Giải các phương trình:
a) \(\dfrac{1}{x-1+\sqrt{x^2-2x+3}}+\dfrac{1}{x-1-\sqrt{x^2-2x+3}}=1;\)
b) \(\dfrac{3}{x+\sqrt{9-x^2}}-\dfrac{2}{3-\sqrt{9-x^2}}=\dfrac{1}{x}.\)
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Những câu hỏi liên quan
27 tháng 7 2023
1) \(\dfrac{x+2\sqrt[]{x}}{\sqrt[]{x}-1}=8\left(1\right)\)
Điều kiện \(\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+2\sqrt[]{x}=8\left(\sqrt[]{x}-1\right)\)
\(\Leftrightarrow x-6\sqrt[]{x}+8=0\left(2\right)\)
Đặt \(t^2=x\Leftrightarrow t=\sqrt[]{x}\)
\(\left(2\right)\Leftrightarrow t^2-6t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x}=2\\\sqrt[]{x}=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\) (thỏa điều kiện)
2) \(\sqrt[]{\dfrac{2x-3}{x-1}}=2\left(1\right)\)
Điều kiện \(\dfrac{2x-3}{x-1}\ge0\Leftrightarrow\left[{}\begin{matrix}x< 1\\x\ge\dfrac{3}{2}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
\(\Leftrightarrow2x-3=4\left(x-1\right)\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\) (thỏa điều kiện)
29 tháng 8 2021
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
KK
2 tháng 4 2017
a) \(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x+\dfrac{x+5}{x+3}=\dfrac{x+5}{x+3}\)
\(\Leftrightarrow x=0\)
b) \(2x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x\left(x-1\right)+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow x+\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}-x\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x\left(x-1\right)}{x-1}\)
\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x^2+x}{x-1}\)
\(\Leftrightarrow x^2-x+3=3x-x^2+x\) ( điều kiện \(x\ne1\) )
\(\Leftrightarrow2x^2-5x+3=0\)
\(\Delta=b^2-4ac\)
\(\Delta=1\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3}{2}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=1\left(loại\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{2}\)
c) \(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\)
\(\Leftrightarrow x^2-4x-2=\sqrt{\left(x-2\right)^2}\) ( điều kiện \(x>2\) )
\(\Leftrightarrow x^2-4x-2=x-2\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\end{matrix}\right.\)
Vậy \(x=5\)
d) \(\dfrac{2x^2-x-3}{\sqrt{2x-3}}=\sqrt{2x-3}\)
\(\Leftrightarrow2x^2-x-3=\sqrt{\left(2x-3\right)^2}\) ( điều kiện \(x>\dfrac{3}{2}\) )
\(\Leftrightarrow2x^2-x-3=2x-3\)
\(\Leftrightarrow2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)
Vậy phương trình vô nghiệm
H9
HT.Phong (9A5)
CTVHS
5 tháng 9 2023
1) \(\sqrt{x^2+1}=\sqrt{5}\)
\(\Leftrightarrow x^2+1=5\)
\(\Leftrightarrow x^2=5-1\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2=2^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=3+1\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=\dfrac{4}{2}\)
\(\Leftrightarrow x=2\left(tm\right)\)
3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))
\(\Leftrightarrow43-x=\left(x-1\right)^2\)
\(\Leftrightarrow x^2-2x+1=43-x\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)
4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))
\(\Leftrightarrow\sqrt{4x-3}=x-2\)
\(\Leftrightarrow4x-3=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-4x+4=4x-3\)
\(\Leftrightarrow x^2-8x+7=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{x}+3=2\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1^2\)
\(\Leftrightarrow x=1\left(tm\right)\)
GH
5 tháng 9 2023
1)
\(\sqrt{x^2+1}=\sqrt{5}\\ \Leftrightarrow x^2+1=5\\ \Leftrightarrow x^2=5-1=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy PT có nghiệm `x=2` hoặc `x=-2`
2)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)
Vậy PT có nghiệm `x=2`
3)
\(ĐKXĐ:x\le43\)
PT trở thành:
\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm `x=-6` hoặc `x=7`
4)
ĐKXĐ: \(x\ge\dfrac{3}{4}\)
PT trở thành:
\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)
Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)
5)
ĐKXĐ: \(x\ge0\)
PT trở thành:
\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\)
Khi đó:
(1)\(\Leftrightarrow3t^2+8t+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)
Vậy PT vô nghiệm.
28 tháng 6 2021
a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)
`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`
`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\)
`b)sqrt{x-3}/sqrt{2x+1}=2`
ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)
`<=>x>=3`
`pt<=>sqrt{x-3}=2sqrt{2x+1}`
`<=>x-3=8x+4`
`<=>7x=7`
`<=>x=1(l)`
`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`
`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`
`<=>|x-1|+|x-2|=3`
`**x>=2`
`pt<=>x-1+x-2=3`
`<=>2x=6`
`<=>x=3(tm)`
`**x<=1`
`pt<=>1-x+2-x=3`
`<=>3-x=3`
`<=>x=0(tm)`
`**1<=x<=2`
`pt<=>x-1+2-x=3`
`<=>=-1=3` vô lý
Vậy `S={0,3}`
a) \(\frac{1}{x-1+\sqrt{x^2-2x+3}}+\frac{1}{x-1-\sqrt{x^2-2x+3}}=1\)
ĐKXĐ : \(x\inℝ\)
\(\Leftrightarrow\frac{x-1-\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}+\frac{x-1+\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}=\frac{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}\)
\(\Rightarrow2x-2=\left[\left(x-1\right)+\left(\sqrt{x^2-2x+3}\right)\right]\left[\left(x-1\right)-\left(\sqrt{x^2-2x+3}\right)\right]\)
\(\Leftrightarrow2x-2=\left(x-1\right)^2-\left(\sqrt{x^2-2x+3}\right)^2\)
\(\Leftrightarrow2x-2=x^2-2x+1-\left(x^2-2x+3\right)\)
\(\Leftrightarrow2x-2=x^2-2x+1-x^2+2x-3\)
\(\Leftrightarrow2x-2=-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy phương trình có nghiệm duy nhất x = 0