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1) đặt đk rùi bình phương 2 vế là ok
2) \(pt\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+2}}{x-x-2}+\frac{\sqrt{x+2}-\sqrt{x+4}}{x+2-x-4}+\frac{\sqrt{x+4}-\sqrt{x+6}}{x+4-x-6}=\frac{\sqrt{10}}{2}-1\)(ĐKXĐ : \(x\ge0\))
<=> \(\frac{\sqrt{x}-\sqrt{x+6}}{-2}=\frac{\sqrt{10}}{2}-1\)
<=> \(\frac{\sqrt{x+6}-\sqrt{x}}{2}=\frac{\sqrt{10}-2}{2}\)
<=> \(\sqrt{x+6}-\sqrt{x}=\sqrt{10}-2\)
<=> \(\sqrt{x+6}+2=\sqrt{10}+\sqrt{x}\)
đến đây bình phương 2 vế rùi giải bình thường nhé
\(x-\frac{1}{x}+\sqrt{x-\frac{1}{x}}+\frac{1}{4}=2x-\frac{5}{x}+\sqrt{2x-\frac{5}{x}}+\frac{1}{4}\)
=>\(\left(\sqrt{x-\frac{1}{x}}+\frac{1}{2}\right)^2=\left(\sqrt{2x-\frac{5}{x}}+\frac{1}{2}\right)^2\)
dễ suy ra đc:\(\sqrt{x-\frac{1}{x}}=\sqrt{2x-\frac{5}{x}}\)
từ đây=>x=?
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
\(x^2+2x\sqrt{x-\frac{1}{x}}+3x+1=0\)
ĐK: \(x-\frac{1}{x}\ge0\)
\(+x=0\text{ thì }pt\text{ thành }0=1\text{ (vô lí)}\)
\(+\text{Xét }x\ne0;\text{ }pt\Leftrightarrow x+2\sqrt{x-\frac{1}{x}}=3+\frac{1}{x}\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)+2\sqrt{x-\frac{1}{x}}-3=0\)
Đặt \(\sqrt{x-\frac{1}{x}}=t\ge0;\text{ }pt\text{ thành }t^2+2t-3=0\)
\(c\text{) }x^2+\sqrt[3]{x^4-x^2}=2x+1\)
\(\Leftrightarrow\left(x^2-1\right)-2x+\sqrt[3]{x^2\left(x^2-1\right)}=0\)
Đặt \(\sqrt[3]{x^2-1}=a;\text{ }\sqrt[3]{x}=b\)
\(pt\text{ trở thành }a^3-2b^3+ab^2=0\Leftrightarrow\left(a-b\right)\left(a^2+ab+2b^2\right)=0\)
\(\Leftrightarrow a=b\text{ hoặc }\left(a+\frac{b}{2}\right)^2+\frac{7b^2}{4}=0\)
\(a=b\text{ thì }\sqrt[3]{x^2-1}=\sqrt[3]{x}\Leftrightarrow x^2-1=x\Leftrightarrow x=\frac{1\pm\sqrt{5}}{2}\)
\(\left(a+\frac{b}{2}\right)^2+\frac{7b^2}{4}=0\Leftrightarrow b=0\text{ và }a+\frac{b}{2}=0\Leftrightarrow a=b=0\)
Suy ra \(\sqrt[3]{x^2-1}=0\text{ và }\sqrt[3]{x}=0\Leftrightarrow x=0\text{ và }x^2-1=0\text{ (vô nghiệm)}\)
\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)
\(\Leftrightarrow\sqrt{x-\frac{1}{x}}-\sqrt{2x-\frac{5}{x}}=x-\frac{4}{x}\)
\(\Leftrightarrow\frac{\frac{4}{x}-x}{\sqrt{x-\frac{1}{x}}+\sqrt{2x-\frac{5}{x}}}=x-\frac{4}{x}\)
\(\Leftrightarrow\left(\frac{4}{x}-x\right).\left(\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{2x-\frac{5}{x}}}+1\right)=0\)
\(\frac{1}{\sqrt{x-\frac{1}{x}}+\sqrt{2x-\frac{5}{x}}}+1>0\Rightarrow\frac{4}{x}-x=0\Rightarrow x=2;x=-2\)
Thử lại, ta có nghiệm \(x=2\) thỏa mãn.
Vậy, \(x=2\).
tui mới hok lớp 6 thui