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a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)
\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)
\(\Leftrightarrow\)\(2^x.15=480\)
\(\Rightarrow\)\(2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5.
a) \(\dfrac{15-x}{2000}+\dfrac{14-x}{2001}=\dfrac{13-x}{2002}+\dfrac{12-x}{2003}\)
\(\Leftrightarrow\dfrac{15-x}{2000}+1+\dfrac{14-x}{2001}+1=\dfrac{13-x}{2002}+1+\dfrac{12-x}{2003}+1\)
\(\Leftrightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}=\dfrac{2015-x}{2002}+\dfrac{2015-x}{2003}\)
\(\Rightarrow\dfrac{2015-x}{2000}+\dfrac{2015-x}{2001}-\dfrac{2015-x}{2002}-\dfrac{2015-x}{2003}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow2015-x=0\)
<=> x=2015
Vậy phương trình có nghiệm là x=2015
b) \(\dfrac{x-5}{2010}+\dfrac{x-4}{2011}=\dfrac{x-2010}{5}+\dfrac{x-2011}{4}\)
\(\Leftrightarrow\dfrac{x-5}{2010}-1+\dfrac{x-4}{2011}-1=\dfrac{x-2010}{5}-1+\dfrac{x-2011}{4}-1\)
\(\Leftrightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}=\dfrac{x-2015}{5}+\dfrac{x-2015}{4}\)
\(\Rightarrow\dfrac{x-2015}{2010}+\dfrac{x-2015}{2011}-\dfrac{x-2015}{5}-\dfrac{x-2015}{4}=0\)
\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x-2015=0\)
=> x=2015
Vậy phương trình có nghiệm x=2015
Câu 1:
\(A=21\left(a+\frac{1}{b}\right)+3\left(b+\frac{1}{a}\right)=21a+\frac{21}{b}+3b+\frac{3}{a}\)
\(=(\frac{a}{3}+\frac{3}{a})+(\frac{7b}{3}+\frac{21}{b})+\frac{62}{3}a+\frac{2b}{3}\)
Áp dụng BĐT Cô-si:
\(\frac{a}{3}+\frac{3}{a}\geq 2\sqrt{\frac{a}{3}.\frac{3}{a}}=2\)
\(\frac{7b}{3}+\frac{21}{b}\geq 2\sqrt{\frac{7b}{3}.\frac{21}{b}}=14\)
Và do $a,b\geq 3$ nên:
\(\frac{62}{3}a\geq \frac{62}{3}.3=62\)
\(\frac{2b}{3}\geq \frac{2.3}{3}=2\)
Cộng tất cả những BĐT trên ta có:
\(A\geq 2+14+62+2=80\) (đpcm)
Dấu "=" xảy ra khi $a=b=3$
Câu 2:
Bình phương 2 vế ta thu được:
\((x^2+6x-1)^2=4(5x^3-3x^2+3x-2)\)
\(\Leftrightarrow x^4+12x^3+34x^2-12x+1=20x^3-12x^2+12x-8\)
\(\Leftrightarrow x^4-8x^3+46x^2-24x+9=0\)
\(\Leftrightarrow (x^2-4x)^2+6x^2+24(x-\frac{1}{2})^2+3=0\) (vô lý)
Do đó pt đã cho vô nghiệm.
a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)
ĐK:\(x\ne0\)
\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)
a/ Đặt \(x^2+x+1=a\Rightarrow x^2+x+2=a+1\)
Pt trở thành \(a\left(a+1\right)-12=0\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
2/ \(\dfrac{x+1}{2014}+1+\dfrac{x+2}{2013}+1=\dfrac{x+3}{2012}+1+\dfrac{x+4}{2011}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}=\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2015=0\) (do \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Rightarrow x=-2015\)
Bài 1:
Ta có:
\(\left(\dfrac{ab}{2}-\dfrac{6ab}{7}\right):\dfrac{5b^2}{14}=\left(\dfrac{7ab}{14}-\dfrac{12ab}{14}\right).\dfrac{14}{5b^2}\)
\(=\dfrac{-5ab}{14}.\dfrac{14}{5b^2}=\dfrac{-a}{b}\)(1)
Thay \(a=\dfrac{2007}{2010};b=\dfrac{2011}{2010}\) vào (1) ta được:
\(\dfrac{-\dfrac{2007}{2010}}{\dfrac{2011}{2010}}=-\dfrac{2007}{2011}\)
Vậy......................
Chúc bạn học tốt!!!
Bài 2:
\(\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -\dfrac{1}{2}.\dfrac{3}{4}:\dfrac{1}{8}+1\)
\(\Rightarrow2.\left(-\dfrac{9}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -3+1\)
\(\Rightarrow-\dfrac{37}{4}< \dfrac{x}{8}< -2\)
\(\Rightarrow\dfrac{-74}{8}< \dfrac{x}{8}< -\dfrac{16}{8}\)
\(\Rightarrow-74< x< -16\)
\(\Rightarrow x\in\left\{-73;-72;-71;....;-18;-17\right\}\)
Vậy..............................
Chúc bạn học tốt!!!
1: \(\Leftrightarrow\left(\dfrac{x+1}{85}+1\right)+\left(\dfrac{x+3}{83}+1\right)=\left(\dfrac{x+5}{81}+1\right)+\left(\dfrac{x+7}{79}+1\right)\)
=>x+86=0
=>x=-86
2: \(\Leftrightarrow\left(\dfrac{x-1}{2015}+1\right)-\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+7}{2007}+1\right)-\left(\dfrac{x+11}{2003}+1\right)\)
=>x+2014=0
=>x=-2014
3: \(\Leftrightarrow3\left(x+4\right)-2\left(x-3\right)=4x\)
=>4x=3x+12-2x+6
=>4x=x+18
=>3x=18
=>x=6
4: \(\Leftrightarrow15x-5\left(x+1\right)=3\left(2x+1\right)\)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
5: \(\Leftrightarrow2\left(2x-7\right)+5\left(x+11\right)=-40\)
=>4x-14+5x+55=-40
=>9x+41=-40
=>x=-9
Bài 2:
\(A=\dfrac{x\left(x^3+1\right)}{x^2-x+1}-\dfrac{x\left(x^3-1\right)}{x^2+x+1}\)
\(=x\left(x+1\right)-x\left(x-1\right)\)
=x^2+x-x^2+x
=2x