giải pt \(\left(x+1\right)^{2010}+\left(x+2\right)^{2010}=2^{-2009}\)
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Lời giải của mình ở đây nhé bạn!
http://olm.vn/hoi-dap/question/424173.html
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đkxđ với mọi x
đặt a=x2+x+1
\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)
<=> \(\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}\)
=> 6a(a+2) +6(a+1)2 =7(a+1)(a+2)
<=> 6a2+12a +6a2 +12a+6 =a2 +21a+14
<=> 12a2 -a2+24a-21a+6-14=0
<=> 11a2+3a-8=0
<=> 11a2 +11a-8a-8=0
<=> (11a2 +11a)-(8a+8)=0
<=> 11a(a+1)-8(a+1)=0
<=> (a+1)(11a-8)=0
=> a=-1 và a=\(\dfrac{8}{11}\)
thay a=x2+x+1 ta đc
x2+x+1=-1
<=> x2+x+2 =0 (vô nghiệm)
và x2+x+\(\dfrac{3}{11}\) =0(vô nghiệm )
vậy pt trên vô nghiệm
c) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)\)ĐKXĐ : x # 0
( 2) <=> \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)
\(< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2\)
\(< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2\)
\(< =>16=\left(x+4\right)^2\)
<=> x2 + 8x = 0
<=> x( x + 8) = 0
<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )
Vậy,....
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ĐKXĐ: \(x\ne2009\) ; \(x\ne2010\)
Đặt : a = \(2009-x\)
b = \(x-2010\)
⇒ a + b = -1 ⇒ a = - ( 1 + b )
⇒ Phương trình đã cho có dạng :
\(\dfrac{a^2+ab+b^2}{a^2-ab+b^2}=\dfrac{19}{49}\)
⇔ \(\dfrac{\left(1+b\right)^2-b\left(1+b\right)+b^2}{\left(1+b\right)^2+b\left(1+b\right)+b^2}\) \(=\dfrac{19}{49}\)
⇔ \(\dfrac{b^2+b+1}{3b^2+3b+1}\) \(=\dfrac{19}{49}\)
⇔ \(49b^2+49b+49=57b^2+57b+19\)
⇔ \(8b^2+8b-30=0\)
⇔ \(4b^2+4b-15=0\)
⇔ \(4b^2-6b+10b-15=0\)
⇔ \(2b\left(2b-3\right)+5\left(2b-3\right)=0\)
⇔ \(\left(2b-3\right)\left(2b+5\right)=0\)
⇔ \(\left[{}\begin{matrix}2b+5=0\\2b-3=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}b=\dfrac{-5}{2}\\b=\dfrac{3}{2}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-5}{2}+2010=2007,5\\x=\dfrac{3}{2}+2010=2011,5\end{matrix}\right.\)
Vậy ......
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Đặt \(x-2009=y\) khi đó phương trình trở thành:
\(\dfrac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\dfrac{19}{49}\)
\(\Leftrightarrow4y^2-4y-15=0\)
\(\Leftrightarrow\left(2y-5\right)\left(2y+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2,5\\y=-1,5\end{matrix}\right.\)
Đổi lại:\(y=x-2009\) ,ta được:
\(\left[{}\begin{matrix}x=2009+2,5=2011,5\\x=2009-1,5=2007,5\end{matrix}\right.\)
Vậy...
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Bài 1: Tìm x biết: $\frac{\left(2009-x\right)^2+\left(2009-x\right..
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Đặt \(2009-x=t\Rightarrow x-2010=-\left(2009-x\right)-1=-t-1\)
Suy ra:
\(\frac{t^2+t\left(-t-1\right)+\left(-t-1\right)^2}{t^2-t\left(-t-1\right)+\left(-t-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{t^2-t\left(t+1\right)+\left(t+1\right)^2}{t^2+t\left(t+1\right)+\left(t+1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{t^2-t^2-t+t^2+2t+1}{t^2+t^2+t+t^2+2t+1}=\frac{19}{49}\)
\(\Leftrightarrow\frac{t^2+t+1}{3t^2+3t+1}=\frac{19}{49}\)
\(\Leftrightarrow49t^2+49t+49=57t^2+57t+19\)
\(\Leftrightarrow8t^2+8t-30=0\)
\(\Leftrightarrow4t^2+4t-15=0\Leftrightarrow4t^2+4t+1=16\)
\(\Leftrightarrow\left(2t+1\right)^2=16\Leftrightarrow\left[{}\begin{matrix}2t+1=-4\\2t+1=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2t=-5\\2t=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\frac{5}{2}\\t=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2009-x=-\frac{5}{2}\\2009-x=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4023}{2}\\x=\frac{4015}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{4015}{2};\frac{4023}{2}\right\}\)
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1. \(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\Rightarrow8x^3-12x^2+6x-1+x^3+6x^2+12x+8=27x^3+27x^2+9x+1\)
\(\Rightarrow-18x^3-33x^2+9x+6=0\)\(\Rightarrow\left(x+2\right)\left(-18x^2+3x+3\right)=0\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)\left(-9x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2};x=-\frac{1}{3}\end{cases}}\)
Vậy \(x=-2;x=\frac{1}{2};x=-\frac{1}{3}\)
2. \(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\Rightarrow\left(\frac{x-1988}{15}-1\right)+\left(\frac{x-1969}{17}-2\right)+\left(\frac{x-1946}{19}-3\right)+\left(\frac{x-1919}{21}-4\right)=0\)
\(\Rightarrow\frac{x-2003}{15}+\frac{x-2003}{17}+\frac{x-2003}{19}+\frac{x-2003}{21}=0\)
\(\Rightarrow x-2003=0\)do \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)
Vậy \(x=2003\)
3. Đặt \(\hept{\begin{cases}2009-x=a\\x-2010=b\end{cases}}\)
\(\Rightarrow\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Rightarrow49a^2+49ab+49b^2=19a^2-19ab+19b^2\)
\(\Rightarrow30a^2+68ab+30b^2=0\Rightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5a=-3b\\3a=-5b\end{cases}}\)
Với \(5a=-3b\Rightarrow5\left(2009-x\right)=-3\left(x-2010\right)\)
\(\Rightarrow-2x=-4015\Rightarrow x=\frac{4015}{2}\)
Với \(3a=-5b\Rightarrow3\left(2009-x\right)=-5\left(x-2010\right)\)
\(\Rightarrow2x=4023\Rightarrow x=\frac{4023}{2}\)
Vậy \(x=\frac{4023}{2}\)hoặc \(x=\frac{4015}{2}\)