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QT
0
TN
12 tháng 6 2017
\(x^4+5x^3-8x-40=0\)
\(\Leftrightarrow x^3\left(x+5\right)-8\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^3-8\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)=0\)
Ta có : \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3\ge3\)\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
12 tháng 6 2017
\(x^4+5x^3-8x-40=0\)
\(\Leftrightarrow x^4+3x^3-10x^2+2x^3+6x^2-20x+4x^2+12x-40=0\)
\(\Leftrightarrow x^2\left(x^2+3x-10\right)+2x\left(x^2+3x-10\right)+4\left(x^2+3x-10\right)=0\)
\(\Leftrightarrow\left(x^2+3x-10\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left(x^2-2x+5x-10\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[x\left(x-2\right)+5\left(x-2\right)\right]\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)\left(x^2+2x+4\right)=0\)
Dễ thấy: \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3>0\) (vô nghiệm)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
NT
0
15 tháng 5 2023
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
1 tháng 3 2019
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
1 tháng 3 2019
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
14 tháng 3 2017
3/(x^2-13x+40)+2/(x^2-8x+15)+1/(x^2-5x+6)+6/5+0
3/(x-8)(x-5)+2/(x-5)(x-3)+1/(x-3)(x-2)+6/5=0
1/(x-8)-1/(x-5)+1/(x-5)-1/(x-3)+1/(x-3)-1/(x-2)+6/5=0
1/(x-8)-1/(x-2)+6/5=0
ban tu giai tiep nhan
m^2x+2x=5-3mx
m^2x+3mx+2x=5
x(m^2+3m+2)=5
khi 0x=5 thi pt vo nghiem
m^2+3m+2=0
(m+1)(m+2)=0
m=-1 hoac m=-2
\(x^4+5x^3-8x-40=0\)
\(\Leftrightarrow x^3\left(x+5\right)-8\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^3-8\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)=0\)
Ta có : \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3\ge3\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}}\)
Chúc bạn học tốt !!!
=x4-16+5x(x2-4)+12(x-2)=0
<=>(x-2)[(x2+4)(x+2)+5x(x+2)+12]=0
<=>(x-2)(x3+7x2+14x+20)=0 <=> x=2 hoặc
x3+7x2+14x+20=0 <=>(x+5)(x2+2x+4)=0 <=>x+5=0(x2+2x+4>0) <=>x=-5
vậy x=2;x=-5