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moi hok lop 6

Bài 2:

Vì a,b là nghiệm PT nên \(\left\{{}\begin{matrix}30a^2-4a=2010\\30b^2-4b=2010\end{matrix}\right.\)

\(\Rightarrow N=\dfrac{a^{2008}\left(30a^2-4a\right)+b^{2008}\left(30b^2-4b\right)}{a^{2008}+b^{2008}}\\ \Rightarrow N=\dfrac{a^{2008}\cdot2010+b^{2008}\cdot2010}{a^{2008}+b^{2008}}=2010\)

Bài 1:

Viét: \(\left\{{}\begin{matrix}x_1+x_2=a\\x_1x_2=a-1\end{matrix}\right.\)

\(M=\dfrac{2x_1^2+x_1x_2+2x_2^2}{x_1^2x_2+x_1x_2^2}=\dfrac{2\left(x_1+x_2\right)^2-3x_1x_2}{x_1x_2\left(x_1+x_2\right)}=\dfrac{2a^2-3a+3}{a^2-a}\)

Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)

Ta có: \(\frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0\)

\(\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow a=b=c=\frac{1}{2}\)

Thay vào tìm x;y;z

Đặt: \(\hept{\begin{cases}\sqrt{x-2009}=a\\\sqrt{y-2010}=b\\\sqrt{z-2011}=c\end{cases}}\)

Ta có: \frac{1}{a}-\frac{1}{a^2}+\frac{1}{b}-\frac{1}{b^2}+\frac{1}{c}-\frac{1}{c^2}-\frac{3}{4}=0a1​−a21​+b1​−b21​+c1​−c21​−43​=0

\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{c^2}-\frac{1}{c}+\frac{3}{4}=0⇔a21​−a1​+b21​−b1​+c21​−c1​+43​=0

\Leftrightarrow\left(\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}\right)+\left(\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}\right)+\left(\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}\right)=0⇔(a21​−a1​+41​)+(b21​−b1​+41​)+(c21​−c1​+41​)=0

\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0⇔(a1​−21​)2+(b1​−21​)2+(c1​−21​)2=0

\Leftrightarrow a=b=c=\frac{1}{2}⇔a=b=c=21​

Thay vào tìm x;y;z

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,....

Giúp mk câu a, c thui nha!! Câu b mk làm đc rùi!!!

Nhã Doanh, ngonhuminh, nguyen thi vang, @hattori heiji, @Phùng Khánh Linh, ...

\(\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{\sqrt{z-2011}-1}{z-2011}=\dfrac{3}{4}\)\(\left(\left\{{}\begin{matrix}x>2009\\y>2010\\z>2011\end{matrix}\right.\right)\)

\(\Leftrightarrow\dfrac{1}{4}-\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{1}{4}-\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{1}{4}-\dfrac{\sqrt{z-2011}-1}{z-2011}=0\)

\(\Leftrightarrow\dfrac{x-2009-4\sqrt{x-2009}+4}{x-2009}+\dfrac{y-2010-4\sqrt{y-2010}+4}{y-2010}+\dfrac{z-2011-4\sqrt{z-2011}+4}{z-2011}=0\)

Nhận xét: \(\left\{{}\begin{matrix}\dfrac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}\ge0\\\dfrac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}\ge0\\\dfrac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(2013;2014;2015\right)\)

Đặt \(a=\sqrt{x-2009};b=\sqrt{y-2010};c=\sqrt{z-2011};a>0;b>0;c>0\)

\(Pt\Leftrightarrow\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)

\(\Leftrightarrow\frac{\left(4a^2-a+1\right)}{a^2}+\frac{\left(4b^2-b+1\right)}{b^2}+\frac{\left(4c^2-c+1\right)}{c^2}=0\)

\(\Leftrightarrow\left(\frac{2a-1}{a}\right)^2+\left(\frac{2b-1}{b}\right)^2+\left(\frac{2c-1}{c}\right)^2=0\)

\(\Rightarrow a=b=c=\frac{1}{2}\Rightarrow\sqrt{x-2009}=\frac{1}{2}\Rightarrow x=2009\frac{1}{4}\)

\(\Rightarrow b=\frac{1}{2}\Rightarrow\sqrt{y-2010}=\frac{1}{2}\Rightarrow y=2010\frac{1}{4}\)

\(\Rightarrow c=\frac{1}{2}\Rightarrow\sqrt{z-2011}=\frac{1}{2}\Rightarrow z=2011\frac{1}{4}\)