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.

64.(75+24+1)

=64.100

=6400

 64 x 75 + 24 x 64 + 64

 64 x 75 + 24 x 64 + 64 + 1

<75 +24+1>x64

100x64

6400

= 64(75+24+1)

=64*100

=6400

64:0,5 + 64:0,25 + 64\(\times4\)

= 64 \(\times\) 2 + 64 \(\times\) 4 +  64 \(\times\) 4

= 64 \(\times\) ( 2 + 4 + 4)

= 64 \(\times\) 10

= 640 

25 x 33 + 25 x 87 + 64 x 73 + 64 x 47

= 25 x ( 33 + 87 ) + 64 x ( 73 + 47 )

= 25 x 120 + 64 x 120

=  120 x ( 15 + 64 )

= 120 x 49

= 5580

25 x 33 + 25 x 87 + 64 x 73 + 64 x 47

= 25 x ( 33 + 87 ) + 64 x ( 73 +47 )

= 25 x  120 + 64 x 120

= 120 x ( 25 + 64 )

= 120 x 89

= 10680

25 x 33 + 25 x 87 + 64 x 73 + 64 x 47

= ( 33 + 87 ) x 25 + ( 73 + 47 ) x 64

= 120 x 25 + 120 x 64

= ( 25 + 64 ) x 120

= 89 x 120

= 10680

@#$%^&* !

=36x(28+82)+64x(69+41)

=36x110+64x110

=110x(26+64)

=110x100

=11000

**** bn hiền

36x28+36x82+64x69+64x41

=36x(28+82)+64x(69+41)

=36x110+64x110

=110x(36+64)

=110x100

=11000

36 x ( 28 + 82 ) + 64 x ( 69 + 41 ) = ( 36 + 64 ) x  (28 + 82)  = 100 x 110 = 11000

a) \(64^x:16^x=256\)

\(\Rightarrow\left(2^6\right)^x:\left(2^4\right)^x=2^8\)

\(\Rightarrow2^{6x}:2^{4x}=2^8\)

\(\Rightarrow2^{6x-4x}=2^8\)

\(\Rightarrow2^{2x}=2^8\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=4\)

b) \(\dfrac{-2401}{7^x}=-7\)

\(\Rightarrow\dfrac{-7^4}{7^x}=-7\)

\(\Rightarrow-7^{4-x}=-7\)

\(\Rightarrow7^{4-x}=7\)

\(\Rightarrow4-x=1\)

\(\Rightarrow x=4-1\)

\(\Rightarrow x=3\)

c) \(\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=\dfrac{64}{-256}\)

\(\Rightarrow\left(-4\right)^x=-4\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^1\)

\(\Rightarrow x=1\)

\(a) 64^x:16^x=256\\\Rightarrow (64:16)^x=256\\\Rightarrow 4^x=4^4\\\Rightarrow x=4\\---\)

\(b,\dfrac{-2401}{7^x}=-7\)

\(\Rightarrow7^x=-2401:\left(-7\right)\)

\(\Rightarrow7^x=343\)

\(\Rightarrow7^x=7^3\)

\(\Rightarrow x=3\)

\(c,\dfrac{64}{\left(-4\right)^x}=-256\)

\(\Rightarrow\left(-4\right)^x=64:\left(-256\right)\)

\(\Rightarrow\left(-4\right)^x=-\dfrac{1}{4}\)

\(\Rightarrow\left(-4\right)^x=\left(-4\right)^{-1}\)

\(\Rightarrow x=-1\)

#\(Toru\)

1/

$(x-1)^{x+10}=(x-1)^{x+8}$

$\Rightarrow (x-1)^{x+10}-(x-1)^{x+8}=0$

$\Rightarrow (x-1)^{x+8}(x^2-1)=0$

$\Rightarrow (x-1)^{x+8}=0$ hoặc $x^2-1=0$

Nếu $(x-1)^{x+8}=0\Rightarrow x-1=0\Rightarrow x=1$

Nếu $x^2-1=0\Rightarrow x^2=1=1^2=(-1)^2\Rightarrow x=1$ hoặc $x=-1$

Vậy $x=1$ hoặc $x=-1$

2/

$1^3+2^3+3^3+...+10^3=(x+1)^2$

Ta có công thức quen thuộc:

$1^3+2^3+...+n^3=(1+2+...+n)^2=\frac{[n(n+1)]^2}{4}$

Bạn có thể xem cm tại đây:

https://diendantoanhoc.org/topic/81694-t%C3%ADnh-t%E1%BB%95ng-s-13-23-33-n3/

Khi đó:

$1^3+2^3+...+10^3=(x+1)^2$

$\Rightarrow \frac{[10(10+1)]^2}{4}=(x+1)^2$

$\Rightarrow 3025=(x+1)^2$

$\Rightarrow x+1=55$ hoặc $x+1=-55$

$\Rightarrow x=54$ hoặc $x=-56$

-33x(57-64)-57x(-33+64)

= -33x57-64-57x(-33)+64

=[-33x(-33)]x(57-57)-(64+64)

=1089 x 0 - 128

=0-128

=-128

THEO MK NGHĨ LÀ ĐÚNG.CHÚC BẠN HỌC TỐT

`@` `\text {Ans}`

`\downarrow`

\(\left(2\cdot x+2\right)^2=64\)

`\Rightarrow`\(\left(2x+2\right)^2=\left(\pm8\right)^2\)

`\Rightarrow`\(\left[{}\begin{matrix}2x+2=8\\2x+2=-8\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=8+2\\2x=-8+2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}2x=10\\2x=-6\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=10\div2\\x=-6\div2\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy, `x \in {5; -3}`

`@` `\text {Kaizuu lv uuu}`

a)16. x² = 64

x² = 64 : 16

x² = 4

x² = 2²

⇒ x = 2

b) (5.x - 2) - 64 = -36

(5.x - 2) = -36 + 64

5.x - 2 = 28

5.x = 28 + 2

5.x = 30

x = 30 : 5

x = 6

c) (2x - 10).(5 - x) = 0

TH1: 2x - 10 = 0

         2x = 0 + 10

         2x = 10

           x = 10 : 2

           x = 5

TH2: 5 - x = 0

              x = 5 - 0

              x = 5

⇒ Vậy x = 5.

cứu tui

\(16.x^2=64\)

\(=>x^2=64:16\)

\(=>x^2=4\)

\(=>x=2\) hoặc \(x=-2\)

_______

\(\left(5x-2\right)-64=-36\)

\(=>5x-2=\left(-36\right)+64\)

\(=>5x-2=28\)

\(=>5x=28+2\)

\(=>5x=30\)

\(=>x=30:5\)

\(=>x=6\)

_______

\(\left(2x-10\right).\left(5-x\right)=0\)

TH1: \(2x-10=0\)

\(=>2x=0+10\)

\(=>2x=10\)

\(=>x=10:2\)

\(=>x=5\)

TH2: \(5-x=0\)

\(x=5-0\)

\(x=5\)

\(=>x=5\)