= 100000

đề bài này sai rồi ! 

\(A=3\left(x^2+y^2\right)-2\left(x^3-y^3\right)\)

\(A=3x^2+3y^2-2\left(x-y\right)\left(x^2+xy+y^2\right)\)

Mà x- y = 1

\(\Rightarrow A=3x^2+3y^2-2\left(x^2+xy+y^2\right)\)

\(\Leftrightarrow A=3x^2+3y^2-2x^2-2xy-2y^2\)

\(\Leftrightarrow A=x^2-2xy+y^2\)

\(\Leftrightarrow A=\left(x-y\right)^2=1^2=1\)

A = 3( x² + y² ) - 2( x³ - y³ )

= 3x² + 3y² - 2( x - y )( x² + xy + y² )

= 3x² + 3y² - 2( x² + xy + y² )

= 3x² + 3y² - 2x² - 2xy - 2y²

= x² - 2xy + y²

= ( x - y )²

= 1² = 1

Đề như thế này phải không ??

\(\left(2x-3\right)^2-\left(2x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-2\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=5\end{cases}}\)

( 2x - 3 )^2 - ( 2x - 3 )( x + 2 ) = 0

Ta có A = (3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)(3⁶⁴ + 1)

=> 8A = 8(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)(3⁶⁴ + 1)

=> 8A = (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)(3⁶⁴ + 1)

=> 8A = (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)(3⁶⁴ + 1)

=> 8A = (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)(3⁶⁴ + 1)

=> 8A = (3¹⁶ - 1)(3¹⁶ + 1)(3³² + 1)(3⁶⁴ + 1)

=> 8A = (3³² - 1)(3³² + 1)(3⁶⁴ + 1)

=> 8A = (3⁶⁴ - 1)(3⁶⁴ + 1)

=> 8A = 3¹²⁸ - 1 (1)

Đặt B = 3¹²⁶

=> 8B = 3¹²⁶ . 8 = 3¹²⁶.(3² - 1) = 3¹²⁸ - 3¹²⁶ (2)

Từ (1)(2) => 8A > 8B 

=> A > B 

Ta có: \(\left(x-7\right)\left(x^2-9x+20\right)\left(x-2\right)=72\)

\(\Leftrightarrow\left(x^2-9x+20\right)\left(x^2-9x+14\right)=72\)

Đặt \(x^2-9x+17=a\) khi đó:

\(PT\Leftrightarrow\left(a+3\right)\left(a-3\right)=72\)

\(\Leftrightarrow a^2-9-72=0\)

\(\Leftrightarrow a^2=81\Rightarrow\orbr{\begin{cases}a=9\\a=-9\end{cases}}\)

Nếu a = 9 khi đó \(x^2-9x+17=9\)

\(\Leftrightarrow x^2-9x+8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=8\end{cases}}\)

Nếu a = -9 khi đó \(x^2-9x+17=-9\)

\(\Leftrightarrow x^2-9x+26=0\)

\(\Leftrightarrow\left(x^2-9x+\frac{81}{4}\right)+\frac{23}{4}=0\)

\(\Leftrightarrow\left(x-\frac{9}{2}\right)^2=-\frac{23}{4}\left(ktm\right)\)

Vậy \(S=\left\{1;8\right\}\)

( x - 7 )( x² - 9x + 20 )( x - 2 ) = 72

⇔ [ ( x - 7 )( x - 2 ) ]( x² - 9x + 20 ) - 72 = 0

⇔ ( x² - 9x + 14 )( x² - 9x + 20 ) - 72 = 0

Đặt t = x² - 9x + 17

⇔ ( t - 3 )( t + 3 ) - 72

⇔ t² - 9 - 72 = 0

⇔ t² - 81 = 0

⇔ ( t - 9 )( t + 9 ) = 0

⇔ ( x² - 9x + 17 - 9 )( x² - 9x + 17 + 9 ) = 0

⇔ ( x² - 9x + 8 )( x² - 9x + 26 ) = 0

⇔ ( x² - 8x - x + 8 )( x² - 9x + 26 ) = 0

⇔ [ x( x - 8 ) - ( x - 8 ) ]( x² - 9x + 26 ) = 0

⇔ ( x - 8 )( x - 1 )( x² - 9x + 26 ) = 0

⇔ x - 8 = 0 hoặc x - 1 = 0 hoặc x² - 9x + 26 = 0

⇔ x = 8 hoặc x = 1 [ x² - 9x + 26 = ( x² - 9x + 81/4 ) + 23/4 = ( x - 9/2 )² + 23/4 ≥ 23/4 > 0 ∀ x ]

( x + 5 )² = ( x + 3 )( x - 7 )

⇔ x² + 10x + 25 = x² - 4x - 21

⇔ x² + 10x - x² + 4x = -21 - 25

⇔ 14x = -46

⇔ x = -23/7

\(\left(x+5\right)^2=\left(x+3\right)\left(x-7\right)\)

\(\Leftrightarrow x^2+10x+25=x^2-7x+3x-21\)

\(\Leftrightarrow x^2+10x+25-x^2+4x+21=0\)

\(\Leftrightarrow14x+46=0\)

\(\Leftrightarrow x=\frac{-23}{7}\)