/ x + 1 / + / x + 2 / + / x + 3 / = 4x - 4 = 9999
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1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
( ) x 9999 x 9999 x 9999 x ... ( có 1000 số 9999 ) x 38 x 45 x 36 x 25 x 58 x 69 x 245 x 246 x 213 x 222 x 546 x 8312 x 999 x 454 x 9999 x 789 x 2563 x 0 x 25 x 24 x 23 x 26 258 x 256 256 x 256 x 24531 x 55 x ( 45 x 1 + 45 x 2 + 45 x 3 + ... + 45 x 1000 + 45 x 1001 + ... + 45 x 9999 ) = 0 nhé
/HT\
( ) x 9999 x 9999 x 9999 x ... ( có 1000 số 9999 ) x 38 x 45 x 36 x 25 x 58 x 69 x 245 x 246 x 213 x 222 x 546 x 8312 x 999 x 454 x 9999 x 789 x 2563 x 0 x 25 x 24 x 23 x 26 258 x 256 256 x 256 x 24531 x 55 x ( 45 x 1 + 45 x 2 + 45 x 3 + ... + 45 x 1000 + 45 x 1001 + ... + 45 x 9999 ) = 0
HT
@@@@@@@
1: =>x^2-x=3-x
=>x^2=3
=>x=căn 3 hoặc x=-căn 3
2: =>x^2-4x+3=x^2-4x+4 và x>=2
=>3=4(vô lý)
3: =>2|x-1|=6
=>|x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2 hoặc x=4
4: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc x=5/3
5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
=>x+2=0
=>x=-2
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
a) \(4x-4y+x^2-2xy+y^2\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
b) \(x^4-4x^3-8x^2+8x\)
\(=x^4+2x^3-6x^3-12x^2+4x^2+8x\)
\(=x^3\left(x+2\right)-6x^2\left(x+2\right)+4x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^3-6x^2+4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
c) \(x^3+x^2-4x-4\)
\(=x^3-2x^2+3x^2-6x+2x-4\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+x+2\right)\)
\(=\left(x-2\right)\left[x\left(x+2\right)+\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)
d) \(x^4-x^2+2x-1\)
\(=x^4-\left(x^2-2x+1\right)\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2\right)^2-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
e)Sửa đề \(x^4+x^3+x^2-1\)
\(=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x-1\right)\)
f) \(x^3-4x^2+4x-1\)
\(=x^3-x^2-3x^2+3x+x-1\)
\(=x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-3x+1\right)\)
`(x - 2)/3 = (x + 1)/4`
`(x - 2) . 4 = (x + 1) . 3`
`<=> 4x - 8 = 3x + 3`
`<=> 4x - 3x = 3 + 8`
`<=> (4 - 3)x = 11`
`=> x = 11`
`=>` `x = 11`
\(|x+1|+|x+2|+|x+3|=4x-4\left(1\right)\)
Vì \(\hept{\begin{cases}|x+1|\ge0;\forall x\\|x+2|\ge0;\forall x\\|x+3|\ge0;\forall x\end{cases}}\)
\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0;\forall x\)
Mà \(|x+1|+|x+2|+|x+3|=4x-4\)
\(\Rightarrow4x-4\ge0\)
\(\Rightarrow4x\ge4\)
\(\Rightarrow x\ge1>0\)
\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)(2)
Thay (2) vào (1) ta được :
\(x+1+x+2+x+3=4x-4\)
\(\Leftrightarrow3x+6=4x-4\)
\(\Leftrightarrow3x-4x=-4-6\)
\(\Leftrightarrow-x=-10\)
\(\Leftrightarrow x=10\)
Vậy x=10