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3 tháng 8 2017
\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)
\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)
PA
2 tháng 8 2017
\(x^4-8x^3+21x^2-24x+9=0\)
\(\Leftrightarrow\left(x^2-3x+3\right)\left(x^2-5x+3\right)=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))
\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)
DT
11 tháng 4 2022
1)
<=> \(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
x= 0
x = 3
2) <=> \(x\left(x-3\right)=4\)
=> \(x=\dfrac{4}{x}+3\)
Y
11 tháng 4 2022
\(2,x^2-3x=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)
\(\Rightarrow\)Pt có 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)
Vậy \(S=\left\{4;-1\right\}\)
\(3,x^4-5x^2+6=0\)
Đặt \(t=x^2\left(t\ge0\right)\)
Pt trở thành
\(t^2-5t+6=0\)
\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)
\(\Rightarrow\)Pt ó 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)
\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)
Vậy \(S=\left\{\pm\sqrt{3}\right\}\)
14 tháng 12 2020
a, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\sqrt{\dfrac{3}{2}}\))
Vì hai vế ko âm, bp 2 vế ta được:
2x2 - 3 = 4x - 3
\(\Leftrightarrow\) 2x2 = 4x
\(\Leftrightarrow\) x2 = 2x
\(\Leftrightarrow\) x2 - 2x = 0
\(\Leftrightarrow\) x(x - 2) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)
Vậy S = {2}
b, \(\sqrt{2x-1}=\sqrt{x-1}\) (x \(\ge\) 1)
Vì hai vế ko âm, bp 2 vế ta được:
2x - 1 = x - 1
\(\Leftrightarrow\) x = 0 (KTM)
Vậy x = \(\varnothing\)
c, \(\sqrt{x^2-x-6}=\sqrt{x-3}\) (x \(\ge\) 3)
Vì hai vế ko âm, bp 2 vế ta được:
x2 - x - 6 = x - 3
\(\Leftrightarrow\) x2 - 2x - 3 = 0
\(\Leftrightarrow\) x2 - 3x + x - 3 = 0
\(\Leftrightarrow\) x(x - 3) + (x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 1) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)
Vậy S = {3}
d, \(\sqrt{x^2-x}=\sqrt{3x-5}\) (x \(\ge\) \(\dfrac{5}{3}\))
Vì hai vế ko âm, bp 2 vế ta được:
x2 - x = 3x - 5
\(\Leftrightarrow\) x2 - 4x + 5 = 0
\(\Leftrightarrow\) x2 - 4x + 4 + 1 = 0
\(\Leftrightarrow\) (x - 2)2 + 1 = 0
Vì (x - 2)2 \(\ge\) 0 với mọi x \(\ge\) \(\dfrac{5}{3}\) \(\Rightarrow\) (x - 2)2 + 1 > 0 với mọi x \(\ge\) \(\dfrac{5}{3}\)
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
Chúc bn học tốt!
19 tháng 9 2021
1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
LA
1
29 tháng 8 2019
\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) ( ĐK : \(x\ge-2\) )
\(\Leftrightarrow\sqrt{x+2-4\sqrt{x+2}+4}+\sqrt{x+2-6\sqrt{x+2}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=1\)
Ta có : \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|=\left|\sqrt{x+2}-2\right|+\left|3-\sqrt{x+2}\right|\)
Áp dụng BĐT : \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow\left|\sqrt{x+2}-2\right|+\left|3-\sqrt{x+2}\right|\ge\left|\sqrt{x+2}-2+3-\sqrt{x+2}\right|=1\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x+2}-2\ge0\\3-\sqrt{x+2}\ge0\end{matrix}\right.\Leftrightarrow2\le x\le7\)
PL
0
11 tháng 1 2023
a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
=>x+1=1 và y-2=1/2
=>x=0 và y=5/2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)
=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6
=>x-2y=9 và 2x-y=12
=>x=5; y=-2
c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
<=>\(9\sqrt{x}=6\)
<=>\(\sqrt{x}=\frac{6}{9}\)
<=>\(x=\left(\frac{6}{9}\right)^2=\frac{36}{81}=\frac{4}{9}\)