TÌM GTNN
\(A=\sqrt{x^2+2x+1}+\sqrt{9x^2-6x+1}\)
\(B=3x^2-6x+1\)
\(C=2x-3\sqrt{x}\left(x>=0\right)\)
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MH
4 tháng 10 2021
c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
4 tháng 10 2021
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
Y
2 tháng 7 2023
Câu b bạn có bị lỗi dấu căn không mà sao nó kéo dài cả 2 vế pt vậy :v
\(a,\sqrt{x^2-6x+9}+x=11\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=11-x\)
\(\Leftrightarrow\left|x-3\right|=11-x\\ TH_1:x\ge3\\ x-3=11-x\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\left(tm\right)\)
\(TH_2:x< 3\\ -x+3=11-x\\ \Leftrightarrow-x+x=11-3\\ \Leftrightarrow0=8\left(VL\right)\)
Vậy \(S=\left\{7\right\}\)
\(c,\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\) \(\left(dk:x\ge-1\right)\)
\(\Leftrightarrow\sqrt{4^2}.\sqrt{\left(x+1\right)}-\sqrt{3^2}.\sqrt{\left(x+1\right)}=4\left(1\right)\)
Đặt \(a=\sqrt{x+1}\left(a\ge0\right)\)
Pt trở thành : \(4a-3a=4\Leftrightarrow a=4\left(tmdk\right)\)
\(\Rightarrow\sqrt{x+1}=4\\ \Rightarrow\left(\sqrt{x+1}\right)^2=16\\ \Rightarrow\left|x+1\right|=16\)
\(TH_1:x\ge-1\\ x+1=16\Leftrightarrow x=15\left(tm\right)\\ TH_2:x< -1\\ -x-1=16\Leftrightarrow x=-17\left(tm\right)\)
Nhưng loại TH2 vì dk ban đầu là \(x\ge-1\)
Vậy \(S=\left\{15\right\}\)
\(d,\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\left(dk:x\ge-1\right)\\ \Leftrightarrow\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{x+1}=0\)
Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\)
Tới đây bạn làm tương tự câu c nha.
26 tháng 10 2023
a: ĐKXĐ: x>=-3/2
\(\sqrt{x^2+4}=\sqrt{2x+3}\)
=>\(x^2+4=2x+3\)
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1(nhận)
b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))
=>\(\sqrt{\left(x-3\right)^2}=2x-1\)
=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)
=>x=4/3(nhận) hoặc x=-2(loại)
c:
Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)
ĐKXĐ: \(x>=-3\)
\(\sqrt{4x+12}=\sqrt{9x+27}-5\)
=>\(2\sqrt{x+3}=3\sqrt{x+3}-5\)
=>\(-\sqrt{x+3}=-5\)
=>x+3=25
=>x=22(nhận)
d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)
=>\(4x^2-6x+1=4x^2-20x+25\)
=>\(-6x+20x=25-1\)
=>\(14x=24\)
=>x=12/7(nhận)
2 tháng 9 2021
a, ĐKXĐ: \(x^2-4x+4\ge0\Rightarrow\left(x-2\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{x^2-4x+4}=1\\ \Rightarrow x-2=1\\ \Rightarrow x=3\)
b,\(ĐKXĐ:1-4x+4x^2\ge0\Rightarrow\left(1-2x\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{1-4x+4x^2}=5\\ \Rightarrow\left|1-2x\right|=5\\ \Rightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
d, ĐKXĐ: \(\left\{{}\begin{matrix}9x^2\ge0\\2x+1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x\ge0\)
\(\sqrt{9x^2}=2x+1\\ \Rightarrow\left|3x\right|=2x+1\\ \Rightarrow\left[{}\begin{matrix}3x=2x+1\\3x=-2x+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
2 tháng 9 2021
c, ĐKXĐ: \(1-2x+x^2\ge0\Rightarrow\left(1-x\right)^2\ge0\left(luônđúng\right)\)
\(\sqrt{1-2x+x^2}-6=0\\ \Rightarrow\left|1-x\right|=6\\ \Rightarrow\left[{}\begin{matrix}1-x=-6\\1-x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
e, \(\left\{{}\begin{matrix}9-6x+x^2\ge0\\x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3-x\right)^2\ge0\left(luônđúng\right)\\x\ge0\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\sqrt{9-6x+x^2}=x\\ \Rightarrow\left|3-x\right|=x\\ \Rightarrow\left[{}\begin{matrix}3-x=-x\\3-x=x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3=0\left(vôlí\right)\\x=1,5\end{matrix}\right.\)
\(A=\sqrt{\left(x+1\right)^2}+\sqrt{\left(3x-1\right)^2}=\left|x+1\right|+\left|3x-1\right|\)
Với \(x\le-1:A=-x-1-3x+1=-4x\)
Để A nhỏ nhất thì x lớn nhất => x = -1 => A = 4
Với -1 < x <= 1/3: \(A=x+1-3x+1=2-2x\)
Để A nhỏ nhất thì x lớn nhất => x = 1/3 => A = 4/3
Với x > 1/3: \(A=x+1+3x-1=4x\)
Do x > 1/3 => A > 4/3
=> A min = 4/3 <=> x = 1/3
\(B=3\left(x^2-2x+\frac{1}{3}\right)=3\left[\left(x^2-2x+1\right)-\frac{2}{3}\right]=3\left(x-1\right)^2-2\)
=> Vì 3(x-1)^2 >= 0 => B >= -2
B min = -2 <=> 3(x-1)^2 = 0 <=> x = 1
\(C=2\left(x-\frac{3}{2}\sqrt{x}\right)=2\left[\left(x-2.\frac{3}{4}\sqrt{x}+\frac{9}{16}\right)-\frac{9}{16}\right]=2\left(\sqrt{x}-\frac{3}{4}\right)^2-\frac{9}{8}\)
=> C >= -9/8
C min = -9/8 <=> căn x = 3/4 => x = 9/16