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Ta có: \(x^5+x+1=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Lại có: \(x^5+x+1=0\)
\(\Rightarrow\left(x^2+x+1\right)\left(x^3-x^2+1\right)=0\)
\(\Rightarrow x^3-x^2+1=0\) (vì \(x^2+x+1>0\))
Đặt \(m=\sqrt[3]{\frac{25+\sqrt{621}}{2}}-\sqrt[3]{\frac{25-\sqrt{621}}{2}}\)
\(\Rightarrow m^3=25+3\sqrt[3]{\frac{25+\sqrt{621}}{2}.\frac{25-\sqrt{621}}{2}}.m\)
\(m^3=25+3m\) (1)
\(n=\frac{1}{3}\left(1-m\right)\Leftrightarrow m=1-3n\) (2)
Từ (1) và (2) suy ra:
\(\left(1-n\right)^3=25+\left(1-3n\right)\)
\(\Leftrightarrow1-9n+27n^2-27n^3=25+3-9n\)
\(\Leftrightarrow27n^3-27n^2+27=0\)
\(\Leftrightarrow n^3-n^2+1=0\)
Vậy \(x=n\) là nghiệm của phương trình \(x^3-x^2+1=0\)
\(\Rightarrow x=n\) cũng là nghiệm của phương trình \(x^5+x+1=0\)
* Nếu \(x>n\) thì \(x^5+x+1>n^5+n+1=0\)
\(\Rightarrow\) Với mọi x > n ko là nghiệm của phương trình.
* Nếu \(x< n\) thì \(x^5+x+1< n^5+n+1=0\)
\(\Rightarrow\) Với mọi x < n ko là nghiệm của phương trình.
(Chúc bạn học giỏi và tíck cho mìk vs nhoa!)
ta có \(3x=1-\sqrt[3]{\frac{25+\sqrt{621}}{2}}-\sqrt[3]{\frac{25-\sqrt{621}}{2}}\)
<=> \(1-3x=\sqrt[3]{\frac{25+\sqrt{621}}{2}}+\sqrt[3]{\frac{25-\sqrt{621}}{2}}\)
<=> \(\left(1-3x\right)^3=\left(\sqrt[3]{\frac{25+\sqrt{621}}{2}}+\sqrt[3]{\frac{25-\sqrt{621}}{2}}\right)^3\)
<=> \(1-9x+27x^2-27x^3=\frac{25+\sqrt{621}}{2}+\frac{25-\sqrt{621}}{2}+3\left(\frac{25+\sqrt{621}}{2}\cdot\frac{25-\sqrt{621}}{2}\right)\left(1-3x\right)\)( vì \(\sqrt[3]{\frac{25+\sqrt{621}}{2}}+\sqrt[3]{\frac{25-\sqrt{621}}{2}}=1-3x\)....phía trên 2 dòng )
<=> \(1-9x+27x^2-27x^3=25+3\cdot1\cdot\left(1-3x\right)\)
<=> \(1-9x+27x^2-27x^3=25+3-9x\)
<=> \(1-9x+27x^2-27x^3=28-9x\)
<=> \(27x^3-27x^2+27=0\)
<=.\(27\left(x^3-x^2+1\right)=0\)
<=. \(x^3-x^2+1=0\)
pt \(x^3-x^2+1=0\) và pt \(x^5+x+1=0\) đều có nghiệm chung
vậy đccm
Lời giải:
Đặt \(\sqrt[3]{4-\sqrt{15}}=m\)
Khi đó \(a=\frac{1}{m}+m\Rightarrow a^3-3a=\frac{1}{m^3}+\frac{3}{m}+3m+m^3-3(\frac{1}{m}+m)\)
\(=\frac{1}{m^3}+m^3=\frac{1}{4-\sqrt{15}}+4-\sqrt{15}=4+\sqrt{15}+4-\sqrt{15}=8(*)\)
Đặt \(\sqrt[3]{\frac{25+\sqrt{621}}{2}}=n; \sqrt[3]{\frac{25-\sqrt{621}}{2}}=p\)
\(\Rightarrow n^3+p^3=25; np=\sqrt[3]{\frac{25^2-621}{4}}=1\)
\(\Rightarrow (n+p)^3=n^3+p^3+3np(n+p)=25+3(n+p)\)
Do đó:
\(b^3-b^2=\frac{1}{27}(1-n-p)^3-\frac{1}{9}(1-n-p)^2\)
\(=\frac{1}{27}[1-3(n+p)+3(n+p)^2-(n+p)^3]-\frac{1}{9}[1-2(n+p)+(n+p)^2]\)
\(=\frac{-2}{27}+\frac{n+p}{9}-\frac{(n+p)^3}{27}\)
\(=\frac{-2}{27}+\frac{n+p}{9}-\frac{25+3(n+p)}{27}=-1(**)\)
Từ \((*);(**)\Rightarrow a^3+b^3-b^2-3a+100=8+(-1)+100=107\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)
\(=\dfrac{5\sqrt{x}-15}{3x-59}\)
2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
\(1,ĐKx\ge5\)
\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)
\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)
\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)
\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)
2a,ĐK \(x\ge0;x\ne9\)
,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)
\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(M=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
đK: \(x\ge0;x\ne25;x\ne9\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right]:\left[\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right]\)
\(=\left[\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right]:\dfrac{25-x-\left(x-9\right)+\left(x-25\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{-\sqrt{x}-3}{\sqrt{x}+5}\)
\(=\dfrac{-5}{\sqrt{x}+5}:\dfrac{\sqrt{x}+5}{-\left(\sqrt{x}+3\right)}=\dfrac{5}{\sqrt{x}+3}\)
Ta có: \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5}{\sqrt{x}+3}\)
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)