Vừa nãy mk làm sai , bây h làm lại r nha

\(\dfrac{a+b}{3a-b}+\dfrac{1}{a+b}.\dfrac{a^2-b^2}{3a-b}\)

\(=\dfrac{a+b}{3a-b}+\dfrac{1}{a+b}.\dfrac{\left(a-b\right)\left(a+b\right)}{3a-b}\)

\(=\dfrac{a+b}{3a-b}+\dfrac{a-b}{3a-b}=\dfrac{2a}{3a-b}\)

\(\dfrac{a+b}{3a-b}+\dfrac{1}{a+b}.\left(\dfrac{a^2-b^2}{3a-b}\right)\)

ĐKXĐ: \(a;b\ne0\)

\(\dfrac{a+b}{3a-b}+\dfrac{1}{a+b}.\dfrac{\left(a-b\right)\left(a+b\right)}{3a-b}\)

\(=\dfrac{a+b}{3a-b}+\dfrac{a-b}{3a-b}\)

\(=\dfrac{a+b-a+b}{3a-b}=\dfrac{2b}{3a-b}\)

Học tốt nha<3

\(5x\left(x-9\right)-2x\left(9-x\right)=0\)

\(\Leftrightarrow5x\left(x-9\right)+20\left(x-9\right)=0\)

\(\Leftrightarrow5\left(x+4\right)\left(x-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=9\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{-4;9\right\}\)

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

ĐKXĐ: \(x\ne\pm3\)

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

\(=\dfrac{3\left(x+2\right)}{\left(3-y\right)\cdot\left(3+y\right)}+\dfrac{1-\left(3-y\right)}{3-y}\)

\(=\dfrac{3\left(x+2\right)}{\left(3-y\right)\left(3+y\right)}+\dfrac{1}{3-y}-1\)

\(=\dfrac{3\left(x+2\right)}{\left(3-y\right)\left(3+y\right)}-\dfrac{3+y}{\left(3-y\right)\left(3+y\right)}-1\)

\(=\dfrac{3x+6-3-y}{\left(3-y\right)\left(3+y\right)}-1\)

\(=\dfrac{3x-y+3}{\left(3-y\right)\left(3+y\right)}-1\)

\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)

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

\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)

\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)

\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)

\(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3+27-8x^3+2=29\)

Vậy biểu thức sau không phụ thuộc vào x

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left(x+y\right)^3+3\left(x+y\right)^2+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)\(=x^3+3x^2y+3xy^2+y^3+3xz^2+3yz^2-x^3-y^3\)\(=3\left(x^2y+xy^2+xz^2+yz^2\right)\)

\(=3\left[xy\left(x+y\right)+z^2\left(x+y\right)\right]\)

\(=3\left(xy+z^2\right)\left(x+y\right)\)

\(25-4x^2-4xy-y^2\)

\(=25-\left(2x+y\right)^2\)

\(=\left(5+2x+y\right)\left(5-2x-y\right)\)

\(a,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)

\(b,\left(x^2-3\right)\left(x^4+3x^2+9\right)=x^6-27\)

\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2\)

\(=x^2+4xy+4y^2-z^2\)

\(d,\left(2x-1\right)\left(4x^2+2x+1\right)=8x^3-1\)

\(e,\left(5+3x\right)^3=125+225x+135x^2+27x^3\)