Bài 1: Tìm min của
\(B=\sqrt{1-2x+x^2}+\sqrt{x^2+6x+9}+\sqrt{x^2+4x+4}\)
tìm min
a)\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)
b)B=\(\sqrt{x-2\sqrt{x-1}}-\sqrt{x+2\sqrt{x-1}}\)
a ) \(A=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
\(\ge\left|x-1+3-x\right|+\left|x-2\right|=\left|x-2\right|+2\ge2\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\\left|x-2\right|=0\end{cases}\Rightarrow x=2}\)(TM)
Vậy \(A_{min}=2\Leftrightarrow x=2\)
b ) \(B=\sqrt{x-2\sqrt{x-1}}-\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}-\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}-\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|-\left|\sqrt{x-1}+1\right|\)
\(\le\left|\sqrt{x-1}-1-\sqrt{x-1}-1\right|=2\)có GTLN là 2
Bài 1: giải p.trình
a,\(\sqrt{x^2-4x+4}=1\)
b,\(\sqrt{1-4x+4x^2}=5\)
c,\(\sqrt{a\left(1-2x+x^2\right)}-6=0\)
d,\(\sqrt{9x^2}=2x+1\)
e,\(\sqrt{9-6x+x^2}=x\)
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: \(\left\{{}\begin{matrix}9x^2\ge0\\2x+1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x\ge0\)
\(\sqrt{9x^2}=2x+1\\ \Rightarrow\left|3x\right|=2x+1\\ \Rightarrow\left[{}\begin{matrix}3x=2x+1\\3x=-2x+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
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) \(\sqrt{x^2-4x+4}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b) \(\sqrt{1-4x+4x^2}=5\)
\(\Leftrightarrow\sqrt{\left(1-2x\right)^2}=5\Leftrightarrow\left|1-2x\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
c) \(\sqrt{x\left(1-2x+x^2\right)}-6=0\)
\(\Leftrightarrow\left(\sqrt{x\left(1-x\right)^2}\right)^2=36\Leftrightarrow x\left(1-x\right)^2=36\)
\(\Leftrightarrow x-2x^2+x^3-36=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+2x+9\right)=0\)
\(\Leftrightarrow x=4\)(do \(x^2+2x+9=\left(x+1\right)^2+8>0\))
d) \(\sqrt{9x^2}=2x+1\)
\(\Leftrightarrow3\left|x\right|=2x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+1\\-3x=2x+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{5}\end{matrix}\right.\)
e) \(\sqrt{9-6x+x^2}=x\left(1\right)\left(đk:x\ge0\right)\)
\(\Leftrightarrow\sqrt{\left(3-x\right)^2}=x\Leftrightarrow\left|3-x\right|=x\)
TH1: \(0\le x\le3\)
\(\left(1\right)\Leftrightarrow3-x=x\Leftrightarrow x=\dfrac{3}{2}\)
TH2: \(x>3\)
\(\left(1\right)\Leftrightarrow x-3=x\Leftrightarrow-3=0\left(vn\right)\)
bài 1 : giải phương trình:
a. \(\sqrt{x+2\sqrt{ }x-1}=2\)
b. \(\sqrt{x^2-4x+4}=\sqrt{4x^212x+9}\)
c.\(\sqrt{x+4\sqrt{ }x-4}=2\)
d. \(\sqrt{x^2-6x+9}=2\)
e. \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)
f. \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)
e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)
c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)
\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)
\(\Leftrightarrow x-4=0\)
hay x=4
a) \(\sqrt{x-1+2\sqrt{x-1}.1+1^2}=2;đk:x\)≥1
⇔\(\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}.1+1^2}=2\left(hđt-1\right)\)
⇔\(\sqrt{\left(\sqrt{x-1}+1\right)^2=2}\)
⇔|\(\sqrt{x-1}+1\)|=2
⇔\(\left[{}\begin{matrix}\sqrt{x+1}-1=2\\\sqrt{x+1-1}=-2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\sqrt{x+1}=3\\\sqrt{x+1}=-1\left(L\right)\end{matrix}\right.\)⇔x+1=9⇔x=10(TM)
→S={10}
Tìm Max của
1) \(5+\sqrt{-4x^2-4x}\)
2) \(\sqrt{x-2}+\sqrt{4-x}\)
3) \(x+\sqrt{2-x^2}\)
4) \(2x+\sqrt{4-2x^2}\)
Tìm Min của
1) \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)
2) \(\sqrt{x\left(x+1\right)\left(x+2\right)\left(x+3\right)+5}\)
a) \(A=5+\sqrt{-4x^2-4x}\)
\(A==5+\sqrt{-4x\left(x+1\right)}\)
Có: \(-4x\left(x+1\right)\le0\)
\(\Rightarrow\sqrt{-4x\left(x+1\right)}=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy: \(Max_A=5\) tại \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(B=\sqrt{x-2}+\sqrt{4-x}\)
ĐKXĐ: \(\hept{\begin{cases}x\ge2\\x\le4\end{cases}}\Rightarrow x\in\left\{2;3;4\right\}\)
Thay \(x=2\Rightarrow\sqrt{2-2}+\sqrt{4-2}=\sqrt{2}\)
Thay \(x=3\Rightarrow\sqrt{3-1}+\sqrt{4-3}=2\)
Thay \(x=4\Rightarrow\sqrt{4-2}+\sqrt{4-4}=\sqrt{2}\)
Vậy: \(Max_B=2\) tại \(x=3\)
Bài 2:
a)\(A=\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}\)
\(=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\)
\(\ge x-1+0+3-x=2\)
Dấu = khi \(\hept{\begin{cases}x-1\ge0\\x-2=0\\x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x=2\\x\le3\end{cases}}\Leftrightarrow x=2\)
Vậy MinA=2 khi x=2
bÀI 2 PHẦN b bạn nhân 2 ngoặc 1 r` đặt ẩn là t =>min...
Giải các phương trình sau:
a) \(\sqrt{x+4\sqrt{x-4}}=2\)
b) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)
c) \(\sqrt{2x^2-2x+1}=2x-1\)
Lời giải:
a. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{(x-4)+4\sqrt{x-4}+4}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-4}+2)^2}=2$
$\Leftrightarrow |\sqrt{x-4}+2|=2$
$\Leftrightarrow \sqrt{x-4}+2=2$
$\Leftrightarrow \sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
b. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$
$\Leftrightarrow |2x-1|=|x-3|$
\(\Rightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)
c.
PT \(\Rightarrow \left\{\begin{matrix} 2x-1\geq 0\\ 2x^2-2x+1=(2x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x^2-2x=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{2}\\ 2x(x-1)=0\end{matrix}\right.\Rightarrow x=1\)
Bài 1 : Tìm GTNN
\(A=\sqrt{x^2+4x+4}+\sqrt{x^2-4x+4}\)
Bài 2 : Giải phương trình
a) \(\sqrt{2+2x-x^2}+\sqrt{-x^2-6x-8}=1+\sqrt{3}\)
b ) \(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{9-\left(3x-1\right)^2}\)
Bài 2 : Tìm GTLN
\(P=\sqrt{x-5}+\sqrt{13-x}\)
Tìm điều kiện xác định
\(\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}=1\)
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
a: ĐKXĐ: \(x\in R\)
b: ĐKXĐ: \(x\in R\)
Giải phương trình:
a) \(\sqrt{x^2+4}=\sqrt{2x+3}\)
b) \(\sqrt{x^2-6x+9}=2x-1\)
c) \(\sqrt{4x+12}=\sqrt{9x+17}-5\)
d) \(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)
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)
a) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
b) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
c) \(\sqrt{x^2+6x-9}-2\sqrt{x^2-2x+1}+\sqrt{x^2}=0\)
a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4
=>2*căn(x+5)=4
=>căn (x+5)=2
=>x+5=4
=>x=-1
b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
=>2*căn x-1=16
=>x-1=64
=>x=65
c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)
TH1: \(x\ge3\)
\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)
TH2: \(2\le x< 3\)
\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)
TH3: \(0\le x< 2\)
\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
TH4: \(x< 0\)
\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)