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Ta có
\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)
Thế vào ta được
\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)
\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)
x=\(\sqrt{\frac{2-\sqrt{3}}{2}}\) =\(\sqrt{\frac{4-2\sqrt{3}}{4}}=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow2x=\sqrt{3}-1\Rightarrow2x+1=\sqrt{3}\Rightarrow\left(2x+1\right)^2=3\Leftrightarrow4x^2+4x+1=3\Leftrightarrow4x^2+4x-2=0\Leftrightarrow2x^2+2x-1=0\)
nên đề bài = \(\left(x^3\left(2x^2+2x-1\right)+1\right)^{2013}+\frac{\left(x\left(2x^2+2x-1\right)-3\right)^{2013}}{x^2\left(2x^2+2x-1\right)-3^{2013}}\)
=\(\left(0+1\right)^{2013}+\frac{\left(0-3\right)^{2013}}{0-3^{2013}}=1+1=2\)
Bài 1:
Đặt \(\left\{\begin{matrix} 5x+3=a\\ 2x+4=b\end{matrix}\right.\) \(\Rightarrow 3x-1=a-b\)
PT trở thành:
\(a^3-b^3=(a-b)^3\)
\(\Leftrightarrow (a-b)(a^2+ab+b^2)=(a-b)^3\)
\(\Leftrightarrow (a-b)[a^2+ab+b^2-(a^2-2ab+b^2)]=0\)
\(\Leftrightarrow 3ab(a-b)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{5}\\x=-2\\5x+3=2x+4\Leftrightarrow x=\dfrac{1}{3}\end{matrix}\right.\)
Thử lại thấy đều thỏa mãn
Vậy \(x\in\left\{\frac{-3}{5};-2;\frac{1}{3}\right\}\)
Bài 2:
\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)
\(\Leftrightarrow \frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1\right)=\frac{x-4}{2010}-1\)
\(\Leftrightarrow \frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}=\frac{x-2014}{2010}\)
\(\Leftrightarrow (x-2014)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\) (1)
Thấy rằng \(2013> 2011; 2012> 2010\Rightarrow \frac{1}{2013}< \frac{1}{2011}; \frac{1}{2012}< \frac{1}{2010}\)
\(\Rightarrow \frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}< 0\) (2)
Từ (1),(2) suy ra \(x-2014=0\Leftrightarrow x=2014\)
Bài 3:
Đặt \(\left\{\begin{matrix} 2x-5=a\\ x-2=b\end{matrix}\right.\Rightarrow x-3=a-b\)
PT trở thành: \(a^3-b^3=(a-b)^3\)
\(\Leftrightarrow (a-b)(a^2+ab+b^2)-(a-b)(a^2-2ab+b^2)=0\)
\(\Leftrightarrow 3ab(a-b)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\\x-3=0\Leftrightarrow x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{2}; 2; 3\right\}\)
6:ĐKXĐ: x>=0; x<>1/25
BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)
=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)
=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)
=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)
7:
ĐKXĐ: x>=0
BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)
=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)
=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)
=>\(-\sqrt{x}-2>=0\)(vô lý)
8:
ĐKXĐ: x>=0; x<>9/4
BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
TH1: 9căn x-14>0 và 2căn x-3<0
=>căn x>14/9 và căn x<3/2
=>14/9<căn x<3/2
=>196/81<x<9/4
TH2: 9căn x-14<0 và 2căn x-3>0
=>căn x>3/2 hoặc căn x<14/9
mà 3/2<14/9
nên trường hợp này Loại
9:
ĐKXĐ: x>=0
\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)
=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)
=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)
=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)
10:
ĐKXĐ: x>=0; x<>1/49
\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)
=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)
=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)
=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)
TH1: 6căn x-1>0 và 7căn x-1>0
=>căn x>1/6 và căn x>1/7
=>căn x>1/6
=>x>1/36
TH2: 6căn x-1<0 và 7căn x-1<0
=>căn x<1/6 và căn x<1/7
=>căn x<1/7
=>0<=x<1/49
Bài 1:
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=4-\sqrt{5}+\sqrt{5}+1=5\)
Bài 2:
a: ĐKXĐ: x>=3
\(\sqrt{x-3}=6\)
=>x-3=36
=>x=36+3=39(nhận)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x-3\right)^2}=12\)
=>\(\left|x-3\right|=12\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 3:
a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)
\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)
\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)
\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)
b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)
\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)
\(=\sqrt{3x-1}+\sqrt{5}\)
d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)