Giải: \(x+4-\sqrt{14x-1}=\frac{\sqrt{10x-9}-1}{x}\)
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Câu hỏi của Nguyễn Thùy Chi - Toán lớp 10 | Học trực tuyến
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b, ĐK \(x\ge-4\)
PT
<=> \(\left(x-\sqrt{x+4}\right)+\left(\sqrt{2x^2-10x+17}-2x+3\right)=0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}+\frac{-2x^2+2x+8}{\sqrt{2x^2-10x+17}+2x-3}=0\)với \(x+\sqrt{x+4}\ne0\)
<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}-\frac{2\left(x^2-x-4\right)}{\sqrt{2x^2-10x+17}+2x-3}=0\)
<=> \(\orbr{\begin{cases}x^2-x-4=0\\\frac{1}{x+\sqrt{x+4}}-\frac{2}{\sqrt{2x^2-10x+17}+2x-3}=0\left(2\right)\end{cases}}\)
Giải (2)
=> \(2x+2\sqrt{x+4}=2x-3+\sqrt{2x^2-10x+17}\)
<=> \(\sqrt{2x^2-10x+17}=2\sqrt{x+4}+3\)
<=> \(2x^2-10x+17=4\left(x+4\right)+9+12\sqrt{x+4}\)
<=> \(x^2-7x-4=6\sqrt{x+4}\)
<=> \(\left(x-6\right)^2+5x-40=6\sqrt{6\left(x-6\right)-5x+40}\)
Đặt x-6=a;\(\sqrt{6\left(x-6\right)-5x+40}=b\)
=> \(\hept{\begin{cases}a^2+5x-40=6b\\b^2+5x-40=6a\end{cases}}\)
=> \(a^2-b^2+6\left(a-b\right)=0\)
<=> \(\orbr{\begin{cases}a=b\\a+b+6=0\end{cases}}\)
+ a=b
=> \(x-6=\sqrt{x+4}\)
=> \(\hept{\begin{cases}x\ge6\\x^2-13x+32=0\end{cases}}\)=> \(x=\frac{13+\sqrt{41}}{2}\)
+ a+b+6=0
=> \(x+\sqrt{x+4}=0\)(loại)
Vậy \(S=\left\{\frac{13+\sqrt{41}}{2};\frac{1+\sqrt{17}}{2}\right\}\)
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ĐKXĐ: ...
\(\Leftrightarrow\left(x-1\right)\left(x+3-\sqrt{14x-15}\right)-\sqrt{10x-19}+1=0\)
\(\Leftrightarrow x^2+2x-2-\left(x-1\right)\sqrt{14x-15}-\sqrt{10x-19}=0\)
\(\Leftrightarrow x-\sqrt{10x-19}+\left(x-1\right)\left(x+2\right)-\left(x-1\right)\sqrt{14x-15}=0\)
\(\Leftrightarrow\frac{x^2-10x+19}{x+\sqrt{10x-19}}+\left(x-1\right)\left(\frac{x^2-10x+19}{x+2+\sqrt{14x+15}}\right)=0\)
\(\Leftrightarrow\left(x^2-10x+19\right)\left(\frac{1}{x+\sqrt{10x-19}}+\frac{x-1}{x+2+\sqrt{14x+15}}\right)=0\)
\(\Leftrightarrow x^2-10x+19=0\)
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a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
Một vài mẹo sử dụng casio FX-570VN.pdf - Google Drive
Bạn kéo xuống mục số 4, khoảng trang 36
Linh Chi
ĐKXĐ: \(x\ge\frac{9}{10}\)
\(\Leftrightarrow x^2+4x+1-x\sqrt{14x-1}-\sqrt{10x-9}=0\)
\(\Leftrightarrow x\left(x+3-\sqrt{14x-1}\right)+x+1-\sqrt{10x-9}=0\)
\(\Leftrightarrow\frac{x\left[\left(x+3\right)^2-\left(14x-1\right)\right]}{x+3+\sqrt{14x-1}}+\frac{\left(x+1\right)^2-\left(10x-9\right)}{x+1+\sqrt{10x-9}}=0\)
\(\Leftrightarrow\frac{x\left(x^2-8x+10\right)}{x+3+\sqrt{14x-1}}+\frac{x^2-8x+10}{x+1+\sqrt{10x-9}}=0\)
\(\Leftrightarrow\left(x^2-8x+10\right)\left(\frac{x}{x+3+\sqrt{14x-1}}+\frac{1}{x+1+\sqrt{10x-9}}\right)=0\)
\(\Leftrightarrow x^2-8x+10=0\) (casio)