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\(B=3-2x+\sqrt{1+4x+4x^2}=3-2x+\sqrt{\left(2x+1\right)^2}=3-2x+\left|2x+1\right|\)
Nếu \(x\ge-\frac{1}{2}\Rightarrow B=3-2x+2x+1=4\)
Nếu \(x< -\frac{1}{2}\Rightarrow B=3-2x-2x-1=4-4x\)
b, x = 2015 tức là \(x>-\frac{1}{2}\)
Vậy với x = 2015 thì B = 4
a)\(A=2x+\sqrt{1-4x+4x^2}\)
\(A=2x+\sqrt{\left(1-2x\right)^2}\)
\(A=2x+\left|1-2x\right|\)
\(A=2x+\left|2x-1\right|\)
\(A=4x-1\)
b)\(x=\frac{1}{4}\Leftrightarrow A=4.\frac{1}{4}-1\)
\(\Leftrightarrow A=-1\)
a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
\(=4-2\sqrt{3}+2\sqrt{3}\)
=4
Thay x=4 vào B, ta được:
\(B=\dfrac{2-4}{2}=-1\)
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 \(A_{min}=-\dfrac{1}{4}\)
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 \(a=\dfrac{1}{4}\)
a, \(B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)ĐK : \(x>0;x\ne1\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b,Ta có \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{2}+1\)
Vậy \(B=\sqrt{2}+1-1=\sqrt{2}\)
a) Ta có: \(B=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}\)
\(=\sqrt{x}-1\)
b) Thay \(x=3+2\sqrt{2}\) vào B, ta được:
\(B=\sqrt{2}+1-1=\sqrt{2}\)
a) \(A=5x-\sqrt{4x^2-4x+1}\)
\(=5x-\sqrt{\left(2x-1\right)^2}\)
\(=5x-\left|2x-1\right|\)
+) Với x < 1/2
A = 5x - [ -( 2x - 1 ) ] = 5x - ( 1 - 2x ) = 5x - 1 + 2x = 7x - 1
+) Với x ≥ 1/2
A = 5x - ( 2x - 1 ) = 5x - 2x + 1 = 3x + 1
b) Với x = -2 < 1/2
=> A = 7.(-2) - 1 = -14 - 1 = -15
\(a,B=3-2x+\sqrt{1+4x+4x^2}\\ =3-2x+\sqrt{\left(2x+1\right)^2}\\ =3-2x+2x+1\\ =4\)
\(b,\) Thay \(x=2015\) ta có:
\(B=4\)
a, B=3-2x+\(\sqrt{4x^2+4x+1}\) =3-2x+\(\sqrt{\left(2x+1\right)^2}\) =3-2x+2x+1=4 b; Ta co B=4=4+0x thay x=2015 =>B=4+0x=4+0.2015=4 vay x=2015=>B=4