Đặt \(\left\{{}\begin{matrix}x-2010=a\\2009-x=b\end{matrix}\right.\)

Theo đề bài ta có:

\(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\)

\(\Leftrightarrow\dfrac{b^2+ab+a^2}{b^2-ab+a^2}=\dfrac{19}{49}\)

\(\Leftrightarrow19\left(b^2-ab+a^2\right)=49\left(b^2+ab+a^2\right)\)
\(\Leftrightarrow19b^2-19ab+19a^2-49b^2-49ab-49a^2=0\)

\(\Leftrightarrow-30a^2-68ab-30b^2=0\)

\(\Leftrightarrow-2\left(15a^2+34ab+15b^2\right)=0\)

\(\Leftrightarrow15a^2+34ab+15b^2=0\)

\(\Leftrightarrow15a^2+25ab+9ab+15b^2=0\)

\(\Leftrightarrow5a\left(3a+5b\right)+3b\left(3a+5b\right)=0\)

\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3a+5b=0\\5a+3b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\left(x-2010\right)+5\left(2009-x\right)=0\\5\left(x-2010\right)+3\left(2009-x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-6030+10045-5x=0\\5x-10050+6027-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x+4015=0\\2x-4023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-4015\\2x=4023\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4015}{-2}=2007,5\\x=\dfrac{4023}{2}=2011,5\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=2007,5\\x=2011,5\end{matrix}\right.\)

Đặt a=(2009-x)²

b=(x-2010)²

Theo đề bài ta có

\(\dfrac{\text{a^2+ab+b^2}}{a^2-ab+b^2}=\dfrac{19}{49}\)

\(\text{49(a^2+ab+b^2)}=19\left(a^2-ab+b^2\right)\)

\(\text{30a^2+68ab+30b^2=0}\)

\(\text{15a^2+34ab+15b^2=0}\)

\(\text{15a^2+9ab+25ab+15b^2=0}\)

\(\text{3a(5a+3b)+5(3b+5a)=0}\)

\(\text{(5a+3b)(3a+5b)=0}\)

\(\left[{}\begin{matrix}3a+5b=0\\3b+5a=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}3\left(2009-x\right)=5\left(x-2010\right)\\5\left(2009-x\right)=3\left(x-2010\right)\end{matrix}\right.\)

\(-8x=-6030-10045\) hay \(8x=-10050-6027\)

\(x\simeq2009\),375 hay \(x\simeq2009,625\)

Đặt

Theo đề bài ta có:

Vậy

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 ......

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...

Cre:Miny.vn

Bài 1: Tìm x biết: $\frac{\left(2009-x\right)^2+\left(2009-x\right..

đặt 2009-x=a,x-2010=b

suy ra a^2+ab+b^2/a^2-ab+b^2=19/49 

suy ra 49(a^2+ab+b^2)=19(a^2-ab+b^2)

49a^2+49ab+49b^2=19a^2-19ab+19b^2

30a^2+68ab+30b^2=0

30a^2+50ab+18ab+30b^2=0

10a(3a+5b)+6b(3a+5b)=0

(3a+5b)(10a+6b)=0

suy ra 3a+5b=0 hoặc 10a+6b=0 

thế vào lại rồi tìm x 

Để ý tử và mẫu là hằng đẳng thức

khó quá đi

Khó quá, ai mà biết được?!

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, ...