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\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)
\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)
\(x^4-8x^3+21x^2-24x+9=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))
\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)
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
\(\left(\sqrt{2x^2-6x+2}-2x+3\right)\left(-\sqrt{2x^2-6x+2}-3x+4\right)=0\)
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\(x=\frac{1}{2}\)
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) \(\sqrt{x-1}-\sqrt{x-4}=1\)
\(\Rightarrow\left(\sqrt{x-1}-\sqrt{x-4}\right)^2=1\)
\(\Rightarrow\left(\sqrt{x-1}\right)^2-2.\sqrt{x-1}.\sqrt{x-4}+\left(\sqrt{x-4}\right)^2=1\)
\(\Rightarrow x-1-2.\sqrt{\left(x-1\right)\left(x-4\right)}+x-4=1\)
\(\Rightarrow2x-2\sqrt{\left(x-1\right)\left(x-4\right)}-6=0\)
\(\Rightarrow2\left(x-\sqrt{\left(x-1\right)\left(x-4\right)}-3\right)=0\)
\(\Rightarrow x-\sqrt{\left(x-1\right)\left(x-4\right)}-3=0\)
....
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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 \(S=\left\{5\right\}\)
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 \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)
x=+-
\(\sqrt{3}\)
<br class="Apple-interchange-newline"><div></div>x4+x2−12 = 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