ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{2}{15}\\y\ne-\dfrac{4}{9}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}9y+6+20x-16=0\\\left(5x-4\right)\left(9y+4\right)=\left(3y+2\right)\left(15x-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}20x+9y=10\\5x+15y=-6\end{matrix}\right.\)

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

sory doan cuoi minh lam sai minh lam lai nhe

pt<=>1/(y-1)(y-2) + 1/(y-2)(y-3) + 1/(y-3)(y-4) + 1(y-4)(y-5)=1/15

=>1/(y-1) -1/(y-2)+1/(y-2)-1/(y-3)+1/(y-3)-1/(y-4)+1/(y-4)-1/(y-5)=1/15

=>1/(y-1) - 1/(y-5)=1/15

=>4/(y-1)(y-5)=1/15

=> (y-1)(y-5)=60

=> y²-6y+5-60=0

=>y²-6y-55=0

=> (y-11)(y+5)=0

=>y=11 hoac y=-5

pt<=> 1/(y-1)(y-2) + 1/(y-2)(y-3) + 1/(y-3)(y-4) + 1/(y-4)(y-5)=1/15

=>1/(y-1)-1/(y-2)+1/(y-2)-1/(y-3)+1/(y-3)-1/(y-4)+1/(y-4)-1/(y-5)=1/15

=>1/(y-1)-1/(y-5)=1/15

=>(y-1)(y-5)=-4.15=-60

=>y²-6y+65=0 k tim dc nghiem

Bài 2: 

a) Ta có: \(\Delta=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=m^2-2m+1+4m^2+8\)

\(=5m^2-2m+9>0\forall m\)

Do đó, phương trình luôn có hai nghiệm phân biệt với mọi m

Bài 1:

ĐKXĐ \(2x\ne y\)

Đặt \(\dfrac{1}{2x-y}=a;x+3y=b\)

HPT trở thành

\(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\4\left(\dfrac{3}{2}-b\right)-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\6-9b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{8}{9}\\a=\dfrac{11}{18}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=\dfrac{8}{9}\\2x-y=\dfrac{18}{11}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-\dfrac{18}{11}\\x+3\left(2x-\dfrac{18}{11}\right)=\dfrac{8}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{82}{99}\\y=\dfrac{2}{99}\end{matrix}\right.\)

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 \(x=3\)

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

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

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)

Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)

\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)

Thay a,b:

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)

d. ĐKXĐ: x khác 1, x khác 3

\(\dfrac{x+5}{x-1}=\dfrac{x+1}{\left(x-3\right)}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\) \(\Leftrightarrow x^2+2x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+1+8=0\)

\(\Leftrightarrow2x-6=0\)

\(\Leftrightarrow x=3\) (loại)

Vậy pt vô nghiệm

Điều kiện \(\left\{{}\begin{matrix}\dfrac{4x-3x^2y-9xy^2}{x+3y}\ge0\\x+3y\ne0\end{matrix}\right.\)

Với \(3y\ge x\), hệ tương đương:

\(\left\{{}\begin{matrix}\left(x^4-2x^2+4\right)\left(x^2+2\right)=6x^5y\\\left(3y-x\right)^2=\dfrac{4x}{x+3y}-3xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^6+8=6x^5y\left(1\right)\\x^3+27y^3=4x\end{matrix}\right.\left(I\right)\)

Vì \(x=0\) thì hệ vô nghiệm nên \(x\ne0\), khi đó:

\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{8}{x^6}=\dfrac{6y}{x}\\1+\dfrac{27y^3}{x^3}=\dfrac{4}{x^2}\end{matrix}\right.\)

Đặt \(\dfrac{3y}{x}=a,\dfrac{2}{x^2}=b\) ta được hệ:

\(\Leftrightarrow\left\{{}\begin{matrix}1+a^3=2b\\1+b^3=2a\end{matrix}\right.\)

Giải hệ này ta được \(a=b\Leftrightarrow\dfrac{3y}{x}=\dfrac{2}{x^2}\Leftrightarrow y=\dfrac{2}{3x}\)

\(\left(1\right)\Leftrightarrow x^6-4x^4+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=\sqrt{1+\sqrt{5}}\\x=-\sqrt{1+\sqrt{5}}\end{matrix}\right.\)

TH1: \(x=\sqrt{2}\Rightarrow y=\dfrac{\sqrt{2}}{3}\)

TH2: \(x=-\sqrt{2}\Rightarrow y=-\dfrac{\sqrt{2}}{3}\)

TH3: \(x=\sqrt{1+\sqrt{5}}\Rightarrow y=\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

TH4: \(x=-\sqrt{1+\sqrt{5}}\Rightarrow y=-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

Đối chiếu với các điều kiện ta được \(\left(x;y\right)=\left(-\sqrt{1+\sqrt{5}};-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\right)\)

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