;5;-8\right\}\)

a.

Với \(y=0\) không phải nghiệm

Với \(y\ne0\Rightarrow\left\{{}\begin{matrix}3x+2=\dfrac{5}{y}\\2x\left(x+y\right)+y=\dfrac{5}{y}\end{matrix}\right.\)

\(\Rightarrow3x+2=2x\left(x+y\right)+y\)

\(\Leftrightarrow2x^2+\left(2y-3\right)x+y-2=0\)

\(\Delta=\left(2y-3\right)^2-8\left(y-2\right)=\left(2y-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-2y+3+2y-5}{4}=-\dfrac{1}{2}\\x=\dfrac{-2y+3-2y+5}{4}=-y+2\end{matrix}\right.\)

Thế vào pt đầu ...

Câu b chắc chắn đề sai