Cho \(A=\left(\frac{5x}{4x^2-1}+\frac{1}{1-2x}-\frac{2}{1+2x}\right):\frac{x-1}{1+4x+4x^2}\)
\(B=\sqrt{4-2\sqrt{3}}+\sqrt{19-8\sqrt{3}}\)
a.Tìm x để A có nghĩa.Rút gọn A,B
b.Tìm x để A=B
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8 tháng 10 2016
Ta có:
x = \(\frac{1}{2}\)\(\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
= \(\frac{1}{2}\)\(\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{1}}\)
= \(\frac{1}{2}\)(\(\sqrt{2}\)-1)
=> 2x = \(\sqrt{2}\)-1
=> (2x)2= ( \(\sqrt{2}\)-1)2
=> 4x2= 2-2\(\sqrt{2}\)+1
=> 4x2= -2( \(\sqrt{2}\)-1)+1
=> 4x2= -4x +1 => 4x2+4x-1=0
Lại có:
A1= (\(4x^5\)+\(4x^4\)- \(x^3\)+1)19
= [ x3( 4x2+4x-1) +1]19
=1
A2=( \(\sqrt{4x^5+4x^4-5x^3+5x+3}\))3
= (\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\))3
= 23=8
A3= \(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\)
= \(\sqrt{2}\)- \(\sqrt{2}\)\(\sqrt{1-\sqrt{2}}\)
Cộng 3 số vào ta được A
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 6 2019
\(x=\frac{1}{2}\left(\sqrt{2}-1\right)\)
\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)
\(\Leftrightarrow4x^2+4x-1=0\)
\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)
\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)
\(\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}\left(4x^2+4x-1\right)+\frac{1}{2}}}=\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}}}=\sqrt{2}-2x=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)
\(M=1+8+1=10\)
11 tháng 12 2018
Bạn ghi lộn đề rồi \(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}\) chứ không phải \(\left(\dfrac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\right)^{2014}\)
11 tháng 12 2018
Ta có \(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2-1}\right)}}=\dfrac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\left|\sqrt{2}-1\right|}{2}=\dfrac{\sqrt{2}-1}{2}\)
Vậy ta có \(x=\dfrac{\sqrt{2}-1}{2}\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow2x+1=\sqrt{2}\Leftrightarrow\left(2x+1\right)^2=2\Leftrightarrow4x^2+4x+1=2\Leftrightarrow4x^2+4x-1=0\)Ta lại có \(\left(4x^5+4x^4-x^3+1\right)^{19}=\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=\left(x^3.0+1\right)^{19}=1^{19}=1\)(1)
\(\left(\sqrt{4x^5+4x^4-5x^3+5x+3}\right)^3=\left(\sqrt{4x^5+4x^4-x^3-4x^3-4x^2+x+4x^2+4x-1+4}\right)^3=\left(\sqrt{x^3\left(4x^2+4x-1\right)-x^2\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\right)^3=\left(\sqrt{x^3.0+x^2.0+0+4}\right)^3=\left(\sqrt{4}\right)^3=2^3=8\left(2\right)\)
\(\left(\dfrac{1-\sqrt{2}x}{\sqrt{2x^2+2x}}\right)^{2014}=\left[\dfrac{1-\sqrt{2}.\dfrac{\sqrt{2}-1}{\sqrt{2}}}{\sqrt{2.\dfrac{3-2\sqrt{2}}{4}+\sqrt{2}-1}}\right]^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}}{2}+\sqrt{2}-1}}\right)^{2014}=\left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{\dfrac{3-2\sqrt{2}+2\sqrt{2}-2}{2}}}\right)^{2014}=\left(\dfrac{\dfrac{\dfrac{1}{\sqrt{2}}}{1}}{\sqrt{2}}\right)^{2014}=1^{2014}=1\left(3\right)\)
Cộng (1),(2),(3) theo vế ta được A=1+8+1=10
Vậy khi x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) thì A=10
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
