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Giair phương trình \(\frac{x+3}{3x}=\sqrt{\frac{1}{9}+\frac{1}{x}\sqrt{\frac{4}{9}+\frac{2}{x^2}}}\)
1.
đặt \(a=\sqrt{2+\sqrt{x}}\),\(b=\sqrt{2-\sqrt{x}}\)\(\left(a,b>0\right)\)
có \(a^2+b^2=4\)
pt thành \(\frac{a^2}{\sqrt{2}+a}+\frac{b^2}{\sqrt{2}-b}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}\left(a^2+b^2\right)-ab\left(a-b\right)=\sqrt{2}\left(\sqrt{2}+a\right)\left(\sqrt{2}-b\right)\)
\(\Leftrightarrow2\sqrt{2}+\sqrt{2}ab-ab\left(a-b\right)-2\left(a-b\right)=0\)
\(\Leftrightarrow\left(ab+2\right)\left(\sqrt{2}-a+b\right)=0\)
vì a,b>o nên \(a-b=\sqrt{2}\)
\(\Rightarrow\sqrt{2+\sqrt{x}}-\sqrt{2-\sqrt{x}}=\sqrt{2}\)
Bình phương 2 vế:
\(4-2\sqrt{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=2\)
\(\Leftrightarrow\sqrt{4-x}=1\)
\(\Rightarrow x=3\)
Máy tính Casio giải ra x = 2
Còn nghiệm nào nữa không thì không biết
..
không ghi lại đề nha
\(\Leftrightarrow\sqrt{\frac{x^4-7}{x^2}}+\sqrt{\frac{x^3-7}{x^2}}=x\) ( * )
ĐKXĐ : \(\hept{\begin{cases}x^4-7\ge0\\x^3-7\ge0\\x^2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^4\ge7\\x^3\ge7\\x\ne0\end{cases}}\)
( * ) \(\Rightarrow\frac{\sqrt{x^4-7}}{\sqrt{x^2}}+\frac{\sqrt{x^3-7}}{\sqrt{x^2}}=x\)
\(\Leftrightarrow\frac{\sqrt{x^4-7}+\sqrt{x^3-7}}{x}=x\)
\(\Leftrightarrow\sqrt{x^4-7}+\sqrt{x^3-7}=x^2\)
\(\Leftrightarrow\left(\sqrt{x^4-7}+\sqrt{x^3-7}\right)^2=x^4\)
\(\Leftrightarrow\left(x^4-7\right)+2\sqrt{\left(x^4-7\right)\left(x^3-7\right)}+\left(x^3-7\right)=x^4\)
\(\Leftrightarrow2\sqrt{x^7-7x^4-7x^3+49}=x^4-x^4+7-x^3+7\)
\(\Leftrightarrow\left(2\sqrt{x^7-7x^4-7x^3+49}\right)^2=\left(14-x^3\right)^2\)
\(\Leftrightarrow4\left(x^7-7x^4-7x^3+49\right)=196-28x^3+x^6\)
\(\Leftrightarrow4x^7-28x^4-28x^3+196=196-28x^3+x^6\)
\(\Leftrightarrow4x^7-x^6-28x^4=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\left(lo\text{ại}\right)\\x=2\left(nh\text{ậ}n\right)\end{cases}}\)
Vậy x = 2
b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)
\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)
Pt trong ngoặc VN suy ra x=2
a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)
\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)
\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)
\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)
pt trong căn vô nghiệm
suy ra x=1; x=-1
Nâng cao và phát triển toán 9 Vũ Hữu Bình tập 2 bài 318a trang 51 :)