(2x+1).(y-5)
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\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
a.(y+5).(y-5)-(y-3)2
=(y2-25)-(y2-6y+9)
=6y-34
\(a;\left(y+5\right).\left(y-5\right)-\left(y-3\right)^2\)
\(=y^2-5^2-\left(y^2-6y+9\right)\)
\(=y^2-25-y^2+6y-9\)
\(=6y-34\)
Đặt \(\dfrac{1}{x+y-1}=a;\dfrac{1}{2x-y+3}=b\)
Hệ phương trình trở thành:
\(\left\{{}\begin{matrix}4a-5b=\dfrac{5}{3}\\3a+b=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12a-15b=5\\12a+4b=\dfrac{28}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-19b=\dfrac{-3}{5}\\3a+b=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{95}\\a=\dfrac{26}{57}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y-1}=\dfrac{26}{57}\\\dfrac{1}{2x-y+3}=\dfrac{3}{95}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y-1=\dfrac{57}{26}\\2x-y+3=\dfrac{95}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{83}{26}\\2x-y=\dfrac{86}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{2485}{78}\\x+y=\dfrac{83}{26}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2485}{234}\\y=\dfrac{83}{26}-\dfrac{2485}{234}=\dfrac{-869}{117}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(\dfrac{2485}{234};\dfrac{-869}{117}\right)\)
y=5 x=8