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13 tháng 1 2017
1. Ta có \(x^3+3x^2+x+3=0\)
\(\Leftrightarrow\left(x^3+3x^2\right)+\left(x+3\right)=0\)
\(\Leftrightarrow x^2\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)
Nếu x+3=0 =>x=-3
Nếu \(x^2+1=0\) =>x\(=\varnothing\) (vì \(x^2+1>0\))
Vậy x=-3
DN
13 tháng 1 2017
2) đặt x^2+x+1 = t
=> x^2 +x +2 =t+1
pt => t(t+1)=2
t^2 + t -2 =0
\(\Rightarrow\left[\begin{matrix}t=1\\t=-2\end{matrix}\right.\)
voi t=1 => x^2 +x+1=1
=> \(\Rightarrow\left[\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
voi t=-2 => x^2+x+1=-2
=> x^2+x+3=0(vo nghiem)
cau 3 lam nhu cau 2
4) pt <=> (x^2-4)(x+3-x+1)=0
ban tu giai not nha
LH
0
12 tháng 10 2015
(x^2+5x+4)(x^2+5x+6)-24
Đặt x^2+5x+5 = a
Do đó (a-1)(a+1)-24
= a^2- 25
= a^2-5^2 =(a-5)(a+5)
= ( x^2+5x+5-5)( x^2+5x+5+5)
= ( x^2+5x)( x^2+5x+10)
NL
Nguyễn Lê Phước Thịnh
CTVHS
13 tháng 8 2020
a) Ta có: \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)(1)
Đặt \(a=x^2+x\)
(1)\(=a^2-14a+24\)
\(=a^2-12a-2a+24\)
\(=a\left(a-12\right)-2\left(a-12\right)\)
\(=\left(a-12\right)\left(a-2\right)\)
\(=\left(x^2+x-12\right)\left(x^2+x-2\right)\)
\(=\left(x^2+4x-3x-12\right)\left(x^2+2x-x-2\right)\)
\(=\left[x\left(x+4\right)-3\left(x+4\right)\right]\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(=\left(x+4\right)\left(x-3\right)\left(x+2\right)\left(x-1\right)\)
b) Ta có: \(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
\(=a^2+4a-12\)
\(=a^2+6a-2a-12\)
\(=a\left(a+6\right)-2\left(a+6\right)\)
\(=\left(a+6\right)\left(a-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+2x-x-2\right)\)
\(=\left(x^2+x+6\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)
\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)
c) Ta có: \(x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
d) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(2)
Đặt \(x^2+5x=b\)
(2)\(=\left(b+4\right)\left(b+6\right)+1\)
\(=b^2+10b+24+1\)
\(=b^2+10b+25\)
\(=\left(b+5\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
e) Ta có: \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)(3)
Đặt \(c=x^2+8x\)
(3)\(=\left(c+7\right)\left(c+15\right)+15\)
\(=c^2+22c+105+15\)
\(=c^2+22c+120\)
\(=c^2+12c+10c+120\)
\(=c\left(c+12\right)+10\left(c+12\right)\)
\(=\left(c+12\right)\left(c+10\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x^2+6x+2x+12\right)\left(x^2+8x+10\right)\)
\(=\left[x\left(x+6\right)+2\left(x+6\right)\right]\left(x^2+8x+10\right)\)
\(=\left(x+6\right)\left(x+2\right)\left(x^2+8x+10\right)\)
Xin lỗi viết sai đề !
(x+3)4-(x-3)4-24.x2.(x-1)=108
<=> [(x+3)2-(x-3)2 ]-24.x2.(x-1)=108
<=> (x+3-x+3).(x+3+x-3)-24.x2.(x-1)=108
<=> 6.2x-24.x2.(x-1)=108
<=> 12x.[1-2x.(x-1)]=108
<=> x.[1-2x.(x-1)]=9
Ta có : 9=9.1=3.3=(-9).(-1)=(-3).(-3)
Thay \(\hept{\begin{cases}x=3\\1-2x.\left(x-1\right)=3\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=ko\left(tm\right)\end{cases}}\left(kotm\right)}\)
Thay \(\hept{\begin{cases}x=-3\\1-2x.\left(x-1\right)=-3\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\x=2\end{cases}}\left(tm\right)\)
Thay \(\hept{\begin{cases}x=1\\1-2x.\left(x-1\right)=9\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\x=ko\left(tm\right)\end{cases}\left(kotm\right)}\)
Thay \(\hept{\begin{cases}x=9\\1-2x.\left(x-1\right)=1\end{cases}\Rightarrow\hept{\begin{cases}x=9\\x=0\end{cases}\left(tm\right)}}\)
Thay \(\hept{\begin{cases}x=-1\\1-2x.\left(x-1\right)=-9\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\x=....\end{cases}\left(tm\right)}\)
Thay \(\hept{\begin{cases}x=-9\\1-2x.\left(x-1\right)=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-9\\x=...\end{cases}\left(tm\right)}}\)
\(\left(x+3\right)^4-\left(x-3\right)^4-24.x^2.\left(x-1\right)=108.\)
\(\Leftrightarrow\)\(\left[\left(x+3\right)^2-\left(x-3\right)^2\right].\left[\left(x+3\right)^2+\left(x-3\right)^2\right].24x^2.\left(x-1\right)=108\)
\(\Leftrightarrow\)\(\left(x+3-x+3\right).\left(x+3+x-3\right).\left[\left(x^2+6x+9\right)+\left(x^2-6x+9\right)\right].24x^2.\left(x-1\right)=108\)
\(\Leftrightarrow\)\(6.2x.\left(x^2+6x+9+x^2-6x+9\right).24x^2.\left(x-1\right)=108\)
\(\Leftrightarrow\)\(12x.\left(2x^2+18\right).24x^2.\left(x-1\right)=108\)
\(\Leftrightarrow\)\(288x^3.\left[2.\left(x^2+9\right)\right].\left(x-1\right)=108\)
\(\Leftrightarrow\)\(\left(x^4-x^3\right).\left(x^2+9\right).2=\frac{3}{8}\)
\(\Leftrightarrow\)\(x^8+9x^4-x^6-9x^3=\frac{3}{16}\)
\(\Leftrightarrow\)\(x^2.\left(x^4-x^3\right)+9.\left(x^4-x^3\right)=\frac{3}{16}\)
\(\Leftrightarrow\)\(\left(x^2+9\right).\left(x^4-x^3\right)=\frac{3}{16}\)
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