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TH
1
NV
4 tháng 9 2016
\(A=4x-\sqrt{4x^2-12x+9}\)
\(=4x-\sqrt{\left(2x-3\right)^2}\)
\(=4x-\left|2x-3\right|\)
Theo đề ta có: \(A=-15\Leftrightarrow4x-\left|2x-3\right|=-15\)
\(\Rightarrow\left|2x-3\right|=4x+15\)
\(\Rightarrow\orbr{\begin{cases}2x-3=4x+15\\2x-3=-4x-15\end{cases}\Rightarrow\orbr{\begin{cases}2x=-18\\6x=-12\end{cases}\Rightarrow}\orbr{\begin{cases}x=-9\\x=-2\end{cases}}}\)
Vậy x = {-2;-9}
5 tháng 7 2021
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
AT
5 tháng 7 2021
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
21 tháng 9 2021
a) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)
b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))
N
1
10 tháng 6 2021
`đk:x-\sqrt{x^2-4x+4}>=0`
`<=>x>=\sqrt{x^2-4x+4}`
`<=>x^2>=x^2-4x+4(x>=0)`
`<=>4x-4>=0`
`<=>4x>=4<=>x>=1`
`b)A=sqrt{x-sqrt{(x-2)^2}}`
`=sqrt{x-|x-2|}`
`x>=2=>|x-2|=x-2`
`=>A=sqrt{x-x+2}=sqrt2`
`1<=x<=2=>|x-2|=2x-`
`=>A=\sqrt{x+x-2}=sqrt{2x-2}`
11 tháng 8 2018
\(a,\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)
\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(=|2x-1|+|2x-3|\)
\(b,\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)
\(=\sqrt{\left(7x-3\right)^2}+\sqrt{\left(7x+3\right)^2}\)
\(=|7x-3|+|7x+3|\)
=.= hok tốt!!
DT
1
\(A=4x-\sqrt{4x^2-12x+9}\)
\(=4x-2x+3\)
\(=2x+3\)
\(A=15\Rightarrow2x+3=15\)
\(2x=12\)
\(x=6\)