đề có đúng như z ko bn:

ta có: \(\frac{1+3y}{15}=\frac{1+6y}{18}\)

\(\Rightarrow\left(1+3y\right).18=\left(1+6y\right).15\)

\(18+54y=15+90y\)

\(54y-90y=15-18\)

\(-36y=-3\)

\(y=-3:-36\)

\(y=\frac{1}{12}\)

ta có: \(\frac{1+6y}{18}=\frac{1+9y}{9x}\)

\(\Leftrightarrow\left(1+6y\right).9x=\left(1+9y\right).18\)

\(9x+54xy=18+162y\)

thay số: \(9x+54.\frac{1}{12}x=18+162.\frac{1}{12}\)

\(9x+\frac{9}{2}x=18+\frac{27}{2}\)

\(x.\left(\frac{9}{2}+9\right)=31\frac{1}{2}\)

\(x.13\frac{1}{2}=31\frac{1}{2}\)

\(x=31\frac{1}{2}:13\frac{1}{2}\)

\(x=45\)

KL: x =45 ; y= 1/12

CHÚC BN HỌC TỐT!!!!

đề như của mình á bạn

Áp dụng dãy tỉ số bằng nhau:

\(\frac{2+3y}{13}=\frac{2+6y}{17}=\frac{2\left(2+3y\right)-\left(2+6y\right)}{2.13-17}=\frac{2}{9}\)

=> \(2+3y=\frac{26}{9}\)=> \(y=\frac{8}{27}\)

\(\frac{2+9y}{8x}=\frac{2+3y}{13}=\frac{2}{9}\)

=> \(9\left(2+9y\right)=2.8x\)

=> \(16x=42\)

=> \(x=\frac{21}{8}\)

thử lại thỏa mãn 

Vậy:...

Ta có : \(\frac{1+3y}{12}=\frac{1+6y}{16}=\frac{1+9y}{4x}\)

\(\Rightarrow\frac{1+3y}{12}=\frac{1+9y}{4x}=\frac{1+3y+1+9y}{12+4x}=\frac{2+12y}{12+4x}\)

\(\Rightarrow\frac{1+6y}{16}=\frac{2.\left(1+6y\right)}{12+4x}\)

Do đó : \(16=\frac{12+4x}{2}\)

Từ đó suy ra : x = 5

ĐK \(y\ne\left\{-\frac{1}{3};\frac{1}{3};3\right\}\)

a. Ta có \(\frac{1}{3y^2-10y+3}=\frac{6y}{9y^2-1}+\frac{2}{1-3y}\)

\(\frac{\Leftrightarrow1}{\left(y-3\right)\left(3y-1\right)}=\frac{6y}{\left(3y+1\right)\left(3y-1\right)}-\frac{2}{3y-1}\)

\(\Leftrightarrow\frac{3y+1}{\left(3y+1\right)\left(3y-1\right)\left(y-3\right)}=\frac{6y\left(y-3\right)-2\left(y-3\right)\left(3y+1\right)}{\left(3y+1\right)\left(3y-1\right)\left(y-3\right)}\)

\(\Leftrightarrow3y+1=-2y+6\Leftrightarrow5y=5\Rightarrow y=1\)

Vậy \(y=1\)

b. Pt \(\Leftrightarrow x-\frac{\frac{x-3}{4}}{2}=3-\frac{\frac{x-3}{6}}{2}\Leftrightarrow x-\frac{x-3}{8}=3-\frac{x-3}{12}\)

\(\Leftrightarrow\left(x-3\right)-\frac{x-3}{8}-\frac{x-3}{12}=0\Leftrightarrow\frac{19}{24}\left(x-3\right)=0\Leftrightarrow x=3\)

Vậy \(x=3\)

ta có: \(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}\)

\(\Rightarrow\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+6y-1-3y}{2x-12}=\frac{3y}{2x-12}\)

\(\Rightarrow\frac{3y}{2x-12}=\frac{1+9y}{5x}=\frac{9y+1-3y}{5x-2x+12}=\frac{1+6y}{3x+12}\)

\(\Rightarrow\frac{1+6y}{3x+12}=\frac{1+6y}{2x}\)

=> 3x + 12 = 2x 

=> 3x - 2x = - 12

x = -12

xog r bn chỉ cần thay x = -12 vào 2 trong 3 p/s bất kì trên là tính đk y

\(\frac{1+3y}{12}=\frac{1+6y}{2x}=\frac{1+9y}{5x}=\frac{1+3y+1+9y}{12+5x}=\frac{2+12y}{12+5x}\)

\(\Rightarrow\frac{1+6y}{2x}=\frac{2+12y}{12+5x}\)

\(\Rightarrow\frac{12+5x}{2}=2x\)

\(\Rightarrow12+5x=4x\)

\(\Rightarrow12=-x\Leftrightarrow x=-12\)

Thay x vô mà tìm y

1+6y/2x=1+9y/5x=1+3y/12

=>1+6y/2=1+9y/5=1+3y/12 (nhân cả 2 vế với x)

=>12*(1+6y)=2*(1+3y)

=>12+72y=2+6y

=>72y-6y=2-12

=>66y=-10

=>y=-10/66

=>y=-5/33

RỒI BẠN TỰ THAY Y VÀO ĐỂ TIM X

HOK TOT

Tự tìm Đkxđ nha.

1/(3y^2 - 10y +3) = 6y/(9y^2 - 1) + 2/(1 - 3y)

=>1/(3y^2 -9y -y +3)=6y/(3y- 1)(3y+ 1)- 2(3y+ 1)/(3y - 1)(3y+ 1)

=>1/(y- 3)(3y -1)=-1/(3y -1)(3y +1)

=>(3y+ 1)/(y- 3)(3y -1)(3y+ 1)=(y -3)/(3y- 1)(3y +1)

=>3y+ 1= y- 3

Đến đây tự làm nha

a)ĐKXĐ:\(\hept{\begin{cases}y\ne3\\y\ne\frac{1}{3}\\y\ne-\frac{1}{3}\end{cases}}\)

\(\frac{1}{3y^2-10y+3}=\frac{6y}{9y^2-1}+\frac{2}{1-3y}\)

\(\Leftrightarrow\frac{1}{\left(y-3\right)\left(3y-1\right)}=\frac{6y}{\left(3y-1\right)\left(3y+1\right)}-\frac{2}{3y-1}\)

\(\Leftrightarrow\frac{3y+1}{\left(y-3\right)\left(3y-1\right)\left(3y+1\right)}=\frac{6y\left(y-3\right)}{\left(3y-1\right)\left(3y+1\right)\left(y-3\right)}-\frac{2\left(3y+1\right)\left(y-3\right)}{\left(3y-1\right)\left(3y+1\right)\left(y-3\right)}\)

\(\Rightarrow6y^2-18y-2\left(3y^2-9y+y-3\right)-3y-1=0\)

\(\Leftrightarrow6y^2-18y-6y^2+18y-2y+6-3y-1=0\)

\(\Leftrightarrow5-5y=0\)

\(\Leftrightarrow5y=5\Leftrightarrow y=1\)(t/m ĐKXĐ)

Vậy....

a)

 \(3y^2-10y+3=3y^2-9y-y+3=3y\left(y-3\right)-\left(y-3\right)=\left(y-3\right)\left(3y-1\right)\)

\(9y^2-1=\left(3y\right)^2-1^2=\left(3y-1\right)\left(3y+1\right)\)

ĐK: \(y\ne3,\frac{1}{3},-\frac{1}{3}\)

pt <=> \(\frac{1}{\left(3y-1\right)\left(y-3\right)}=\frac{6y}{\left(3y-1\right)\left(3y+1\right)}-\frac{2}{3y-1}\)

<=> \(\frac{1}{y-3}=\frac{6y}{3y+1}-2\)

<=> \(\frac{3y+1}{\left(y-3\right)\left(3y+1\right)}=\frac{6y\left(y-3\right)}{\left(3y+1\right)\left(y-3\right)}-\frac{2\left(3y+1\right)\left(y-3\right)}{\left(3y+1\right)\left(y-3\right)}\)

<=> 3y+1=6y(y-3)-2(3y+1)(y-3)

<=> \(3y+1=6y^2-18y-6y^2+16y+6\)

<=> 5y=5 <=> y=1 ( thỏa mãn )

vậy y=1