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ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
giải pt sau \(\left(\dfrac{x+1}{x-2}\right)^2-3\left(\dfrac{2x-4}{x-4}\right)^2+\dfrac{x+1}{x-4}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ne\left\{2;4\right\}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x+1}{x-2}=a\\\dfrac{x-2}{x-4}=b\end{matrix}\right.\) \(\Rightarrow\dfrac{x+1}{x-4}=ab\)
Phương trình trở thành:
\(a^2-12b^2+ab=0\)
\(\Leftrightarrow a^2+4ab-3ab-12b^2=0\)
\(\Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x+1}{x-2}-\dfrac{3\left(x-2\right)}{x-4}=0\\\dfrac{x+1}{x-2}+\dfrac{4\left(x-2\right)}{x-4}=0\end{matrix}\right.\)
Bạn tự quy đồng và hoàn thành phần còn lại nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
bai 1
1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0
<=>(2x)^2-5^2=0
<=>(2x+5)*(2x-5)=0
<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
(4x - 3)2 - (2x + 1)2 = 0
\(\Leftrightarrow\) (4x - 3 - 2x - 1)(4x - 3 + 2x + 1) = 0
\(\Leftrightarrow\) (2x - 4)(6x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
3x - 12 - 5x(x - 4) = 0
\(\Leftrightarrow\) 3x - 12 - 5x2 + 20x = 0
\(\Leftrightarrow\) -5x2 + 23x - 12 = 0
\(\Leftrightarrow\) 5x2 - 23x + 12 = 0
\(\Leftrightarrow\) 5x2 - 20x - 3x + 12 = 0
\(\Leftrightarrow\) 5x(x - 4) - 3(x - 4) = 0
\(\Leftrightarrow\) (x - 4)(5x - 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-4=0\\5x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy ...
(8x + 2)(x2 + 5)(x2 - 4) = 0
\(\Leftrightarrow\) (8x + 2)(x2 + 5)(x - 2)(x + 2) = 0
Vì x2 \(\ge\) 0 \(\forall\) x nên x2 + 5 > 0 \(\forall\) x
\(\Rightarrow\) (8x + 2)(x - 2)(x + 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}8x+2=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy ...
Chúc bn học tốt!
a) Ta có: \(\left(4x-3\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3-2x-1\right)\left(4x-3+2x+1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(6x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\dfrac{1}{3}\right\}\)
b) Ta có: \(3x-12-5x\left(x-4\right)=0\)
\(\Leftrightarrow3\left(x-4\right)-5x\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{4;\dfrac{3}{5}\right\}\)
c) Ta có: \(\left(8x+2\right)\left(x^2+5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow2\left(4x+1\right)\left(x^2+5\right)\left(x-2\right)\left(x+2\right)=0\)
mà \(2>0\)
và \(x^2+5>0\forall x\)
nên \(\left(4x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\x=2\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{4};2;-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)+x^3\left(x+1\right)+2x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+x^2+x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy....
b) \(x^4+3x^3-2x^2+x-3=0\)
\(\Leftrightarrow x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0\)
\(\Leftrightarrow x^3\left(x-1\right)+4x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+4x^2+2x+3\right)=0\)
...
\(\Leftrightarrow x=1\)
p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))
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a)Ta có \(\left(2x+1\right)\left(x^2+2\right)=0\)<=>
2x+1=0<=>x=\(-\frac{1}{2}\)
hoặc \(x^2+2=0\)<=>\(x^2=-2\)(Vô lí)
Vậy tập nghiệm của pt S=(\(-\frac{1}{2}\))
b)\(\left(x^2+4\right)\left(7x-3\right)=0\)
<=>\(\left[{}\begin{matrix}x^2+4=0\\7x-3=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x^2=-4\\x=\frac{3}{7}\end{matrix}\right.\)
\(x^2=-4\) vô lí
Vậy ..........
c)\(\left(x^2+x+1\right)\left(6-2x\right)=0\)
<=>\(\left[{}\begin{matrix}x^2+x+1=0\\6-2x=0\end{matrix}\right.\)
Vì \(x^2+x+1>0\)(dễ dàng c/m)
=>6-2x=0=>x=3
Vậy...
d)\(\left(8x-4\right)\left(x^2+2x+2\right)=0\)
<=>8x-4=0,x=\(\frac{1}{2}\)
hoặc \(x^2+2x+2=0\)(vô lí)
Vậy .....
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(x^4-2x^3+3x^2-2x+1=0\)
\(\Leftrightarrow\left(x^4-2x^3+x^2\right)+\left(x^2-2x+1\right)+x^2=0\)
\(\Leftrightarrow x^2\left(x-1\right)^2+\left(x-1\right)^2+x^2=0\)
\(\Leftrightarrow\) (x - 1)2 = 0 và x2 = 0
\(\Leftrightarrow\) x - 1 = 0 và x = 0
\(\Leftrightarrow\) x = 1 và x = 0 (vô lí)
Vậy phương trình vô nghiệm.
2. \(\left(x^2-4\right)^2=8x+1\)
\(\Leftrightarrow x^4-8x^2+16=8x+1\)
\(\Leftrightarrow x^4-8x^2-8x+15=0\)
\(\Leftrightarrow x^4-x^3+x^3-x^2-7x^2+7x-15x+15=0\)
\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)-7x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-7x-15\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2+4x^2-12x+5x-15\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)+4x\left(x-3\right)+5\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2+4x+5\right)=0\)
\(\Leftrightarrow\) x - 1 = 0 hoặc x - 3 = 0 hoặc x2 + 4x + 5 = 0
1) x - 1 = 0 \(\Leftrightarrow\) x = 1
2) x - 3 = 0 \(\Leftrightarrow\) x = 3
3) \(x^2+4x+5=0\left(\text{loại vì }x^2+4x+5=\left(x+2\right)^2+1>0\forall x\right)\)
Vậy tập nghiệm của pt là S = {1;3}.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)
<=> \(\left(x+1\right)^2.\left(x-2\right)^2.\left(x-4\right)^2+\frac{x+1}{x-4}.\left(x-2\right)^2.\left(x-4\right)^2-\frac{3\left(2x-4\right)^2}{\left(x-4\right)^2}.\left(x-2\right)^2.\left(x-4\right)^2\)\(=0.\left(x-2\right)^2.\left(x-4\right)^2\)
<=> \(\left(x+1\right)^2.\left(x-4\right)^2+\left(x+1\right).\left(x-2\right)^2.\left(x-4\right)^2-3\left(2x-4\right)^2.\left(x-2\right)^2=0\)
<=> \(-\left(x-3\right)\left(5x-4\right)\left(2x^2-9x+16\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
Mà vì: \(2x^2-9x+16\ne0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)