Đề còn cho gì nữa ko

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

\(\frac{4x+12xy}{48x}=\frac{3x+18xy}{48x}=\frac{12+108y}{48x}\)

\(\Rightarrow4x+12xy=3x+18xy=108y+12\)

\(\Rightarrow18xy+3x-4x-12xy-108y-12=0\)

\(\Rightarrow6xy+x-108y-12=0\)

Đến đây mk chịu

bạn tham khảo ở đây nha : Kết quả tìm kiếm | Học trực tuyến

          \(-\left(2x+1\right)\left(3y-2\right)+6y\left(x-1\right)=15-\left(1+9y\right)\)

\(\Leftrightarrow-\left(6xy-4x+3y-2\right)+6xy-6y=15-1-9y\)

\(\Leftrightarrow-6xy+4x-3y+2+6xy-6y-14+9y=0\)

\(\Leftrightarrow4x-12=0\)

\(\Leftrightarrow4x=12\)

\(\Leftrightarrow x=3\)

Chúc bạn học tốt!!!
Tk cho mk nha... Cảm ơn nhìu...

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

\(\Leftrightarrow\dfrac{1}{3y^2-10y}=\dfrac{6y-2\left(3y+1\right)}{\left(3y-1\right)\left(3y+1\right)}\)

\(\Leftrightarrow\dfrac{1}{3y^2-10y}=\dfrac{-2}{9y^2-1}\)

\(\Leftrightarrow9y^2-1=-6y^2+20y\)

\(\Leftrightarrow15y^2-20y-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{10+\sqrt{115}}{15}\\y=\dfrac{10-\sqrt{115}}{15}\end{matrix}\right.\)

\(a,A+B-C=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)

\(=\left(16x^4-15x^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)+\left(-9y^4-6y^4-17y^4\right)-1\)

\(=x^4-10x^3y-x^2y^2-32y^4-1\)

\(b,A-C+B=A+B-C\) ( giống câu a )

\(a,\)

\(A+B+C\)

\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-\left(5x^3y+3x^2y^2+17y^4+1\right)\)

\(=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-5x^3y-3x^2y^2-17y^4-1\)

\(=\left(16x^4-15x^4\right)+\left(-9y^4-6y^4-17y^4\right)+\left(-8x^3y+3x^3y-5x^3y\right)+\left(7x^2y^2-5x^2y^2-3x^2y^2\right)-1\)

\(=x^4-32y^4-10x^3y-x^2y^2-1\)

\(b,\)

\(A-C+B=A+B-C=x^4-32y^4-10x^3y-x^2y^2-1\)

a: A+B-C

=16x^4-8x^3y+7x^2y^2-9y^4-15x^4+3x^3y-5x^2y^2-6y^4-C

=x^4-5x^3y+2x^2y^2-15y^4-5x^3y-3x^2y^2-17y^4-1

=x^4-10x^3y-x^2y^2-32y^4-1

b: A-C+B=A+B-C=x^4-10x^3y-x^2y^2-32y^4-1