Phương trình chính tắc của elip là: c) \(\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{25}} = 1\).

a) Không là PTCT vì a =b =8

b) Không là PTCT

d) Không là PTCT vì a =5 < b =8.

\(A=1.\left(x+y\right)\left(x^2+y^2\right)...\left(x^{64}+y^{64}\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)...\left(x^{64}+y^{64}\right)\)

\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)...\left(x^{64}+y^{64}\right)\)

\(=\left(x^4-y^4\right)...\left(x^{64}+y^{64}\right)\)

\(=...=\left(x^{64}-y^{64}\right)\left(x^{64}+y^{64}\right)=x^{128}-y^{128}\)

ta có : x(x+y)=36

          y(x+y)=64

         \(\Rightarrow y\left(x+y\right)+x\left(x+y\right)=\left(x+y\right)\left(x+y\right)=\left(x+y\right)^2=64+36=100\)

             \(\Rightarrow x+y=\sqrt{100}=10\)

                 thế vào ta có \(10y=64\Leftrightarrow y=\frac{64}{10}=6,4\)

                                      \(10x=36\Leftrightarrow x=\frac{36}{10}=3,6\)

                                             vậy y=6,4 ; x=3,6

a: \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

=>\(\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)

=>\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)

=>\(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)

Đặt \(\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}=k\)

=>x=k; y=2k; z=3k

\(x^2+y^2+z^2=14\)

=>\(k^2+4k^2+9k^2=14\)

=>\(14k^2=14\)

=>\(k^2=1\)

=>k=1 hoặc k=-1

TH1: k=1

=>\(x=k=1;y=2k=2\cdot1=2;z=3k=3\cdot1=3\)

TH2: k=-1

=>\(x=k=-1;y=2k=2\cdot\left(-1\right)=-2;z=3k=3\cdot\left(-1\right)=-3\)

b: \(\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\)

=>\(\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{3}\right)^3=\left(\dfrac{z}{4}\right)^3\)

=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

=>x=2k; y=3k; z=4k

\(x^2+2y^2-3z^2=-650\)

=>\(\left(2k\right)^2+2\cdot\left(3k\right)^2-3\cdot\left(4k\right)^2=-650\)

=>\(4k^2+18k^2-3\cdot16k^2=-650\)

=>\(-26\cdot k^2=-650\)

=>\(k^2=25\)

=>\(\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\)

TH1: k=5

=>\(x=2\cdot5=10;y=3\cdot5=15;z=4\cdot5=20\)

TH2: k=-5

=>\(x=2\cdot\left(-5\right)=-10;y=3\cdot\left(-5\right)=-15;z=4\cdot\left(-5\right)=-20\)

64 x y - y x 13 -y = 43

y x (64-13-1)=43

y x 50 =43 

y=43:50

y=43/50

bbbbbbbbbb

y \(\times\) \(\dfrac{16}{64}\) + y \(\times\) \(\dfrac{25}{100}\) + y \(\times\) \(\dfrac{1}{4}\) + y \(\times\) \(\dfrac{15}{60}\) -  \(\dfrac{13}{15}\) = \(\dfrac{17}{15}\)

y \(\times\) ( \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)) - \(\dfrac{13}{15}\) = \(\dfrac{17}{15}\)

y                             = \(\dfrac{17}{15}\) + \(\dfrac{13}{15}\)

y = \(\dfrac{30}{15}\)

y = 2

Cái này bằng rủ em chơi oẳn tù tì ,em ra lá còn anh ra hôn má em

me nhìn công thức của cô Hoài moà đâu đầu nun á

Áp dụng tc dtsbn:

\(\dfrac{x}{y}=\dfrac{5}{3}\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x+y}{5+3}=\dfrac{64}{8}=8\\ \Leftrightarrow\left\{{}\begin{matrix}x=40\\y=24\end{matrix}\right.\)