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bạn tham khảo thêm cách này nha Shonogeki No Soma
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\\x\ne-1\end{cases}}\)
Đặt \(a=\left(x-1\right)^3;b=x^3;c=\left(x+1\right)^3\)
pt đã cho đc viết lại thành
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}a=-b\\b=-c\\c=-a\end{cases}}\) (kí hiệu [..] mới đúng nha)
- TH1: a = -b hay \(\left(x-1\right)^3=-x^3\) \(\Leftrightarrow2x^3-3x^2+3x-1=0\) \(\Leftrightarrow x=\frac{1}{2}\) (Nhận)
- TH2: b = -c hay \(\left(x+1\right)^3=-x^3\) \(\Leftrightarrow2x^3+3x^2+3x+1=0\) \(\Leftrightarrow x=-\frac{1}{2}\) (Nhận)
- TH3: c = -a hay \(\left(x+1\right)^3=-\left(x-1\right)^3\) \(\Leftrightarrow x=0\) (Loại)
KL: \(S=\left\{\frac{1}{2};-\frac{1}{2}\right\}\)
\(\frac{1}{\left(x-1\right)^3}+\frac{1}{\left(x+1\right)^3}+\frac{1}{x^3}=\frac{1}{3x\left(x^2+2\right)}\)
\(\Leftrightarrow4x^8+15x^6+12x^4+8x^2-6=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x^2+3\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{2}\end{cases}}\)
Đặt \(y=\left(x-3\right)\sqrt{\frac{x+1}{x-3}}\)
Suy ra pt trở thành \(y^2+y+3=0\)
Mà : \(y^2+y+3=\left(y^2+y+\frac{1}{4}\right)+\frac{11}{4}=\left(y+\frac{1}{2}\right)^2+\frac{11}{4}>0\)
Do đó pt trên vô nghiệm.
b)\(\frac{1}{x+\sqrt{x^2+x}}+\frac{1}{x-\sqrt{x^2+x}}=x\)
\(\Leftrightarrow\frac{x-\sqrt{x^2+x}}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}+\frac{x+\sqrt{x^2+x}}{\left(x-\sqrt{x^2+x}\right)\left(x+\sqrt{x^2+x}\right)}-\frac{x\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
\(\Leftrightarrow\frac{x-\sqrt{x^2+x}+x+\sqrt{x^2+x}-x^2}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
\(\Leftrightarrow\frac{-x^2+2x}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
\(\Leftrightarrow\frac{-x\left(x+2\right)}{\left(x+\sqrt{x^2+x}\right)\left(x-\sqrt{x^2+x}\right)}=0\)
Dễ thấy: x=0 ko là nghiệm nên \(x+2=0\Rightarrow x=-2\)
c)\(\sqrt{2x+4}-2\sqrt{2-x}=\frac{12x-8}{\sqrt{9x^2+16}}\)
\(\Leftrightarrow\frac{\left(2x+4\right)-4\left(2-x\right)}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}\)
\(\Leftrightarrow\frac{2\left(3x-2\right)}{\sqrt{2x+4}+2\sqrt{2-x}}=\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}\)
\(\Leftrightarrow\frac{2\left(3x-2\right)}{\sqrt{2x+4}+2\sqrt{2-x}}-\frac{4\left(3x-2\right)}{\sqrt{9x^2+16}}=0\)
\(\Leftrightarrow\left(3x-2\right)\left(\frac{2}{\sqrt{2x+4}+2\sqrt{2-x}}-\frac{4}{\sqrt{9x^2+16}}\right)=0\)
\(\Leftrightarrow x=\frac{2}{3}\)
`(1+\frac{1}{x})^3.(1+x)^3=16`
`<=>(2+x+\frac{1}{x})^3=16`
`<=>2+x+\frac{1}{x}=\root{3}{16}`
`<=>x+\frac{1}{x}-(\root{3}{16}-2)=0`
`=>x^2-(\root{3}{16}-2)+1=0`
`<=>x^2-2.x.\frac{\root{3}{16}-2}{2}+\frac{\root{\frac{3}{2}}{16}-2}{4}+(1-\frac{\root{\frac{3}{2}}{16}-2}{4})=0`
`<=>(x-\frac{\root{3}{16}-2}{2})^2+(1-\frac{\root{\frac{3}{2}}{16}-2}{4})>0` (vô nghiệm)
Vậy phương trình vô nghiệm
ĐKXĐ: ...
\(\dfrac{\left(1+x\right)^3\left(1+x\right)\left(x^2-x+1\right)}{x^3}=16\)
\(\Leftrightarrow\left(\dfrac{\left(1+x\right)^2}{x}\right)^2\left(\dfrac{x^2-x+1}{x}\right)=16\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}+2\right)^2\left(x+\dfrac{1}{x}-1\right)=16\)
Đặt \(x+\dfrac{1}{x}+2=t\)
\(\Rightarrow t^2\left(t-3\right)=16\Rightarrow t^3-3t^2-16=0\)
\(\Leftrightarrow\left(t-4\right)\left(t^2+t+4\right)=0\)
\(\Leftrightarrow t=4\Rightarrow x+\dfrac{1}{x}+2=4\)
\(\Rightarrow x^2-2x+1=0\)
\(\Rightarrow x=1\)