Tìm x biết:
a) \(\sqrt{25x^2-10x+1}=x+2\)
b) \(\sqrt{4x^2+12x+9}=7\)
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26 tháng 10 2021
a: \(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
26 tháng 10 2021
a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
LD
20 tháng 1 2019
a.
\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)
\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)
\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)
b.
\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)
\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)
\(\Leftrightarrow x^2-8=5x+1\)
\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)
............................
tương tự ..
3 tháng 1 2023
c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)
=>x-5=0 hoặc x+5=1
=>x=-4 hoặc x=5
d: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=7/2 hoặc x=-3/2
e: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)
=>x-2=0 hoặc 3 căn x+2=1
=>x=2 hoặc x+2=1/9
=>x=-17/9 hoặc x=2
5 tháng 8 2021
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
26 tháng 10 2021
1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
2) ĐKXĐ: \(x\ge3\)
\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)
\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)
4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
1 tháng 11 2017
a)
\(\sqrt{4x-4}-\sqrt{9x-9}+\sqrt{25x-25}=4+\sqrt{16x-16}\\ \Leftrightarrow2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}+5\sqrt{x-1}=4\\ \Leftrightarrow0\sqrt{x-1}=4\\ \Rightarrow kh\text{ô}ng\:c\text{ó}\:gi\text{á}\:tr\text{ị}\:x\:th\text{õa}\:m\text{ãn}\)
b)
\(•\sqrt{7-x}+\sqrt{x-5}\le\sqrt{2.\left(7-x+x-5\right)}=2\\ •x^2-12x+38=\left(x-6\right)^2+2\ge2\)
ta thấy \(VT\le2\:v\text{à}\:VP\ge2\) nên \(VT=VP=2\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}7-x=x-5\\x-6=0\end{matrix}\right.\Rightarrow x=6\)
vậy nghiệm của phương trình trên là x=6
T
10 tháng 8 2021
Làm a, c là tiêu biểu thôi, bài b đơn giản.
a) \(\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}=\sqrt{x-1}-1\)
ĐKXĐ: $x\ge 1.$ Do $VT\ge 0 \Rightarrow VT\ge 0 \to x\ge 2.$
Ta có \(VT=\sqrt{\left[\sqrt{x-1}-1\right]^2}=\left|\sqrt{x-1}-1\right|=VP\) (vì \(\sqrt{x-1}-1=VP\ge0.\))
Vậy phương trình có vô số nghiệm.
c) Ta có:
\(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=2\)
ĐKXĐ: $x\ge 1.$
Ta có: \(VT=\sqrt{\left(\sqrt{x-1}+1\right)^2}=\left|\sqrt{x-1}+1\right|=\sqrt{x-1}+1.\)
(vì $\sqrt{x-1}+1>0\forall x\ge 1.$)
Ta có: \(\sqrt{x-1}+1=2\Rightarrow x=2.\) (thỏa mãn)
10 tháng 8 2021
b: Ta có: \(\sqrt{36x^2-12x+1}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=6\\6x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
18 tháng 8 2020
c)\(C=5+\sqrt{-4x^2-4x}\)
\(C=5+\sqrt{1-\left(4x^2+4x+1\right)}\)
\(C=5+\sqrt{1-\left(2x+1\right)^2}\)
Ta có: \(-\left(2x+1\right)^2\le0\)
\(\sqrt{1-\left(2x+1\right)^2}\le1\)
\(\sqrt{1-\left(2x+1\right)^2}+5\le6\Leftrightarrow C\le6\)
Vậy \(C_{max}=6\) khi \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
f) \(F=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(F=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(F=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x+1+3-2x\right|=4\)
\(F_{min}=4\) khi \(\left(2x-1\right)\left(3-2x\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)
Mấy còn lại tương tự =)))
2 tháng 10 2021
a) \(\sqrt{\left(2x-3\right)^2}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)
\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)
\(\Leftrightarrow5\sqrt{x+2}=20\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
HP
2 tháng 10 2021
a. \(\sqrt{\left(2x-3\right)^2}=7\)
<=> \(\left|2x-3\right|=7\)
<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\) ĐK: \(x\ge-2\)
<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)
<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)
<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)
<=> \(5\sqrt{x+2}=20\)
<=> \(\sqrt{x+2}=4\)
<=> \(\left(\sqrt{x+2}\right)^2=4^2\)
<=> \(\left|x+2\right|=16\)
<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)
c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\) ĐK: \(x\ge3\)
<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)
<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)
a) \(\sqrt{25x^2-10x+1}=x+2\)
<=> \(\sqrt{\left(5x-1\right)^2}=x+2\)
<=> \(\left|5x-1\right|=x+2\)
TH1: 5x - 1 \(\ge\)0 <=> x \(\ge\)1/5
Khi đó pt trở thành: 5x - 1 = x + 2
<=> 4x = 3 <=> x = 3/4 (tm)
TH2: 5x - 1 < 0 <=> x < 1/5
Khi đó pt trở thành: 1 - 5x = x + 2
<=> -6x = 1 <=> x = -1/6 (tm)
Vậy S = {3/4; -1/6}
b) \(\sqrt{4x^2+12x+9}=7\)
<=> \(\sqrt{\left(2x+3\right)^2}=7\)
<=> \(\left|2x+3\right|=7\)
TH1: 2x + 3 \(\ge\)0 <=> x \(\ge\)-3/2
Khi đó pt trở thành: 2x + 3 = 7 <=> 2x = 4 <=> x = 2 (Tm)
TH2: 2x + 3 < 0 <=> x < -3/2
Khi đó pt trở thành: -2x - 3 = 7
<=> -2x = 10 <=> x = -5 (tm)
Vậy S = {-5; 2}