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TN
4 tháng 7 2017
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
TN
4 tháng 7 2017
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
TN
30 tháng 7 2017
a)Đk:\(0\le x\le1\)
\(\sqrt{x}+\sqrt{1-x}+\sqrt{x+1}=2\)
\(pt\Leftrightarrow\sqrt{x}+\sqrt{1-x}-1+\sqrt{x+1}-1=0\)
\(\Leftrightarrow\sqrt{x}+\frac{1-x-1}{\sqrt{1-x}+1}+\frac{x+1-1}{\sqrt{x+1}-1}=0\)
\(\Leftrightarrow\frac{x}{\sqrt{x}}-\frac{x}{\sqrt{1-x}+1}+\frac{x}{\sqrt{x+1}-1}=0\)
\(\Leftrightarrow x\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{1-x}+1}+\frac{1}{\sqrt{x+1}-1}\right)=0\)
\(\Rightarrow x=0\)
b)\(\frac{3x+3}{\sqrt{x}}=4+\frac{x+1}{\sqrt{x^2-x+1}}\)
\(pt\Leftrightarrow\frac{3x+3}{\sqrt{x}}-6=\frac{x+1}{\sqrt{x^2-x+1}}-2\)
\(\Leftrightarrow\frac{3x+3-6\sqrt{x}}{\sqrt{x}}=\frac{x+1-2\sqrt{x^2-x+1}}{\sqrt{x^2-x+1}}\)
\(\Leftrightarrow\frac{\frac{\left(3x+3\right)^2-36x}{3x+3+6\sqrt{x}}}{\sqrt{x}}=\frac{\frac{\left(x+1\right)^2-4\left(x^2-x+1\right)}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\)
\(\Leftrightarrow\frac{\frac{9x^2+18x+9-36x}{3x+3+6\sqrt{x}}}{\sqrt{x}}=\frac{\frac{x^2+2x+1-4x^2+4x-4}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\)
\(\Leftrightarrow\frac{\frac{9x^2-18x+9}{3x+3+6\sqrt{x}}}{\sqrt{x}}-\frac{\frac{-3x^2+6x-3}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\frac{\frac{9\left(x-1\right)^2}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{3\left(x-1\right)^2}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow3\left(x-1\right)^2\left(\frac{\frac{3}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{1}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}\right)=0\)
Dêx thấy: \(\frac{\frac{3}{3x+3+6\sqrt{x}}}{\sqrt{x}}+\frac{\frac{1}{x+1+2\sqrt{x^2-x+1}}}{\sqrt{x^2-x+1}}>0\forall....\)
\(\Rightarrow3\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
24 tháng 9 2016
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
13 tháng 11 2016
xy - 2x - 3y + 1 = 0
<=> x(y - 2) = 3y - 1
<=> \(=\frac{3y-1}{y-2}=3+\frac{5}{y-2}\)
Để x nguyên thì (y - 2) phải là ước của 5 hay
(y - 2) = (1, 5, - 1, - 5)
Giải tiếp sẽ ra
19 tháng 7 2017
a)
\(\frac{\sqrt{5x-4}}{\sqrt{x+1}}=2\Rightarrow2\sqrt{x+1}=\sqrt{5x-4}\)
\(\Leftrightarrow4\left(x+1\right)=5x-4\)(bình phương 2 vế)
\(\Leftrightarrow4x+4=5x-4\)
\(\Leftrightarrow x=8\)
b)
\(\sqrt{\frac{2x-1}{x+1}}=2\Leftrightarrow\frac{\sqrt{2x-1}}{\sqrt{x+1}}=2\)
\(\Rightarrow2\left(\sqrt{2x-1}\right)=\sqrt{x+1}\)(tích chéo)
\(\Leftrightarrow4\left(2x-1\right)=x+1\)
\(\Leftrightarrow8x-4=x+1\)
\(\Leftrightarrow x=\frac{5}{7}\)
17 tháng 11 2019
\(\frac{\sqrt{5x-4}}{\sqrt{x+1}}=2\)
\(\Leftrightarrow\frac{5x-4}{x+1}=4\)
\(\Leftrightarrow5x-4=4\left(x+1\right)\)
\(\Leftrightarrow5x-4=4x+4\)
\(\Leftrightarrow5x-4x=4+4\)
\(\Leftrightarrow x=8\)
\(\Rightarrow x=8\)
30 tháng 3 2020
\(\sqrt{\sqrt{2}-1-x}+\sqrt[4]{x}=\frac{1}{\sqrt[4]{2}}\)
ĐKXĐ: Tự tìm nhé.
\(\left(\sqrt{\sqrt{2}-1-x};\sqrt[4]{x}\right)\rightarrow\left(b;a\right)\)
Phương trình <=> \(\hept{\begin{cases}a+b=\frac{1}{\sqrt[4]{2}}\\a^4+b^2=\sqrt{2}-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}b=\frac{1}{\sqrt[4]{2}}-a\\a^4+b^2=\sqrt{2}-1\left(2\right)\end{cases}}\)
(2) <=> \(a^4+a^2-\frac{2}{\sqrt[4]{2}}a+\frac{1}{\sqrt{2}}-\sqrt{2}+1=0\)
\(\Leftrightarrow\sqrt{2}a^4+\sqrt{2}a^2-2\sqrt[4]{2}a+\sqrt{2}-1=0\)
\(\Leftrightarrow\left(a^2-a+\frac{\sqrt{2}-\sqrt[4]{2}}{\sqrt{2}}\right)\left(\sqrt{2}a^2+\sqrt{2}a+2\sqrt{2}+\sqrt[4]{2}-\sqrt{2}\right)=0\)
\(\Leftrightarrow a^2-a+\frac{\sqrt{2}-\sqrt[4]{2}}{\sqrt{2}}=0\)( vì \(\Leftrightarrow\sqrt{2}a^2+\sqrt{2}a+2\sqrt{2}+\sqrt[4]{2}-\sqrt{2}>0\))
Tự làm tiếp nhé
30 tháng 3 2020
ĐK: \(x\ge\frac{1}{2}\)
\(\sqrt{\frac{x+7}{x+1}}+8=2x^2+\sqrt{2x-1}\)
\(\Leftrightarrow\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)+2\left(2-x\right)\left(2+x\right)=\left(\sqrt{2x-1}-\sqrt{3}\right)\)
\(\Leftrightarrow\frac{2\left(2-x\right)}{\sqrt{\left(x+7\right)\left(x+1\right)}+\sqrt{3}\left(x+1\right)}+2\left(2-x\right)\left(2+x\right)=\frac{2\left(x-2\right)}{\sqrt{2x-1}+\sqrt{3}}\)
\(\Leftrightarrow\frac{2\left(2-x\right)}{\sqrt{\left(x+7\right)\left(x+1\right)}+\sqrt{3}\left(x+1\right)}+2\left(2-x\right)\left(2+x\right)+\frac{2\left(2-x\right)}{\sqrt{2x-1}+\sqrt{3}}=0\)
\(\Leftrightarrow\left(2-x\right)\left[\frac{2}{\sqrt{\left(x+7\right)\left(x+1\right)}+\sqrt{3}\left(x+1\right)}+2\sqrt{2+x}+\frac{2}{\sqrt{2x-1}+\sqrt{3}}\right]=0\)
\(\Leftrightarrow x=2\)( \(\frac{2}{\sqrt{\left(x+7\right)\left(x+1\right)}+\sqrt{3}\left(x+1\right)}+2\left(2+x\right)+\frac{2}{\sqrt{2x-1}+\sqrt{3}}>0\))
KL:...
16 tháng 7 2019
\(\sqrt{x+2\sqrt{x-1}=2}\)
\(\Leftrightarrow\sqrt{x-1+2.\sqrt{x-1}.\sqrt{1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(x-1+1\right)^2}=2\)
\(\Leftrightarrow\sqrt{x^2}=2\)
\(\Leftrightarrow x=2\)
Các câu kia lm tương tự........
7 tháng 1 2019
\(x^4+2x^3=4x+4\)
\(x^4+2x^3+x^2-x^2-4x-4=0\)
\(x^2\left(x^2+2x+1\right)-\left(x^2+4x+4\right)=0\)
\(\left[x\left(x+1\right)\right]^2-\left(x+2\right)^2=0\)
\(\left(x^2+x-x-2\right)\left(x^2+x+2\right)=0\)
\(\left(x^2-2\right)\left(x^2+x+2\right)=0\)
\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x^2+x+2\right)=0\)
tự làm nốt nhé~
SN
7 tháng 1 2019
\(b,\frac{1}{x^2}+\sqrt{x+2}=\frac{1}{x}+\sqrt{2x+1}\)(1)
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x+2\ge0\\2x+1\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne0\\x\ge\frac{-1}{2}\end{cases}}\)
\(\left(1\right)\Leftrightarrow1+x^2\sqrt{x+2}=x+x^2\sqrt{2x+1}\)
\(\Leftrightarrow\left(1-x\right)+x^2\frac{1-x}{\sqrt{x+2}+\sqrt{2x+1}}=0\)
\(\Leftrightarrow\left(1-x\right)\left(1+\frac{x^2}{\sqrt{x+2}+\sqrt{2x+1}}\right)=0\)(2)
Vì\(\hept{\begin{cases}x\ne0\\x\ge\frac{-1}{2}\end{cases}}\Rightarrow1+\frac{x^2}{\sqrt{x+2}+\sqrt{2x+1}}>0\)
Nên từ (2) => Phương trình đã cho có nghiệm x = 1 (TMĐKXĐ)
Điều kiện: \(x\ge0\)
\(\sqrt{x^2+x+\frac{1}{4}}=x\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=x\Leftrightarrow\left|x+\frac{1}{2}\right|=x\)
Với \(x\ge0\) thì \(x+\frac{1}{2}=x\Rightarrow0=\frac{1}{2}\) (vô lý)
Vậy \(x\in\varnothing\)