Giải phương trình:
\(\sqrt{14-x}-\sqrt{x-4}=\sqrt{x-1}\)
K
Khách
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26 tháng 9 2021
\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)
\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)
\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)
5 tháng 9 2023
a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 7 2021
a.
ĐKXĐ: \(x\ne-1\)
\(x^2+5x+2=\left(2x+2\right)\sqrt{x^2+x+2}\)
\(\Leftrightarrow\left(x^2+x+2\right)-2\left(x+1\right)\sqrt{x^2+x+2}+4x=0\)
Đặt \(\sqrt{x^2+x+2}=t>0\)
\(\Rightarrow t^2-2\left(x+1\right)t+4x=0\)
\(\Leftrightarrow t\left(t-2x\right)-2\left(t-2x\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\\\sqrt{x^2+x+2}=2x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\\x^2+x+2=4x^2\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 7 2021
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-5x+14-4\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
AH
Akai Haruma
Giáo viên
6 tháng 10 2021
Lời giải:
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow (x^2-6x+9)+[(x+1)-4\sqrt{x+1}+4]=0$
$\Leftrightarrow (x-3)^2+(\sqrt{x+1}-2)^2=0$
Vì $(x-3)^2; (\sqrt{x+1}-2)^2\geq 0$ với mọi $x\geq -1$
Do đó để tổng của chúng $=0$ thì:
$(x-3)^2=(\sqrt{x+1}-2)^2=0$
$\Leftrightarrow x=3$ (tm)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 10 2021
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\) \(\Leftrightarrow x=3\)
NK
1
H
21 tháng 10 2023
\(3\sqrt{x-2}-\sqrt{4x-8}+4\sqrt{\dfrac{9x-18}{4}}=14\left(x\ge0;x\ne2\right)\\ \Leftrightarrow3\sqrt{x-2}-\sqrt{4\left(x-2\right)}+4\cdot\dfrac{1}{2}\sqrt{9\left(x-2\right)}=14\\ \Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+6\sqrt{x-2}=14\\ \Leftrightarrow7\sqrt{x-2}=14\\ \Leftrightarrow\sqrt{x-2}=2\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\)
5 tháng 7 2021
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
HQ
Hà Quang Minh
Giáo viên
30 tháng 9 2023
a) \(\sqrt {2{x^2} - 14} = x - 1\quad \left( 1 \right)\)
ĐK: \(x - 1 \ge 0\,\, \Leftrightarrow \,\,x \ge 1.\)
\( \Rightarrow \) TXĐ: \(D = \left[ {1; + \infty } \right)\)
\(\begin{array}{l}\left( 1 \right)\,\, \Leftrightarrow \,\,{\left( {\sqrt {2{x^2} - 14} } \right)^2} = {\left( {x - 1} \right)^2}\\ \Leftrightarrow \,\,2{x^2} - 14 = {x^2} - 2x + 1\\ \Leftrightarrow \,\,{x^2} + 2x - 15 = 0\\ \Leftrightarrow \,\,\left[ {\begin{array}{*{20}{c}}{x = 3}\\{x = - 5}\end{array}} \right.\end{array}\)
Nhận thấy \(x = 3\) thỏa mãn điều kiện
Vậy nghiệm của phương trình \(\left( 1 \right)\) là: \(x = 3\)
b) \(\sqrt { - {x^2} - 5x + 2} = \sqrt {{x^2} - 2x - 3} \quad \left( 2 \right)\)
ĐK: \(\left\{ {\begin{array}{*{20}{c}}{ - {x^2} - 5x + 2 \ge 0}\\{{x^2} - 2x - 3 \ge 0}\end{array}} \right.\,\, \Leftrightarrow \,\,\frac{{ - 5 - \sqrt {33} }}{2} \le x \le - 1.\)
\( \Rightarrow \) TXĐ: \(D = \left[ {\frac{{ - 5 - \sqrt {33} }}{2}; - 1} \right].\)
\(\begin{array}{l}\left( 2 \right)\,\, \Leftrightarrow \,\,{\left( {\sqrt { - {x^2} - 5x + 2} } \right)^2} = {\left( {\sqrt {{x^2} - 2x - 3} } \right)^2}\\ \Leftrightarrow \,\, - {x^2} - 5x + 2 = {x^2} - 2x - 3\\ \Leftrightarrow \,\,2{x^2} + 3x - 5 = 0\\ \Leftrightarrow \,\,\left[ {\begin{array}{*{20}{c}}{x = 1}\\{x = - \frac{5}{2}}\end{array}} \right.\end{array}\)
Nhận thấy \(x = - \frac{5}{2}\) thỏa mãn điều kiện
Vậy nghiệm của phương trình \(\left( 2 \right)\) là: \(x = - \frac{5}{2}\)
28 tháng 10 2023
c: \(x^2-6\sqrt{x^2+5}+x=2\sqrt{x-1}-14\)
=>\(x^2-4-6\left(\sqrt{x^2+5}-3\right)+x-2-2\sqrt{x-1}+2=0\)
=>\(\left(x-2\right)\left(x+2\right)-6\cdot\dfrac{x^2+5-9}{\sqrt{x^2+5}+3}+\left(x-2\right)-2\cdot\dfrac{x-1-1}{\sqrt{x-1}+1}=0\)
=>\(\left(x-2\right)\left(x+2\right)-\dfrac{6}{\sqrt{x^2+5}+3}\cdot\left(x-2\right)\left(x+2\right)+\left(x-2\right)-2\cdot\dfrac{x-2}{\sqrt{x-1}+1}=0\)
=>\(\left(x-2\right)\left[\left(x+2\right)-\dfrac{6}{\sqrt{x^2+5}+3}\cdot\left(x+2\right)+1-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)
=>x-2=0
=>x=2
d: \(x^2-\sqrt{\left(x^2-8\right)\left(x-2\right)}+x=\sqrt{x^2-8}+\sqrt{x-2}+9\)
=>\(x^2-9-\sqrt{\left(x^2-8\right)\left(x-2\right)}+x-\sqrt{x^2-8}-\sqrt{x-2}=0\)
=>\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\sqrt{x^3-2x^2-8x+16}+x-3+1-\sqrt{x^2-8}+2-\sqrt{x-2}=0\)
=>\(\left(x-3\right)\left(x+3\right)+\left(x-3\right)-\sqrt{x^3-2x^2-8x+16}+1+\dfrac{1-x^2+8}{1+\sqrt{x^2-8}}+1-\sqrt{x-2}=0\)
=>\(\left(x-3\right)\left(x+4\right)-\dfrac{x^3-2x^2-8x+16-1}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2-8}+1}+\dfrac{1-x+2}{1+\sqrt{x-2}}=0\)
=>\(\left(x-3\right)\left(x+4\right)-\dfrac{x^3-2x^2-8x+15}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2-8}+1}-\dfrac{x-3}{1+\sqrt{x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)-\dfrac{\left(x-3\right)\left(x^2+x-5\right)}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{\left(x-3\right)\left(x+3\right)}{\sqrt{x^2-8}+1}-\dfrac{x-3}{1+\sqrt{x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x+4\right)-\dfrac{x^2+x-5}{\sqrt{x^3-2x^2-8x+16}+1}-\dfrac{x+3}{\sqrt{x^2-8}+1}-\dfrac{1}{\sqrt{x-2}+1}\right]=0\)
=>x-3=0
=>x=3
`\sqrt{14-x}-\sqrt{x-4}=\sqrt{x-1}` `ĐK: 4 <= x <= 14`
`<=>\sqrt{14-x}=\sqrt{x-1}+\sqrt{x-4}`
`<=>14-x=x-1+x-4+2\sqrt{(x-1)(x-4)}`
`<=>14-x=2x-5+2\sqrt{x^2-5x+4}`
`<=>2\sqrt{x^2-5x+4}=19-3x` với `4 <= x <= 19/3`
`<=>4(x^2-5x+4)=361-114x+9x^2`
`<=>4x^2-20x+16=361-114x+9x^2`
`<=>5x^2-94x+345=0`
`<=>5x^2-25x-69x+345=0`
`<=>(x-5)(5x-69)=0`
`<=>x=5` hoặc `x=69/5`
(t/m) (ko t/m)
Vậy `S={5}`