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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
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left[4\left(x-1\right)^2\right]-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(2x-2\right)^2-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(2x-2-x-1\right)\left(2x-2+x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
c) |2x - 1| = x + 2
<=> 2x - 1 = +(x + 2) hoặc -(x + 2)
* 2x - 1 = x + 2
<=> 2x - x = 2 + 1
<=> x = 3
* 2x - 1 = -(x + 2)
<=> 2x - 1 = x - 2
<=> 2x - x = -2 + 1
<=> x = -1
Vậy.....
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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}.
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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)
\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)
<=> \(\left(x+1\right)^2.\left(x-2\right)^2.\left(x-4\right)^2+\frac{x+1}{x-4}.\left(x-2\right)^2.\left(x-4\right)^2-\frac{3\left(2x-4\right)^2}{\left(x-4\right)^2}.\left(x-2\right)^2.\left(x-4\right)^2\)\(=0.\left(x-2\right)^2.\left(x-4\right)^2\)
<=> \(\left(x+1\right)^2.\left(x-4\right)^2+\left(x+1\right).\left(x-2\right)^2.\left(x-4\right)^2-3\left(2x-4\right)^2.\left(x-2\right)^2=0\)
<=> \(-\left(x-3\right)\left(5x-4\right)\left(2x^2-9x+16\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
Mà vì: \(2x^2-9x+16\ne0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
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ĐK: x khác 0
Đặt: \(\frac{x^4}{2x^2+1}=t>0\Rightarrow\frac{2x^2+1}{x^4}=\frac{1}{t}\)
Ta có phương trình: \(t+\frac{1}{t}=2\Leftrightarrow t^2-2t+1=0\Leftrightarrow\left(t-1\right)^2=0\Leftrightarrow t=1\)
Với t = 1 ta có: \(\frac{x^4}{2x^2+1}=1\)<=> \(x^4-2x^2-1=0\Leftrightarrow\orbr{\begin{cases}x^2=1+\sqrt{2}\\x^2=1-\sqrt{2}\left(loai\right)\end{cases}}\)
khi đó: \(x=\pm\sqrt{1+\sqrt{2}}\)tm
Vậy....
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a/ \(2x-3=5x+2\)
\(\Leftrightarrow5x-2x=-3-2\)
\(\Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\)
Vậy..
b. \(2x\left(x-1\right)=2x+2\)
\(\Leftrightarrow2x^2-4x-2=0\)
\(\Leftrightarrow x^2-2x-1=0\)
\(\Leftrightarrow\left(x-1+\sqrt{2}\right)\left(x-1-\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{matrix}\right.\)
Vậy...
c/ ĐKXĐ : \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{6}{\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2+4x+4-x^2=6x-12\)
\(\Leftrightarrow2x-16=0\)
\(\Leftrightarrow x=8\)
Vậy..
\(\sqrt{\left(x+1\right)^2}=\sqrt{4.\left(x^2-2x+1\right)^2}^{ }\)
\(\Rightarrow x+1=2.\left(x^2-2x+1\right)^{ }\)
\(\Leftrightarrow\)x+1-2x^2 +2x -1=0
\(\Leftrightarrow\)-2x^2 +3x =0
\(\Leftrightarrow\)x.(-2x+3)=0\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)