\(\left|\frac{2x-1}{3}\right|=x+1\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2x-1}{3}=x+1\\\frac{2x-1}{3}=-x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+3=2x-1\\-3x-3=2x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2x=-3-1\\-3x-2x=3-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\-5x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=-\frac{2}{5}\end{cases}}\)

\(\left|2x-\frac{1}{3}\right|=x+1\)

\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=x+1\\2x-\frac{1}{3}=-x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\frac{1}{3}\\x=\frac{4}{9}\end{cases}}\)

a) \(5x\left(x-1\right)-3x\left(x-1\right)\)

\(=2x\left(x-1\right)\)

b) \(x\left(x+y\right)-5x-5y\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x-5\right)\left(x+y\right)\)

c) \(x\left(x-y\right)+y\left(y-x\right)\)

\(=\left(x-y\right)\left(x-y\right)\)

\(=\left(x-y\right)^2\)

d) \(x^2+xy+x=x\left(x+y+1\right)\)

a. 5x ( x - 1 ) - 3x ( x - 1 )

= ( 5x - 3x ) ( x - 1 )

b. x ( x + y ) - 5x - 5y

= x ( x + y ) - 5 ( x + y )

= ( x - 5 ) ( x + y )

c. x ( x - y ) + y ( y - x )

= x ( x - y ) - y ( x - y )

= ( x - y )²

 d. x² + xy + x 

= x ( x + y + 1 )

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