Tìm x: x^4 + 5x^3 + 10x^2 + 5x - 21 = 0
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ND
1
MR
1
KT
22 tháng 7 2018
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
A
0
29 tháng 8 2021
\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
29 tháng 8 2021
a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)
Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)
b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy ...
d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
NH
2
15 tháng 5 2021
`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`
15 tháng 5 2021
2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`
B
1
\(x^4+5x^3+10x^2+5x-21=0\)
\(\Leftrightarrow x^4-x^3+6x^3-6x^2+16x^2-16x+21x-21=0\)
\(\Leftrightarrow x^3\left(x-1\right)+6x^2\left(x-1\right)+16x\left(x-1\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+6x^2+16x+21\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+3x^2+9x+21\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)+9\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2+3x+9\right)=0\)
<=> x-1=0 <=> x=1
x+3=0 <=> x=-3
\(x^2+3x+9=x^2+2.\frac{3}{2}x+\frac{9}{4}+\frac{27}{4}=\left(x+\frac{3}{2}\right)^2+\frac{27}{4}>0\)
vậy nghiệm của pt là x=1; x=-3