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\(A=3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)
\(=3x^2+3y^2-2\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=3x^2+3y^2-2.1\left(x^2-xy+y^2\right)\)
\(=3x^2+3y^2-2x^2+2xy-2y^2\)
\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)
\(B=x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left[\left(x+y\right)^2-2xy\right]+6x^2y^2.1\)
\(=x^3+y^3+3xy\left(x+y\right)^2-6x^2y^2+6x^2y^2\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=x^2-xy+y^2+3xy\)
\(=x^2+2xy+y^2=\left(x+y\right)^2=1^2=1\)
P/s: Ko chắc lắm.
\(A=x^3+y^3+6xy-3x-3y+1\)
\(A=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)+6xy+1\)
\(A=\left(x+y\right)\left(x^2+2xy+y^2-2xy-xy\right)-3\left(x+y\right)+6xy+1\)
\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left(x+y\right)+6xy+1\)
\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy-3\right]+6xy+1\)
Thay x+y=2 vào biểu thức, ta có:
\(A=2\left(2^2-3xy-3\right)+6xy+1\)
\(A=2\left(1-3xy\right)+6xy+1\)
\(A=2-6xy+6xy+1\)
\(A=3\)
\(B=x^2-y^2+4y+1\)
\(B=\left(x-y\right)\left(x+y\right)+4y+1\)
\(B=2\left(x-y\right)+4y+1\)
\(B=2x-2y+4y+1\)
\(B=2x+2y+1\)
\(B=2\left(x+y\right)+1=2.2+1=5\)
\(N=x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)
\(N=x^3+y^3+6x^2y^2+3xy\left[\left(x+y\right)^2-2xy\right]\)
\(N=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2+3xy-6x^2y^2\)
\(N=x^2-xy+y^2+3xy\)
\(N=\left(x+y\right)^2\)
\(N=1\)
\(x^3+y^3+6x^2y^2\left(x+y\right)+3xy\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+6x^2y^2\left(x+y\right)+3xy\left[\left(x+y\right)^2-2xy\right]\)
\(=x^2-xy+y^2+6x^2y^2+3xy-6x^2y^2\)( Do \(x+y=1\))
\(=\left(x+y\right)^2-2xy-xy+3xy+6x^2y^2-6x^2y^3\)
\(=\left(x+y\right)^2=1^2=1\)
\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)
\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)
\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)
\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)
\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)
\(=\frac{2x+1}{x-3}\)
b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)
thay \(x=-\frac{3}{2}\) vào P tâ đc: \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)
c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)
\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)
\(\Leftrightarrow4x+2=x^2-3x\)
\(\Leftrightarrow x^2-7x-2=0\)
\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)
\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)
bạn tự giải nốt nhé!!
d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)
\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
bạn tự làm nốt nhé
a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)
\(=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}:\dfrac{-2}{2x+1}=\dfrac{-2\left(x-3\right)\left(2x+1\right)}{-2\left(x+3\right)\left(x-3\right)}=\dfrac{2x+1}{x+3}\)
b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
a) A = -1; b) B = ( x + y ) 3 =1.