\(x^4+x^2-12\)

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18 tháng 2 2020

x=+-

\(\sqrt{3}\)

<br class="Apple-interchange-newline"><div></div>x4+x212 = 0 

\(\Leftrightarrow x^4-3x^2+4x^2-12=0\)

\(\Leftrightarrow x^2\left(x^2-3\right)+4\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2-4\right)=0\)

Đến đây giải nốt nha 

6 tháng 6 2018

@Akai Haruma , @phynit giải dùm em vs ạ

3 tháng 8 2017

\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)

\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)

2 tháng 8 2017

\(x^4-8x^3+21x^2-24x+9=0\)

\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)

\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))

\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

NV
13 tháng 8 2020

ĐKXĐ: \(-\frac{16}{3}\le x\le4\)

\(\Leftrightarrow3x^2-12x+36=12\sqrt{4-x}+3\sqrt{3x+16}\)

\(\Leftrightarrow3x^2-9x+4\left(6-x-3\sqrt{4-x}\right)+\left(x+12-3\sqrt{3x+16}\right)=0\)

\(\Leftrightarrow3\left(x^2-3x\right)+\frac{4\left(x^2-3x\right)}{6-x+3\sqrt{4-x}}+\frac{x^2-3x}{x+12+3\sqrt{3x+16}}=0\)

\(\Leftrightarrow\left(x^2-3x\right)\left(3+\frac{4}{6-x+3\sqrt{4-x}}+\frac{1}{x+12+3\sqrt{3x+16}}\right)=0\)

\(\Leftrightarrow x^2-3x=0\)

b2

\(\left(\sqrt{2x^2-6x+2}-2x+3\right)\left(-\sqrt{2x^2-6x+2}-3x+4\right)=0\)

14 tháng 8 2017

Dự đoán \(\frac{1}{2}\)là nghiệm của phương trình ( casio :v)

Áp dụng AM-GM:\(2VF=3.\sqrt[3]{4.8x\left(4x^2+3\right)}\le4+8x+4x^2+3=4x^2+8x+7\)

và \(4x^2+8x+7\le8x^4+2x^2+6x+8\)vì nó tương đương \(\left(2x-1\right)^2\left(2x^2+2x+1\right)\ge0\)

Do đó \(VT\ge VF\)

Dấu = xảy ra khi\(x=\frac{1}{2}\)

12 tháng 6 2018

a) Ta có: \(\sqrt{x-2}+\sqrt{4x-8}=12\)

\(\Rightarrow\sqrt{x-2}+\sqrt{4}.\sqrt{x-2}=12\)

\(\Rightarrow\sqrt{x-2}\left(2+1\right)=12\)

\(\Rightarrow\sqrt{x-2}=4\)

\(\Rightarrow x-2=4^2\Rightarrow x=18\)

b) \(\sqrt{x-1}-\sqrt{x-4}=1\)

\(\Rightarrow\left(\sqrt{x-1}-\sqrt{x-4}\right)^2=1\)

\(\Rightarrow\left(\sqrt{x-1}\right)^2-2.\sqrt{x-1}.\sqrt{x-4}+\left(\sqrt{x-4}\right)^2=1\)

\(\Rightarrow x-1-2.\sqrt{\left(x-1\right)\left(x-4\right)}+x-4=1\)

\(\Rightarrow2x-2\sqrt{\left(x-1\right)\left(x-4\right)}-6=0\)

\(\Rightarrow2\left(x-\sqrt{\left(x-1\right)\left(x-4\right)}-3\right)=0\)

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

....

20 tháng 2 2018

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20 tháng 2 2018

gì đây?

21 tháng 9 2018

a)\(\dfrac{2}{x^2-1}+\dfrac{1}{x+1}=2\) Điều kiện:x#1,-1

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

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

\(\Leftrightarrow\dfrac{1}{x-1}=2\)

\(\Leftrightarrow1=2\left(x-1\right)\)

\(\Leftrightarrow2x=3\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

b)\(1-\dfrac{12}{x^2-4}=\dfrac{3}{x+2}\) Điều kiện:x#2,-2

\(\Leftrightarrow\dfrac{x^2-4-12}{x^2-4}=\dfrac{3}{x+2}\)

\(\Leftrightarrow x^2-16=3\left(x-2\right)\)

\(\Leftrightarrow x^2-16-3x+6=0\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{5\right\}\)

23 tháng 11 2022

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

b: =>(x^2-x-3)(x^2+x-4)=0

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)