Question

tìm a, b biế a+b = 10 a² + b² = 52

biết x+y = 1 tính x³ +y³ +3xy

cho a+b= 13 a.b = 9 tính a³ + y³

cho x+y= 7 và x²+y² = 33 . tính a³ +y³

cho x,y là 2 số khác nhau sao cho x² - y = y² - x. tính A = x² + 2 xy +y ² - 3x -3y

Answer 1

a , Ta có: \(a+b=10\Rightarrow a=10-b\)

\(a+b=10\Rightarrow\left(a+b\right)^2=100\)

\(\Leftrightarrow a^2+2ab+b^2=100\)

\(\Leftrightarrow2ab=100-\left(a^2+b^2\right)=100-52=48\Rightarrow ab=24\)\(\Leftrightarrow\left(10-b\right)b=24\Leftrightarrow10b+b^2-24=0\)

\(\Leftrightarrow b^2+10b+25-49=0\)

\(\Leftrightarrow\left(b+5\right)^2=49\Rightarrow\left[{}\begin{matrix}b+5=7\\b+5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=2\\b=-12\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=8\\a=-2\end{matrix}\right.\)b, ta có:

\(\left(x+y\right)=1\Rightarrow\left(x+y\right)^3=1\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\)

\(\Leftrightarrow x^3+3xy\left(x+y\right)+y^3=1\)

\(\Leftrightarrow x^3+3xy+y^3=1\)

c, \(a+b=13\Rightarrow\left(a+b\right)^2=169\)

\(\Leftrightarrow a^2+2ab+b^2=169\Rightarrow a^2+b^2=169-2ab=169-2.9=151\)\(\Rightarrow a^3+b^3=\left(a+b\right)\left(a^2+b^2+ab\right)=13.\left(151+9\right)=2080\)

Answer 2

\(d,x+y=7\Rightarrow\left(x+y\right)^2=49\)

\(\Leftrightarrow x^2+2xy+y^2=49\Rightarrow2xy=49-\left(x^2+y^2\right)=16\Rightarrow xy=8\)

\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=7.\left(33-8\right)=175\)