giải pt
x4+3x3+4x2+3x+1=0
2x-3/3 - x-3/6=4x+3/5 -17
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B
19 tháng 1 2022
Ta có : x4+3x3+4x2+3x+1=0
⇔ ( x4 + x3 ) + ( 2x3 + 2x2 ) + ( 2x2 + 2x ) + ( x + 1 ) = 0
⇔ x3 ( x + 1 ) + 2x2 ( x + 1 ) + 2x ( x+1 ) + ( x + 1 ) =0
⇔ ( x + 1 ) ( x3 + 2x2 + 2x + 1 ) = 0
⇔ ( x + 1 ) [ ( x3 + 1 ) + ( 2x2 + 2x ) ] = 0
⇔ ( x + 1 ) [ (x + 1 ) ( x2 - x +1 ) + 2x ( x + 1 ) ] =0
⇔ ( x +1 ) ( x + 1 ) ( x2 + x +1 ) =0
⇒ \(\left[{}\begin{matrix}x+1=0\\x^{2^{ }}+x+1=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=-1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(VoLy\right)\end{matrix}\right.\)
Vậy x = -1
D
19 tháng 1 2022
x4+3x3+4x2+3x+1=0
⇔(x4+2x3+x2)+(x3+2x2+1)+(x2+2x+1)=0
⇔x2(x2+2x+1)+x(x2+2x+1)+(x2+2x+1)=0
⇔x2(x+1)2+x(x+1)2+(x+1)2=0
⇔(x+1)2(x2+x+1)=0
Vì x2+x+1=x2+x+\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\)=(x+\(\dfrac{1}{2}\))2+\(\dfrac{3}{4}\)>0 nên phương trình đã cho tương đương:
(x+1)2=0 ⇔(x+1)(x+1)=0 ⇔x=-1.
TE
1
22 tháng 8 2019
a, \(x^4-4x^3-6x^2-4x+1=0\)(*)
<=> \(x^4+4x^2+1-4x^3-4x+2x^2-12x^2=0\)
<=> \(\left(x^2-2x+1\right)^2=12x^2\)
<=>\(\left(x-1\right)^4=12x^2\) <=> \(\left[{}\begin{matrix}\left(x-1\right)^2=\sqrt{12}x\\\left(x-1\right)^2=-\sqrt{12}x\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x^2-2x+1-\sqrt{12}x=0\left(1\right)\\x^2-2x+1+\sqrt{12}x=0\left(2\right)\end{matrix}\right.\)
Giải (1) có: \(x^2-2x+1-\sqrt{12}x=0\)
<=> \(x^2-2x\left(1+\sqrt{3}\right)+\left(1+\sqrt{3}\right)^2-\left(1+\sqrt{3}\right)^2+1=0\)
<=> \(\left(x-1-\sqrt{3}\right)^2-3-2\sqrt{3}=0\)
<=> \(\left(x-1-\sqrt{3}\right)^2=3+2\sqrt{3}\) <=> \(\left[{}\begin{matrix}x-1-\sqrt{3}=\sqrt{3+2\sqrt{3}}\\x-1-\sqrt{3}=-\sqrt{3+2\sqrt{3}}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(ktm\right)\\x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(tm\right)\end{matrix}\right.\)
=> \(x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)
Giải (2) có: \(x^2-2x+1+\sqrt{12}x=0\)
<=> \(x^2-2x\left(1-\sqrt{3}\right)+\left(1-\sqrt{3}\right)^2-\left(1-\sqrt{3}\right)^2+1=0\)
<=> \(\left(x+\sqrt{3}-1\right)^2=3-2\sqrt{3}\) .Có VP<0 => PT (2) vô nghiệm
Vậy pt (*) có nghiệm x=\(-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)
3 tháng 6 2023
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
28 tháng 4 2023
a: =>17x-5x-15-2x-5=0
=>10x-20=0
=>x=2
b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)
=>11x+23=-2x-16
=>13x=-39
=>x=-3(nhận)
c: =>5x+7>=3x-3
=>2x>=-10
=>x>=-5
d: =>5(3x-1)=-2(x+1)
=>15x-5=-2x-2
=>17x=3
=>x=3/17
e: =>4x^2-1-4x^2-3x-2=0
=>-3x-3=0
=>x=-1
g: =>7x-5-8x+2-7<0
=>-x-10<0
=>x+10>0
=>x>-10
ND
Nguyễn Đức Trí
VIP
22 tháng 8 2023
a) \(...=P\left(x\right)=2x^4-x^4+3x^3+4x^2-3x^2+3x-x+3\)
\(P\left(x\right)=x^4+3x^3+x^2+2x+3\)
\(...=Q\left(x\right)=x^4+x^3+3x^2-x^2+4x+4-2\)
\(Q\left(x\right)=x^4+x^3+2x^2+4x+2\)
b) \(P\left(x\right)+Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)+\left(x^4+x^3+2x^2+4x+2\right)\)
\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)
\(P\left(x\right)-Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)-\left(x^4+x^3+2x^2+4x+2\right)\)
\(\)\(\Rightarrow P\left(x\right)-Q\left(x\right)=x^4+3x^3+x^2+2x+3-x^4-x^3-2x^2-4x-2\)
\(\Rightarrow P\left(x\right)-Q\left(x\right)=2x^3-x^2-2x+1\)
NT
1
21 tháng 6 2022
a: \(\dfrac{3x-7}{2}+\dfrac{x-1}{3}=-16\)
\(\Leftrightarrow3\left(3x-7\right)+2\left(x-1\right)=-96\)
\(\Leftrightarrow9x-21+2x-2=-96\)
=>11x=-73
hay x=-73/11
b: \(x-\dfrac{x-1}{3}=\dfrac{2x+1}{5}\)
=>15x-5(x-1)=3(2x+1)
=>15x-5x+5=6x+3
=>10x+5=6x+3
=>4x=-2
hay x=-1/2
c: \(\dfrac{2x-1}{3}-\dfrac{5x+2}{7}=x+13\)
=>14x-7-15x-6=21(x+13)
=>21x+273=-x-13
=>22x=-286
hay x=13
x + 3x + 4x + 3x + 1 = 0
⇒x + x + 2x + 2x + 2x + 2x + x + 1 = 0
⇒x x + 1 + 2x x + 1 + 2x x + 1 + x + 1 = 0 ⇒ x + 1 x + x + x + x + x + 1 = 0 ⇒ x + 1 x x + 1 + x x + 1 + x + 1 = 0 ⇒ x + 1 x + 1 x + x + 1 = 0 ⇒ x + 1 x + x + 1 = 0 ⇒ x + 1 = 0 vix̀ + x + 1 ≠ 0 ⇒x + 1 = 0 ⇒x = −1 vậy pt có No ......... 3 2x − 3 − 6 x − 3 = 5 4x + 3 − 17 ⇔ 30 10 2x − 3 − 30 5 x − 3 = 30 6 4x + 3 − 30 17.30 ⇔20x − 30 − 5x + 15 = 24x + 18 − 510 ⇔20x − 5x − 24x = 18 − 510 + 30 − 15
⇔− 9x = −477 ⇔x = 53
vậy pt có No........
\(x^4+3x^3+4x^2+3x+1=0\)
\(\Rightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(\Rightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3+x^2+x^2+x+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)\left(x^2+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2=0\left(vìx^2+x+1\ne0\right)\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
vậy pt có No .........
\(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)
\(\Leftrightarrow\frac{10\left(2x-3\right)}{30}-\frac{5\left(x-3\right)}{30}=\frac{6\left(4x+3\right)}{30}-\frac{17.30}{30}\)
\(\Leftrightarrow20x-30-5x+15=24x+18-510\)
\(\Leftrightarrow20x-5x-24x=18-510+30-15\)
\(\Leftrightarrow-9x=-477\)
\(\Leftrightarrow x=53\)
vậy pt có No........