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IL
2
26 tháng 7 2018
I don't now
or no I don't
..................
sorry
KT
26 tháng 7 2018
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow\)\(x^4-x^3-7x^2+7x-6x+6=0\)
\(\Leftrightarrow\)\(x^3\left(x-1\right)-7x\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-1\right)\left(x^3-7x-6\right)=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\)
đến đây lm tiếp
DL
4
PA
26 tháng 1 2017
x3 - 4x2 - 8x + 8 = 0
x3 + 2x2 - 6x2 - 12x + 4x + 8 = 0
\(x^2\left(x+2\right)-6x\left(x+2\right)+4\left(x+2\right)=0\)
\(\left(x+2\right)\left(x^2-6x+4\right)=0\)
\(x+2=0\)
\(x=-2\)
27 tháng 7 2018
Ukm
It's very hard
l can't do it
Sorry!
27 tháng 7 2018
a) \(x^4-x^3-7x^2+x+6=0\)
\(\Leftrightarrow x^4+2x^3-3x^3-6x^2-x^2-2x+3x+6=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x^2\left(x+2\right)-x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x^2-x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-3\right)=0\). Làm nốt
b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-3\right)^2+\left|y+3\right|=0\)
Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)
\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
c) \(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)^2-2.\left(2x^2+x\right).2+4-1=0\)
\(\Leftrightarrow\left(2x^2+x-2\right)^2=1\Leftrightarrow\orbr{\begin{cases}2x^2+x-2=1\\2x^2+x-2=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)
Từ đó tính đc x
d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)
\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)
\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
Đặt \(x^2+5x+5=a\), khi đó pt có dạng:
\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
15 tháng 11 2019
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
KN
15 tháng 11 2019
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
PN
0
13 tháng 7 2015
1a/ x3+x2+x+1=0
x2(x+1).(x+1)=0
=> x2(x+1)=0 x =1
hoặc =>[
x+1=0 x=-1
b/(x+2)2=x+2
x2+2.x.2+22 =x+2
x+x+4x+4=x+2
6x+4=x+2
....
c/(x+1)(6x2+2x)+(x-1)(6x2+2x)=0
x2-12 + (6x2+2x)2=0
=> x2-1 = 0 x=1
hoặc => [
(6x2+2x)2=0 x= 0
VQ
3
4 tháng 8 2019
\(a,-2x\left(2-3x\right)+3\left(-5+7x-6x^2\right)=-4\)
\(\Rightarrow-4x+6x^2-15+21x-18x^2=-4\)
\(\Rightarrow-12x^2+17x-11=0\)
\(\Rightarrow12x^2-17x+11=0\)
\(\Rightarrow9x^2-2.3.\frac{17}{6}x+\left(\frac{17}{6}\right)^2-\left(\frac{17}{6}\right)^2+11=0\)
\(\Rightarrow\left(3x-\frac{17}{6}\right)^2+\frac{107}{36}=0VN\)
Không có gt x thỏa mãn
\(b,-3x\left(-1+3x-4x^2\right)+6x^2\left(-2x+3\right)=0\)
\(\Rightarrow3x-9x^2+12x^3-12x^3+18x^2=0\)
\(\Rightarrow9x^2+3x=0\)
\(\Rightarrow3x\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\3x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\3x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}}\)
a) Thieu de :v
\(x^3-7x-6=0\)
\(\Leftrightarrow x^3-4x-3x-6=0\)
\(\Leftrightarrow x\left(x^2-4\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-3x+x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x-3\right)+\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\)Hoặc x + 2 = 0
Hoặc x - 3 = 0
Hoặc x + 1 = 0
\(\Leftrightarrow\)Hoặc x = -2 Hoặc x = 3 Hoặc x = -1
Vậy tập nghiệm của phương trình là ; \(S=\left\{-2;3;-1\right\}\)