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

mình vừa lên lớp 9 , chưa học phương trình bậc 2 

a)2x³ + 7x² - x - 12 =0

=>2x³+x²-4x+6x²+3x-12=0

=>x(2x²+x-4)+3(2x²+x-4)=0

=>(x+3)(2x²+x-4)=0

=>x+3=0 hoặc 2x²+x-4=0

Xét x+3=0 <=>x=-3

Xét 2x²+x-4=0 ta dùng delta

\(\Delta=1^2-\left(-4\left(2.4\right)\right)=33>0\)

=>pt có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{33}}{4}\)

b)- x^3 + x^2 + 7x + 2 =0

=>-x³+3x²+x-2x²+6x+2=0

=>-x(x²-3x-1)+(-2)(x²-3x-1)=0

=>-(x+2)(x²-3x-1)=0

=>-(x+2)=0 hoặc x²-3x-1=0

Xét -(x+2)=0 <=>x=-2

Xét x²-3x-1=0 theo delta ta có:

\(\Delta=\left(-3\right)^2-\left(-4\left(1.1\right)\right)=13>0\)

=>pt cũng có 2 nghiệm phân biệt

\(\Rightarrow x_{1,2}=\frac{3\pm\sqrt{13}}{2}\)

xài hóc ne đi

ĐKXĐ: \(x\ge3\)

\(x-3\sqrt{x-3}-3=0\Rightarrow x-3-3\sqrt{x-3}=0\)

\(\Rightarrow\sqrt{x-3}\left(\sqrt{x-3}-3\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x-3}=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=12\end{matrix}\right.\)

ĐK: `x>=3`

`x-3\sqrt(x-3)-3=0`

`<=>(x-3)-3\sqrt(x-3)=0`

`<=>\sqrt(x-3) (\sqrt(x-3)-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x-3}=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=12\end{matrix}\right.\)

Vậy...

ĐK: \(x\ge3\)

\(PT\Leftrightarrow\sqrt{x-3}\left(\sqrt{x-3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x-3=9=>x=12\left(tm\right)\end{matrix}\right.\)

Vậy...

 `x^3+1/x^3=6(x+1/x)(x ne 0)`

`<=>(x+1/x)(x^2-1+1/x^2)=6(x+1/x)`

`<=>(x+1/x)(x^2-1+1/x^2-6)=0`

`<=>((x^2+1)/x)(x^2+1/x^2-7)=0`

`(x^2+1)/x ne 0(AA x)`

`=>x^2+1/x^2-7=0`

`=>x^2+2+1/x^2-9=0`

`<=>(x+1/x)^2-3=0`

`<=>(x+1/x+3)(x+1/x-3)=0`

`+)x+1/x+3=0`

`<=>(x^2+3x+1)/x=0`

`<=>x^2+3x+1=0`

`<=>x^2+3x+9/4=5/4`

`<=>(x+3/2)^2=5/4`

`<=>x=(+-\sqrt{5}-3)/2`

`+)x+1/x-3=0`

`<=>(x^2-3x+1)/x=0`

`<=>x^2-3x+1=0`

`<=>x^2-3x+9/4=5/4`

`<=>(x-3/2)^2=5/4`

`<=>x=(+-\sqrt{5}+3)/2`

Vậy `S={(\sqrt{5}-3)/2,(-\sqrt{5}-3)/2,(\sqrt{5}+3)/2,(-\sqrt{5}+3)/2}`

ĐK: \(x\ne0\)

\(PT\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(x^2-1+\dfrac{1}{x^2}\right)=6\left(x+\dfrac{1}{x}\right)\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=0\\x^2+\dfrac{1}{x^2}-7=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(loai\right)\\x^4-7x^2+1=0\left(1\right)\end{matrix}\right.\)

Giải (1): \(\Leftrightarrow x^2=\dfrac{7\pm3\sqrt{5}}{2}\) \(\Rightarrow x=\pm\sqrt{\dfrac{7\pm3\sqrt{5}}{2}}\)

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

`a,(x+\sqrt{3})+4(x^2-3)=0`

`<=>(x+\sqrt{3})+4(x-\sqrt{3})(x+\sqrt{3})=0`

`<=>(x+\sqrt{3})[4(x-\sqrt{3}+1]=0`

`<=>(x+\sqrt{3})(4x-4\sqrt{3}+1)=0`

`<=>` \(\left[ \begin{array}{l}x+\sqrt{3}=0\\4x-4\sqrt{3}+1=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\sqrt{3}\\4x=4\sqrt{3}-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\sqrt{3}\\x=\sqrt{3}-\dfrac{1}{4}\end{array} \right.\) 

Vậy phương trình có tập nghiệm `S={-\sqrt{3},\sqrt{3}-1/4}`

\(\Leftrightarrow\left(x+\sqrt{3}\right)+4\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{3}\right)\left(1+4x-4\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\dfrac{4\sqrt{3}-1}{4}\end{matrix}\right.\)

a) Ta có: \(\left(x+\sqrt{3}\right)+4\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{3}\right)+4\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(x+\sqrt{3}\right)\left(1+4x-4\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=0\\4x+1-4\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\4x=4\sqrt{3}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=\dfrac{4\sqrt{3}-1}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{3};\dfrac{4\sqrt{3}-1}{4}\right\}\)

\(\Leftrightarrow\left(x^3+3x^2.3+3.x.3^2+3^3\right)-\left(x^3-3x^2+3x-1\right)=0\\ \Leftrightarrow\left(x^3-x^3\right)+\left(9x^2+3x^2\right)+\left(27x-3x\right)+\left(27+1\right)=0\\ \Leftrightarrow12x^2+24x+28=0\\ \Leftrightarrow x^2+2x+\dfrac{7}{3}=0\\ \Leftrightarrow\left(x^2+2x+1\right)+\dfrac{4}{3}=0\\\Leftrightarrow\left(x+1\right)^2+\dfrac{4}{3}=0\\ \Leftrightarrow\left(x+1\right)=-\dfrac{4}{3}\left(vô.lí\right)\)

=> Pt vô nghiệm