Question

[nguyen tu uyen]

( x² + 4x + 3). ( x² + 12x + 35) = 9

Answer 1

[l҉o҉n҉g҉ d҉z҉]

( x² + 4x + 3 )( x² + 12x + 35 ) = 9

<=> ( x² + x + 3x + 3 )( x² + 5x + 7x + 35 ) = 9

<=> [ x( x + 1 ) + 3( x + 1 ) ][ x( x + 5 ) + 7( x + 5 ) ] = 9

<=> ( x + 1 )( x + 3 )( x + 5 )( x + 7 ) = 9

<=> [ ( x + 1 )( x + 7 ) ][ ( x + 3 )( x + 5 ) ] = 9

<=> ( x² + 8x + 7 )( x² + 8x + 15 ) = 9

<=> ( x² + 8x + 7 )( x² + 8x + 15 ) - 9 = 0

Đặt t = x² + 8x + 7 

Phương trình tương đương với :

t( t + 8 ) - 9 = 0

<=> t² + 8t - 9 = 0

<=> t² - t + 9t - 9 = 0

<=> t( t - 1 ) + 9( t - 1 ) = 0

<=> ( t - 1 )( t + 9 ) = 0

<=> \(\orbr{\begin{cases}t-1=0\\t+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=1\\t=-9\end{cases}}\)

Với t = 1

=> x² + 8x + 7 = 1

<=> x² + 8x + 7 - 1 = 0 

<=> x² + 8x + 6 = 0 (1)

\(\Delta'=b'^2-ac=4^2-1\cdot6=10\)

\(\Delta'>0\)nên (2) có hai nghiệm phân biệt :

\(\hept{\begin{cases}x_1=\frac{-b'+\sqrt{\Delta'}}{a}=-4+\sqrt{10}=\sqrt{10}-4\\x_2=\frac{-b-\sqrt{\Delta'}}{a}=-4-\sqrt{10}=-\sqrt{10}-4\end{cases}}\)

Với t = -9

=> x² + 8x + 7 = -9

<=> x² + 8x + 7 + 9 = 0

<=> x² + 8x + 16 = 0

<=> ( x + 4 )² = 0

<=> x + 4 = 0

<=> x = -4

Vậy S = { \(\pm\sqrt{10}-4;-4\)}

Answer 2

[ミ★Ƙαї★彡]

PT <=> \(x^4+16x^3+86x^2+176x+96=0\)

\(\left(x^2+8x+6\right)\left(x+4\right)^2=0\)

TH1 : \(\Delta=8^2-4.6=64-24=40\)

\(x_1=\frac{-8-\sqrt{40}}{2};x_2=\frac{-8+\sqrt{40}}{2}\)

TH2 : \(x=-4\)

Vậy \(\left\{x=-4\right\}\)