Bài làm:

ADTCDTSBN ta có:

\(\frac{x}{3}=\frac{y}{7}=\frac{y-x}{7-3}=\frac{12}{4}=3\)

\(\Rightarrow\hept{\begin{cases}x=9\\y=21\end{cases}}\)

theo tính chất dãy tỉ số bằng nhau \(\frac{x}{3}=\frac{y}{7}=\frac{y-x}{7-3}=\frac{12}{4}=3\)

\(\hept{\begin{cases}\frac{x}{3}=3\Leftrightarrow x=3.3=9\\\frac{y}{7}=3\Leftrightarrow y=7.3=21\end{cases}}\)

vậy x=9 và y=21

Bài làm:

Đặt \(\frac{x}{7}=\frac{y}{5}=\frac{z}{3}=k\)

=> \(\hept{\begin{cases}x=7k\\y=5k\\z=3k\end{cases}}\)

Mà \(yz=135\Leftrightarrow15k^2=135\Leftrightarrow k^2=9\Rightarrow k=\pm3\)

=> \(\hept{\begin{cases}x=\pm21\\y=\pm15\\z=\pm9\end{cases}}\)

Đặt \(\frac{x}{7}=\frac{y}{5}=\frac{z}{3}=k\Rightarrow\hept{\begin{cases}x=7k\\y=5k\\z=3k\end{cases}}\)

Khi đó yz = 135

<=> 5k.3k = 135

=> 15.k² = 135

=> k² = 9

=> k = \(\pm\)3

Nếu k = 3 => x = 21 ; y = 15 ; z = 9

Nếu k = -3 => x = -21 ; y = -15 ; z = -9

Vậy các cặp (x;y;z) thỏa mãn bài toán là (21 ; 15 ; 9) ; (-21 ; - 15 ; -9)

5 . x - 200 = 2⁴ .4

5 . x - 200 = 16 . 4

5. x  - 200 = 64

5 . x = 64 + 200

5 . x = 264

     x = 52,8.

5.x-200=16.4

5.x-200=64

5.x=64+200

5.x=264

x=264:5

x=52,8

vậy x=52,8

Ta cóa : \(20x^6-\left(8-40y\right)x^3+25y^2-5\)

\(=20x^6-8x^3+40x^3y+25y^2-5\)

\(=16x^6+40x^3y+25y^2+4x^6-8x^3+4-9\)

\(=\left(4x^3+5y\right)^2+4\left(x^3-1\right)^2-9\)

Ta thấy ngay \(\left(4x^3+5y\right)^2\ge0;4\left(x^3-1\right)^2\ge0\)

\(\Rightarrow\left(4x^3+5y\right)^2+4\left(x^3-1\right)^2-9\ge-9\)

\(\Rightarrow M=\frac{6}{20x^6-\left(8-40y\right)x^3+25y^2-5}\le\frac{6}{-9}=-\frac{2}{3}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}4x^3+5y=0\\x^3-1=0\end{cases}\Leftrightarrow x=1;y=-\frac{4}{5}}\)

làm ơn giúp mk

- 6x² - 9xy + 15x

= - 3x ( 3y + 2x - 5 )

2x ( x - 3 ) + y ( x - 3 ) + ( 3 - x )

= ( 2x + y ) ( x - 3 ) - ( x - 3 )

= ( 2x + y + 1 ) ( x - 3 )

|x + 4/15| - |-3,75| = -|- 2,15|

=> |x + 4/15| - 3,75 = 2,15

=> |x + 4/15| = 59/10

=> \(\orbr{\begin{cases}x+\frac{4}{15}=\frac{59}{10}\\x+\frac{4}{15}=-\frac{59}{10}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{169}{30}\\x=\frac{-185}{30}\end{cases}}\)

\(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\frac{4}{15}\right|=-2,15+3,75\)

\(\left|x+\frac{4}{15}\right|=1,6\)

\(\left|x+\frac{4}{15}\right|=\frac{8}{5}\)

\(\Rightarrow\hept{\begin{cases}x+\frac{4}{15}=\frac{8}{5}\\x+\frac{4}{15}=\frac{-8}{5}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{8}{5}-\frac{4}{15}=\frac{4}{3}\\x=\frac{-8}{5}-\frac{4}{15}=\frac{-28}{15}\end{cases}}}\)

Ke \(MK\perp BC\Rightarrow MK=\frac{AH}{2}=\frac{BM}{2}\) (tinh chat duong trung binh)

Xet tam giac MBK co \(\frac{MK}{MB}=\frac{1}{2}\Rightarrow\widehat{MBK}=\widehat{MBC}=30^0\)