Tìm GTLN
a)\(\sqrt{3-2x^2}\)
b)\(\sqrt{-9x^2+6x+3}\)
c)\(5+\sqrt{-4x^2-4x}\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
2.tìm x
a)\(\sqrt{x^2-6x+9}\)
b)\(\sqrt{x^2-2x+1}\)
c)\(\sqrt{4x+12}-3\sqrt{x+3}+7\sqrt{9x+27}=20\)
d)\(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
a) \(\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)
\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x
⇒x∈\(R\)
b) \(\sqrt{x^2-2x+1}\)
\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)
\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x
⇒x∈\(R\)
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)
Tìm x biết:
a)\(\sqrt{9x^2}=6\)
b)\(\sqrt{\left(x-2\right)^2}=5\)
c)\(\sqrt{x^2-6x+9}=3\)
d)\(\sqrt{x^2+4x+4}-2x=3\)
`a)sqrt{9x^2}=6`
`<=>|3x|=6`
`<=>|x|=2`
`<=>x=+-2`
`b)sqrt{(x-2)^2}=5`
`<=>|x-2|=5`
`**x-2=5`
`<=>x=7`
`**x-2=-5`
`<=>x=-3`
`c)sqrt{x^2-6x+9}=3`
`<=>\sqrt{(x-3)^2}=3`
`<=>|x-3|=3`
`**x-3=3`
`<=>x=6`
`**x-3=-3`
`<=>x=0`
`d)sqrt{x^2+4x+4}-2x=3`
`<=>sqrt{(x+2)^2}=3+2x`
`<=>|x+2|=2x+3(x>=-3/2)`
`**x+2=2x+3`
`<=>x=-1(tm)`
`**x+2=-2x-3`
`<=>3x=-5`
`<=>x=-5/3(l)`
Sử dụng công thức:`sqrtA^2=|A|`
ĐKXĐ : \(x\in R\)
a, \(\sqrt{9x^2}=\left|3x\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy ..
b, \(\sqrt{\left(x-2\right)^2}=\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy ...
c, \(\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)
Vậy ..
d, \(\sqrt{x^2+4x+4}-2x=\sqrt{\left(x+2\right)^2}-2x=\left|x+2\right|-2x=3\)
\(\Leftrightarrow\left|x+2\right|=2x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+2=2x+3\\x+2=-2x-3\end{matrix}\right.\\2x+3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{3}{2}\\\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-\dfrac{5}{3}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
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 \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)
GIẢI PHƯƠNG TRÌNH:
a)\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
b)\(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
c)\(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=1+\sqrt{3}\)
d)\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
a) giải pt ra ta được : x=-1
b) giải pt ra ta được : x=2
c)giải pt ra ta được : x vô ngiệm
d)giải pt ra ta được : x=vô ngiệm
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
tìm gtln
a) \(B=1-\sqrt{x^2-2x+2}\)
b) \(C=1+\sqrt{4x-x^2-2}\)
a: \(B=1-\sqrt{\left(x-1\right)^2+1}\)
(x-1)^2+1>=1
=>\(\sqrt{\left(x-1\right)^2+1}>=1\)
=>\(B< =0\)
Dấu = xảy ra khi x=1
b:
ĐKXĐ: -(x+2)^2+2>=0
=>-(x+2)^2>=2
=>(x+2)^2<=2
=>\(-\sqrt{2}-2< =x< =\sqrt{2}-2\)
\(-x^2+4x-2=-\left(x^2-4x+2\right)\)
\(=-\left(x^2-4x+4-2\right)=-\left(x-2\right)^2+2< =2\)
=>\(0< =\sqrt{4x-x^2-2}< =\sqrt{2}\)
=>1<=C<=căn 2+1
\(C_{max}=\sqrt{2}+1\Leftrightarrow x=2\)
Giải phương trình:
a. \(\sqrt{x^2-4}-x^2+4=0\)
b. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
c. \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=1+\sqrt{3}\)
d. \(\sqrt{9x^2+6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
Đăng 1 lúc mà nhiều thế. Lần sau đăng 1 câu thôi b.
b/ \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)
Ta có: \(VT\ge1+2+\sqrt{5}=3+\sqrt{5}\)
Dấu = xảy ra khi \(x=2\)
c/ \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=\sqrt{3-\left(x-1\right)^2}+\sqrt{1-\left(x+3\right)^2}\)
\(\le1+\sqrt{3}\)
Dấu = không xảy ra nên pt vô nghiệm
Câu d làm tương tự
\(a,\sqrt{x^2-4}-x^2+4=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)
\(\Leftrightarrow x^2-4=\left(x-4\right)^2\)
\(\Leftrightarrow x^2-4-x^4+8x^2-16=0\)
\(\Leftrightarrow-x^4-7x^2-20=0\)
\(\Leftrightarrow-\left(x^4+7x^2+\frac{49}{4}\right)-\frac{31}{4}=0\)
\(\Leftrightarrow-\left(x^2+\frac{7}{2}\right)^2=\frac{31}{4}\)
\(\Leftrightarrow\left(x^2+\frac{7}{2}\right)=-\frac{31}{4}\)
\(\Rightarrow\)pt vô nghiệm
a/ĐK: \(x\ge2\)
\(PT\Leftrightarrow x^2-4=\sqrt{x^2-4}\)
Đặt \(x^2-4=t\Rightarrow x^2=t+4\)
Thay vào,phương trình đã cho tương đương với:
\(t=\sqrt{t}\Leftrightarrow t^2=t\Rightarrow\orbr{\begin{cases}t=1\\t=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-4=1\\x^2-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=5\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{5}\\x=2\end{cases}}\) (t/m)