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1) \(x^2+x-6=x\left(x-2\right)+3\left(x-2\right)=\left(x+3\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3) \(x^2+2x-48=\left(x-6\right)\left(x+8\right)\)
4) \(x^2-2x-48=\left(x-8\right)\left(x+6\right)\)
5) \(x^2+x-42=\left(x-6\right)\left(x+7\right)\)
6) \(x^2-x-42=\left(x-7\right)\left(x+6\right).\)
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A = ( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 48
= ( x2 + 7x + 10 ) ( x2 + 7x + 12 ) - 48
Đặt x2 + 7x + 10 = t
=> A = t. ( t + 2 ) - 48
= t2 + 2t + 1 - 49
= ( t + 1 )2 - 72
= ( t + 1 - 7 ) ( t + 1 + 7 )
= ( t - 6 ) ( t + 8 )
Thay t = x2 + 7x + 10
=> A = ( x2 + 7x + 4 )( x2 + 7x + 18 )
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a: \(5x\left(2x+3\right)+6x+9\)
\(=5x\left(2x+3\right)+\left(6x+9\right)\)
\(=5x\left(2x+3\right)+3\left(2x+3\right)\)
\(=\left(2x+3\right)\left(5x+3\right)\)
b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(3x+48+5\right)\)
=(x+4)(3x+53)
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mk sửa đề 1 chút
\(x^6-x^5+3x^4-16x^2+16x-48=0\)
\(\Leftrightarrow\left(x^6-16x^2\right)-\left(x^5-16x\right)+\left(3x^4-48\right)=0\)
\(\Leftrightarrow x^2\left(x^4-16\right)-x\left(x^4-16\right)+3\left(x^4-16\right)=0\)
\(\Leftrightarrow\left(x^4-16\right)\left(x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2+4\right)\left(x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\left(x^2-x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x^2+4>0\\x^2-x+3=0\left(vonghiem\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy.............
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`20((x-2)/(x+1))^2-5((x+2)/(x-1))^2+48(x^2-4)/(x^2-1)=0(x ne +-1)`
Đặt `(x-2)/(x+1)=a,(x+2)/(x-1)=b`
`pt<=>20a^2-5b^2+48ab=0`
`<=>20a^2+48ab-5b^2=0`
`<=>20a^2-2ab+50ab-5b^2=0`
`<=>2a(a-10b)+5b(10a-b)=0`
`<=>(a-10b)(2a+5b)=0`
Đến đây dễ rồi bạn tự giải tiếp.
ĐKXĐ: x \(\ne\)\(\pm\)1
Ta có: \(20\left(\dfrac{x-2}{x+1}\right)^2-5\left(\dfrac{x+2}{x-1}\right)^2+48\cdot\dfrac{x^2-4}{x^2-1}=0\)
Đặt: \(\dfrac{x-2}{x+1}=a\) ; \(\dfrac{x+2}{x-1}=b\)
=> ab = \(\dfrac{x^2-4}{x^2-1}\)
Do đó, ta có pt mới: 20a2 - 5b2 + 48ab = 0
<=> 20a2 + 50ab - 2ab - 5b2 = 0
<=> (10a - b)(2a + 5b) = 0
<=> \(\left[{}\begin{matrix}10a=b\\2a=-5b\end{matrix}\right.\)
TH1: 10a = b => \(10\cdot\dfrac{x-2}{x+1}=\dfrac{x+2}{x-1}\)
<=> 10(x - 2)(x - 1) = (x + 2)(x + 1)
<=> 10x2 - 30x + 20 = x2 + 3x + 2
<=> 9x2 - 33x + 18 = 0
<=> 9x2 - 27x - 6x + 18 = 0
<=> (9x - 6)(x - 3) = 0
<=> \(\left[{}\begin{matrix}x=3\\x=\dfrac{2}{3}\end{matrix}\right.\)(tm)
TH2: \(2a=-5b\)=> \(2\cdot\dfrac{x-2}{x+1}=-5\cdot\dfrac{x+2}{x-1}\)
=> (2x - 4)(x - 1) = (-5x - 10)(x + 1)
<=> 2x2 - 6x + 4 = -5x2 - 15x - 10
<=> 7x2 + 9x + 14 = 0
=> pt vn
x=\(\frac{76}{5}\)
đúng thì ****!!
\(\frac{48}{x+4}+\frac{48}{x-4}=5\)
\(\Leftrightarrow\frac{48\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{48\left(x+4\right)}{\left(x+4\right)\left(x-4\right)}=\frac{5\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\)
\(\Rightarrow48\left(x-4\right)+48\left(x+4\right)=5\left(x-4\right)\left(x+4\right)\)
\(\Leftrightarrow48x-192+48x+192-5\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow96x-5\left(x^2-4\right)=0\)
\(\Leftrightarrow96x-5x^2+20=0\)
Giải phương trình bậc hai tìm x