(x + 1)(x + 3)(x + 5)(x + 7) = - 15
<=> (x + 1)(x + 7)(x + 3)(x + 5) = -15
<=> (x^2 + 8x + 7)(x^2 + 9x + 15) = -15
đặt x^2 + 8x + 11 = a
<=> (a + 4)(a - 4) = -15
<=> a^2 - 16 + 15 = 0
<=> a^2 - 1 = 0
<=> (a - 1)(a + 1) = 0
<=> a = 1 hoặc a = -1
thay vào tìm x
Cách kia phân tích loằng ngoằng lắm , e lm cách này ko bt đúng ko nha !
\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)=-15\)
Th1 : \(x+1=-15\Leftrightarrow x=-16\)
Th2: \(x+3=-15\Leftrightarrow x=-18\)
Th3 : \(x+5=-15\Leftrightarrow x=-20\)
Th4: \(x+7=-15\Leftrightarrow x=-22\)
\(\left(x+1\right).\left(x+3\right).\left(x+5\right).\left(x+7\right)=-15\)
\(\Rightarrow\left(x^2+3x+x+3\right).\left(x^2+7x+5x+35\right)=-15\)
\(\Rightarrow\left(x^2+4x+3\right).\left(x^2+12x+35\right)=-15\)
\(\Rightarrow x^4+12x^3+35x^2+4x^3+48x^2+105x+3x^2+36x+105=-15\)
\(\Rightarrow\left(x^4+35x^2+48x^2+3x^2\right)+\left(12x^3+4x^3\right)+105x+36x+105=-15\)
\(\Rightarrow x^2.\left(x^2+35+48+3\right)+x^3.\left(12+4\right)+141x+105=-15\)
\(\Rightarrow x^2.\left(x^2+86\right)+16x^3+141x=-15-105\)
\(\Rightarrow x^4+86x^2+16x^3+141x=-120\)
\(\Rightarrow x.\left(x^3+86x+16x^2+141\right)=-120\)
\(\Rightarrow x\inƯ\left(-120\right)=\left\{\pm1;\pm2;................;\pm60;\pm120\right\}\)
Vậy \(x\in\left\{\pm1;\pm2;..............;\pm60;\pm120\right\}\)
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