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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) \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}=\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)
b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)
c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2a-1}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{4a^2-3a+5+2a^2-2a-a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)
d) \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x-3y}{x\left(x+3y\right)}\)
e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(=\frac{3x-2}{\left(x-1\right)^2}-\frac{6}{\left(x-1\right)\left(x+1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)
\(=\frac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)}-\frac{\left(3x-2\right)\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}\)
\(=\frac{3x^3+6x^2+3x+2x^2+4x+2-6x^2+6-3x^3+6x^2-3x+2x^2-4x+2}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)
f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)
\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)
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 \(x=3\)
ap dung cong thuc: a/b = c/d <=> ad= bc <=> c = ad/b
A = (4x2-7x+3)(x2+2x+1)/(x2-1)
a) \(\frac{-x^2y^5}{-x^2y^5}=1\)
b)\(\frac{-\left(x^7y^5z\right)^2}{-\left(xy^3z\right)^2}=\frac{x^{14}y^{10}z^2}{x^2y^6z^2}=x^7.y^4\)Thế vào ta được 1.(-10)^4=10000 cái khi nãy làm lộn
câu a cả tử và mẫu đều giống nhau nên kết quả là 1
b) chia ra ta được x6y2. Thế vào thì ra 1.102=100
