Giải phương trình sau
\(4\sqrt{1-x}=x+6-3\sqrt{1-x^2}+5\sqrt{1+x}\)
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15 tháng 7 2023
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
15 tháng 9 2021
\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)
\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)
\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)
NH
0
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.
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
1.
ĐKXĐ: \(x< 5\)
\(\Leftrightarrow\sqrt{\dfrac{42}{5-x}}-3+\sqrt{\dfrac{60}{7-x}}-3=0\)
\(\Leftrightarrow\dfrac{\dfrac{42}{5-x}-9}{\sqrt{\dfrac{42}{5-x}}+3}+\dfrac{\dfrac{60}{7-x}-9}{\sqrt{\dfrac{60}{7-x}}+3}=0\)
\(\Leftrightarrow\dfrac{9x-3}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{9x-3}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}=0\)
\(\Leftrightarrow\left(9x-3\right)\left(\dfrac{1}{\left(5-x\right)\left(\sqrt{\dfrac{42}{5-x}}+3\right)}+\dfrac{1}{\left(7-x\right)\left(\sqrt{\dfrac{60}{7-x}}+3\right)}\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2021
b.
ĐKXĐ: \(x\ge2\)
\(\sqrt{\left(x-2\right)\left(x-1\right)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}-\sqrt{x-2}+\sqrt{x+3}-\sqrt{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)-\sqrt{x+3}\left(\sqrt{x-1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x-2}-\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{x-2}-\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-2=x+3\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=2\)
23 tháng 10 2021
\(1,PT\Leftrightarrow2x-1=5\Leftrightarrow x=3\\ 2,\Leftrightarrow x-5=9\Leftrightarrow x=14\\ 3,ĐK:x\ge1\\ PT\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\Leftrightarrow x=50\left(tm\right)\\ 4,\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)
1 tháng 9 2023
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 9 2023
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)