\(\sqrt{4x^2}-20x+25+2x=5\)
\(\sqrt{1-2x}+36x^2=5\)
\(\sqrt{4x^2-20x+25x+2x}=5\)
\(\sqrt{x-2}\sqrt{x-1}=\sqrt{x-1-1}\)
5. giải phương trình
a.\(\sqrt{\left(x-3\right)^2}=3-x\)
b.\(\sqrt{4x^2-20x+25}+2x=5\)
c.\(\sqrt{1-12x+36x^2}=5\)
a: Ta có: \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3\le0\)
hay \(x\le3\)
b: Ta có: \(\sqrt{4x^2-20x+25}+2x=5\)
\(\Leftrightarrow\left|2x-5\right|=5-2x\)
\(\Leftrightarrow2x-5\le0\)
hay \(x\le\dfrac{5}{2}\)
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
a.
\(\sqrt{x+2\sqrt{x-1}}=2\)
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)
b.
\(\sqrt{4x^2-20x+25}=5-2x\)
\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)
\(\Leftrightarrow\left|5-2x\right|=5-2x\)
\(\Leftrightarrow5-2x\ge0\)
\(\Leftrightarrow x\le\dfrac{5}{2}\)
c.
ĐKXĐ: \(x\ge3\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\Rightarrow x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=3\end{matrix}\right.\)
d.
ĐKXĐ: \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
\(\Rightarrow x^2-x=3-x\)
\(\Leftrightarrow x^2=3\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) (thỏa mãn)
1) \(\sqrt{x^2}=2x-5\)
2) \(\sqrt{25x^2-10x+1}=2x-6\)
3) \(\sqrt{25-10x+x^2}=2x-5\)
4) \(\sqrt{1-2x+x^2}=2x-1\)
5) \(\sqrt{4x^2+4x+1}=-x-3\)
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: \(x\ge\dfrac{1}{2}\)
\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)
Giải các phương trình sau:
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{1-12x+36x^2}=5\)
\(\sqrt{x^2+x}=x\)
\(\sqrt{x^2-4x+3}=x-2\)
\(\sqrt{1-x^2}=x-1\)
\(a,\sqrt{4x^2-20x+25}+2x=5\)
\(\Rightarrow\sqrt{\left(2x-5\right)^2}+2x=5\)
\(\Rightarrow4x=10\Rightarrow x=\frac{5}{2}\)
\(b,\sqrt{1-12x+36x^2}=5\)
\(\Rightarrow6x-1=5\)
\(\Rightarrow6x=6\Rightarrow x=1\)
\(c,\sqrt{x^2+x}=x\)
\(\Rightarrow x^2+x=x^2\)
\(\Rightarrow x=0\)
\(c,\Rightarrow\left(x-2\right)^2-1=\left(x-2\right)^2\)
\(\Rightarrow-1=0\) (vô lý)
=> PT vô nghiệm
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(x+\sqrt{5-4x}=0\)
\(\sqrt{1-2x^2}=x-1\)
a: ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow\sqrt{x-1}=1\)
hay x=2
c: Ta có: \(\sqrt{1-2x^2}=x-1\)
\(\Leftrightarrow1-2x^2=x^2-2x+1\)
\(\Leftrightarrow-3x^2+2x=0\)
\(\Leftrightarrow-x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)
Tìm GTLN:
a) A= \(\sqrt{3-2x^2}\)
b) B= \(\sqrt{-9x^2+6x+3}\)
c) B= \(5+\sqrt{-4x^2-4x}\)
d) C= \(\sqrt{-x^2+x+\frac{3}{4}}\)
e) D= \(\sqrt{x^2+2x+1}+\sqrt{x^2-2x+1}\)
g) G= \(\sqrt{25x^2-20x+4}+\sqrt{25x^2}\)
f) F= \(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
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 =)))
a)\(\sqrt{X^2-3X+2}=3-X\)
b)\(\sqrt{4x^2-20x+25}+2x=5\)
c)\(\sqrt{\left(3-2x\right)^2}=4\)
a
ĐK:
\(3-x\ge0\\ \Leftrightarrow x\le3\)
\(\sqrt{x^2-3x+2}=3-x\\ \Leftrightarrow x^2-3x+2=\left(3-x\right)^2=9-6x+x^2\\ \Leftrightarrow x^2-3x+2-9+6x-x^2=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\dfrac{7}{3}\left(nhận\right)\)
Thử lại: \(\sqrt{\left(\dfrac{7}{3}\right)^2-3.\dfrac{7}{3}+2}=\dfrac{2}{3}>0\)
Vậy phương trình có nghiệm duy nhất \(x=\dfrac{7}{3}\)
b
\(\sqrt{4x^2-20x+25}=\sqrt{\left(2x\right)^2-2.2x.5+5^2}=\sqrt{\left(2x-5\right)^2}=\left|2x-5\right|\)
Phương trình trở thành:
\(\left|2x-5\right|+2x=5\) (1)
Với \(x< \dfrac{5}{2}\) thì (1) \(\Leftrightarrow5-2x+2x=5\Leftrightarrow5=5\)
=> Với \(x< \dfrac{5}{2}\) thì phương trình có nghiệm với mọi x \(< \dfrac{5}{2}\) (I)
Với \(x\ge\dfrac{5}{2}\) thì (1)
\(\Leftrightarrow2x-5+2x=5\\ \Leftrightarrow2x-5+2x-5=0\\ \Leftrightarrow4x=10\\ \Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\left(nhận\right)\left(II\right)\)
Từ (I), (II) kết luận phương trình có nghiệm với mọi \(x\le\dfrac{5}{2}\)
c
\(\Leftrightarrow\left|3-2x\right|=4\) (1)
Nếu \(x\le\dfrac{3}{2}\) thì (1)
\(\Leftrightarrow3-2x=4\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x=-\dfrac{1}{2}\left(nhận\right)\)
Nếu \(x>\dfrac{3}{2}\) thì (1)
\(\Leftrightarrow2x-3=4\\ \Leftrightarrow2x=7\\ \Leftrightarrow x=\dfrac{7}{2}\left(nhận\right)\)
Vậy phương trình có 2 nghiệm \(S=\left\{-\dfrac{1}{2};\dfrac{7}{2}\right\}\)
a: =>x^2-3x+2=x^2-6x+9 và x<=3
=>3x=7 và x<=3
=>x=7/3(loại)
b: =>|2x-5|=5-2x
=>2x-5<=0
=>x<=5/2
c: =>|2x-3|=4
=>2x-3=4 hoặc 2x-3=-4
=>x=-1/2 hoặc x=7/2
\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(\sqrt{x^2+x+1}=x+1\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{x^2-2x+1}=4\)
\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)
do \(x^2+x+1=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(\Rightarrow\sqrt{x^2+x+1}>0\forall x\)
voi dk \(x\ge-1\) ta co
\(x^2+x+1=x^2+2x+1\Rightarrow x=0\)(tm)
b,\(\sqrt{4x^2-20x+25}+2x=5\)
\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}+2x=5\)
\(\Leftrightarrow\left|2x-5\right|+2x=5\)
th1 \(2x-5\ge0\Leftrightarrow x\ge\frac{5}{2}\) ta co\(2x-5+2x=5\Leftrightarrow4x=10\Rightarrow x=2.5\left(tm\right)\)
th2 \(2x-5< 0\Leftrightarrow x< \frac{5}{2}\) \(5-2x+2x=5\Leftrightarrow5=5\)
\(\Rightarrow\) dung voi moi \(x< \frac{5}{2}\)
kl \(x\le\frac{5}{2}\)
c, \(\left|x-1\right|=4\) \(\Rightarrow\orbr{\begin{cases}x-1=4\left(x\ge1\right)\\x-1=-4\left(x< 1\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-3\left(tm\right)\end{cases}}}\)
d.\(\sqrt{3\left(x^2+2x+1\right)+4}+\sqrt{5\left(x^2+2x+1\right)+16}\)
=\(\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}\ge\sqrt{4}+\sqrt{16}=6\)
ma \(-x^2-2x+5=-\left(x^2+2x+1\right)+6=-\left(x+1\right)^2+6\le6\)
dau = xay ra \(\Leftrightarrow x=-1\)
Giải các phương trình:
a) \(\sqrt{\left(x-3\right)^2}=3-x\)
b) \(\sqrt{4x^2-20x+25}+2x=5\)
c) \(\sqrt{1-12x+36x^2}=5\)