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a) (x2 - 1)(x2 + 4x + 3) = 192
=> (x - 1)(x + 1)(x2 + x + 3x + 3) - 192 = 0
=> (x - 1)(x + 1)(x + 1)(x + 3) - 192 = 0
=> [(x - 1)(x + 3)](x + 1)2 - 192 = 0
=> (x2 + 2x - 3)(x2 + 2x + 1) - 192 = 0
Đặt x2 + 2x - 3 = k
=> k(k + 4) - 192 = 0
=> k2 + 4k - 192 = 0
=> k2 + 16k - 12k - 192 = 0
=> k(k + 16) - 12(k + 16) = 0
=> (k - 12)(k + 16) = 0
=> (x2 + 2x - 3 - 12)(x2 + 2x - 3 + 16) = 0
=> (x2 + 2x - 15)(x2 + 2x + 13) = 0
=> x2 + 5x - 3x - 15 = 0 (do x2 + 2x + 13 \(\ne\)0)
=> x(x + 5) - 3(x + 5) = 0
=> (x - 3)(x + 5) = 0
=> \(\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\) => \(\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
a) \(\left(x^2-1\right)\left(x^2+4x+3\right)=192\)
\(\Leftrightarrow x^4+4x^3+3x^2-x^2-4x-3=192\)
\(\Leftrightarrow x^4+4x^3+2x^2-4x-3=192\)
\(\Leftrightarrow x^4+4x^3+2x^2-4x-3-192=0\)
\(\Leftrightarrow x^4+4x^3+2x^2-4x-195=0\)
\(\Leftrightarrow\left(x^3+7x^2+23x+65\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2+2x+13\right)\left(x+5\right)\left(x-3\right)=0\)
mà \(x^2+2x+13\ne0\) nên:
\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)
a)
\(\left(x^2-1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left[\left(x+2\right)^2-1\right]=\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)\)
\(\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)
dặt x^2+2x-1=t(*)
(a) \(\Leftrightarrow\left(t-2\right)\left(t+2\right)=192\) \(\Leftrightarrow t^2-4=192\Rightarrow t^2=196\Rightarrow\left\{\begin{matrix}t=-14\\t=14\end{matrix}\right.\)
Thay t vào (*) => x (tự làm)
a) (x-1)(x+1)(x+1)(x+3)=192. \(\Leftrightarrow\) (x+1)2(x-1)(x+3)=192 \(\Leftrightarrow\) (x2+2x+1) (x2+2x-3)=192 Đặt x2+2x+1=t thì x2+2x-3=t-4 ta có t(t-4)=192 \(\Leftrightarrow\) t2-4t-192=0 \(\Leftrightarrow\) t=-12 hoặc t=16 Với t=-12 thì (x+1)2=-12 ( vô lí ) Với t=16 thì (x+1)2=16 \(\Leftrightarrow\) x=-5 hoặc x=3 b) x\(^5\)+x4-2x4-2x3+5x3+5x2-2x2-2x+x+1=0 \(\Leftrightarrow\) x4(x+1)-2x3(x+1)+5x2(x+1)-2x(x+1)+(x+1)=0 \(\Leftrightarrow\) (x+1)(x4-2x3+5x2-2x+1)=0 \(\Leftrightarrow\) x=-1 ( CM x4-2x3+5x2-2x+1 vô nghiệm ) c) x4-x3-2x3+2x2+2x2-2x-x+1=0 \(\Leftrightarrow\) x3(x-1)-2x2(x-1)+2x(x-1)-(x-1)=0 \(\Leftrightarrow\) (x-1)(x3-2x2+2x-1)=0 \(\Leftrightarrow\) (x-1)(x-1)(x2-x+1)=0 \(\Leftrightarrow\) x-1=0 ( vì x2-x+1=(x-\(\frac{1}{2}\))2+\(\frac{3}{4}\)>0 với mọi x) \(\Leftrightarrow\) x=1
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=192\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=192\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3-192=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-195=0\)
\(\Leftrightarrow\left(x^2+2x-15\right)\left(x^2+2x+13\right)=0\)
=>(x+5)(x-3)=0
=>x=3 hoặc x=-5
\(y\left(y-4\right)=192\Leftrightarrow y^2-4y+4=196\)\(\Leftrightarrow\left(y-2\right)^2=196=14^2\)
\(\orbr{\begin{cases}y-2=14\\y-2=-14\end{cases}\Rightarrow\orbr{\begin{cases}y=16\\y=-12\left(loai\right)\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)=4\\\left(x+1\right)=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)
( x2 - 1 ).( x2 + 4x + 3 ) = 192
\(\Leftrightarrow\) ( x - 1 ).( x + 1 ) .( x2 + 3x + x + 3 ) = 192
\(\Leftrightarrow\) ( x - 1 ).( x + 1 ).[ x.( x + 3 )+ ( x + 3 ) ] = 192
\(\Leftrightarrow\) ( x - 1 .( x + 1 ).( x + 1 ).( x + 3 ) -192 = 0
\(\Leftrightarrow\) ( x + 1 )2.( x - 1 ).( x +3 ) - 192 = 0
Đặt : x + 1 = a
Khi đó phương trình trở thành :
\(\Rightarrow\) a2.( a - 2 ).( a + 2 ) - 192 = 0
\(\Leftrightarrow\)a2.( a2 - 4 ) - 192 = 0
\(\Leftrightarrow\) a4 - 4a2 - 192 = 0
\(\Leftrightarrow\) ( a4 - 4a2 + 4 ) - 4 - 192 = 0
\(\Leftrightarrow\) ( a2 - 2 )2 - 196 = 0
\(\Leftrightarrow\)( a2 - 2 )2 - 142 = 0
\(\Leftrightarrow\)( a2 - 2 - 14 ).( a2 - 2 + 14 ) = 0
\(\Leftrightarrow\)( a2 - 16 ).( a2 + 12 ) = 0
\(\Leftrightarrow\) ( a - 4 ).( a + 4 ).( a2 + 12 ) = 0
\(\Leftrightarrow\) \(\orbr{\begin{cases}\left(a-4\right).\left(a+4\right)=0\\a^2+12=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(a-4\right).\left(a+4\right)=0\\a^2=-12\left(vl\right)\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}a-4=0\\a+4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}a=4\\a=-4\end{cases}}\)
Với a = 4 Với a = -4
\(\Rightarrow\) x + 1 = 4 \(\Rightarrow\) x + 1 = -4
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = -5
Vậy phương trình có nghiệm là x = 3 , x = -5
(x² - 1)(x² + 4x + 3) = 192
<=> (x - 1)(x + 1)(x + 1)(x + 3) = 192
<=> (x - 1)(x + 3)(x + 1)² = 192
<=> (x² + 2x - 3)(x² + 2x + 1) = 192
Đặt t = x² + 2x + 1 => x² + 2x - 3 = t - 4
ta có pt: (t - 4)t = 192
<=> t² - 4t - 192 = 0
<=> t = - 12 hoặc t = 16
*t = x² + 2x + 1 = -12: vn
*t = x² + 2x + 1 = 16
<=> (x+1)² = 16
<=> x = -5 hoặc x = 3
Mãi mãi có một tương lai tươi sáng
Ta có :\(\frac{x^7+x^6+x^5+x^4+x^3+x^2+1}{x^2-1}\)
\(=\frac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+\left(x+1\right)}{x^2-1}\)
\(=\frac{\left(x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(x^6+x^4+x^2+1\right)}{\left(x-1\right)}\)
\(\left(x^2-1\right)\left(x+1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)^2\left(x+3\right)\)
=192 đâu hả anh