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Ta có : \(x^3=\left(9+4\sqrt{5}\right)+\left(9-4\sqrt{5}\right)+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\)
\(\left(\sqrt[3]{9-4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(\Leftrightarrow x^3=18+30\)
\(\Leftrightarrow x^3-3x-18x=0\)
Ta có :
\(x^3=\left(9+4\sqrt{5}\right)+\left(9-4\sqrt{5}\right)+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\)\(\left(\sqrt[3]{9-4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(\Leftrightarrow x^3=18+3x\)
\(\Leftrightarrow x^3-3x-18x=0\)
nghiệm a si đa quá ._.
\(\sqrt{2}x^2+x-1=0\)
\(\Delta=1^2-\left(4\sqrt{2}-1\right)=\sqrt{32}+1\)
\(\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{\sqrt{32}+1}}{2\sqrt{2}}\).....
\(\sqrt{x-3}+\sqrt{5-x}=x^2-8x+18.\)
ĐK: \(3\le x\le5\)
\(PT\Leftrightarrow\sqrt{x-3}-1+\sqrt{5-x}-1=x^2-8x+18-2\)
\(\Leftrightarrow\frac{x-3-1}{\sqrt{x-3}-1}+\frac{5-x-1}{\sqrt{5-x}+1}=\left(x-4\right)^2\)
\(\Leftrightarrow\frac{x-4}{\sqrt{x-3}+1}+\frac{4-x}{\sqrt{5-x}+1}=\left(x-4\right)^2\)
\(\Leftrightarrow\left(x-4\right)^2-\frac{x-4}{\sqrt{x-3}+1}+\frac{x-4}{\sqrt{5-x}+1}=0\)
\(\Leftrightarrow\left(x-4\right).\left(x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\left(TM\right)\\x-4-\frac{1}{\sqrt{x-3}-1}+\frac{1}{\sqrt{5-x}+1}=0\end{cases}}\) (Vô nghiệm)
Vậy pt có nghiệm x-4
chứng minh: \(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\) là nhiệm của phưng trình \(x^3-3x-18=0\)
Ta có :
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow x^3=\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)^3\)
\(=18+3\sqrt[3]{\left(9+4\sqrt{5}\right)^2\left(9-4\sqrt{5}\right)}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)^2}\)
\(=18+3\sqrt{\left(9+4\sqrt{5}\right)\left(9^2-4\sqrt{5}^2\right)}+3\sqrt{\left(9-4\sqrt{5}\right)\left(9^2-4\sqrt{5}^2\right)}\)
\(=18+3\sqrt[3]{9+4\sqrt{5}}+3\sqrt[3]{9-4\sqrt{5}}=18+3x\)
⇔ x3 - 3x - 18 = 0 ⇒ đpcm
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}=m\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(2-\sqrt{x-4}\right)^2}=m\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|+\left|2-\sqrt{x-4}\right|=m\)
mà \(\left|\sqrt{x-4}+2\right|+\left|2-\sqrt{x-4}\right|\)
\(\ge\left|\sqrt{x-4}+2+2-\sqrt{x-4}\right|=4\)
\(\Rightarrow m\ge4\) thì pt trên có no
Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?
Câu 1:ĐK \(x\ge\frac{1}{2}\)
\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)
Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)
=> \(x=1\)(TM ĐKXĐ)
Vậy x=1
bấm máy tính
được thế đã tốt =))) phải giải hẳn ra cơ