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562=3136
tích mình nhé
ai nghĩ mình giỏi ấn vào
\(73\cdot67=\left(70+3\right)\left(70-3\right)=70^2-3^2=4900-9=4891\)
a) Theo đầu bài ta có:
\(x+y=2\Rightarrow x=2-y\)
\(x^2+y^2=10\)
\(\Rightarrow\left(2-y\right)^2+y^2=10\)
\(\Rightarrow4+y^2-4y+y^2=10\)
\(\Rightarrow2y^2-4y=6\)
\(\Rightarrow2\left(y^2-2y\right)=6\)
\(\Rightarrow y\left(y-2\right)=3\)
Mà \(\hept{\begin{cases}y-\left(y-2\right)=2\\y+\left(y-2\right)=k\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}\\y-2=\frac{k-2}{2}\end{cases}}}\)( với k là hằng số )
\(\Rightarrow y\left(y-2\right)=\frac{k+2}{2}\cdot\frac{k-2}{2}\)
\(\Rightarrow\frac{\left(k+2\right)\left(k-2\right)}{4}=3\)
\(\Rightarrow k^2-4=12\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=4;-4\)
- Nếu k = 4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=3\\x=2-y=-1\end{cases}\Rightarrow x^3+y^3=-1+27=26}\)
- Nếu k = -4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=-1\\x=2-y=3\end{cases}\Rightarrow x^3+y^3=27+-1=26}\)
Vậy x3 + y3 = 26
a, \(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+2xy+y^2=4\Rightarrow10+2xy=4\Rightarrow xy=-3\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2.13=26\)
vậy............
b, \(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\)
\(\Rightarrow xy=\frac{a^2-b}{2}\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)
Vậy....
\(77^2+23+77.46\)
\(=77.77+23+77.46\)
\(=77.\left(77+46\right)+23\)
\(=77.123+23\)
\(=9471+23=9494\)
\(77^2+23+77.46\)
\(=77.77+23+77.46\)
\(=77.\left(77+46\right)+23\)
\(=77.123+23\)
\(=9471+23\)
\(=9494\)
\(\left(27x^3-1\right):\left(3x-1\right)=\left[\left(3x\right)^3-1^3\right]:\left(3x-1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right):\left(3x-1\right)\)
\(=9x^2+3x+1\)
\(\frac{27x^3-1}{3x-1}\)
\(=\frac{\left(3x\right)^3-1}{3x-1}\)
\(=\frac{\left(3x-1\right)\left(9x^2+3x+1\right)}{3x-1}\)
\(=9x^2+3x+1\)
53.47=(50+3).(50-3)=502+32
=2500+9=2509
53.47 = (50 + 3)(50 - 3) = 502 - 32 = 2500 - 9 = 2491