Giải phương trình sau :
a) \(\frac{x^2-2x+1}{x^2-2x+2}+\frac{x^2-2x+2}{x^2-2x+3}=\frac{7}{6}\)
b) \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
2x(x+2)^2-8x^2=2(x-2)(x^2+2x+4)
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}\)
giải phương trình sau
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\) O
Để em trình bày dễ hiểu có chú thíck lun cho chụy :)
Ta có :
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\) ( cộng 3 phân số đầu cho 3, trừ phân số cuối cho 3 )
\(\Leftrightarrow\)\(\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\) ( quy đồng )
\(\Leftrightarrow\)\(\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\) ( vì tổng lớn hơn 0 nên khác 0 )
Nên \(x+2010=0\)
\(\Rightarrow\)\(x=-2010\) ( chuyển vế )
Vậy \(x=-2010\)
Chúc chụy học tốt ~
x + 2 x + 3 x + 4 x + 2028
▬▬▬ + ▬▬▬ + ▬▬▬▬ + ▬▬▬▬▬ = 0
2008 2007 2006 6
<=> 2007.2006.6.x + 2.2007.2006.6 + 2008.2006.6x + 3.2008.2006.6 + 2008.2007.6x + 4.2008.2007.6 + 2008.2007.2006x + 2028.2008.2007.2006 = 0
<=> ( 2007.2006.6 + 2008.2006.6 + 2008.2007.6 + 2008.2007.2006 )x = -( 2.2007.2006.6 + 3.2008.2006.6 + 4.2008.2007.6 + 2028.2008.2007.2006 )
<=> x = -( 2.2007.2006.6 + 3.2008.2006.6 + 4.2008.2007.6 + 2028.2008.2007.2006 ) / ( 2007.2006.6 + 2008.2006.6 + 2008.2007.6 + 2008.2007.2006 ) = -2010
cộng 1 vào 3 phân thức đầu và trừ 3 ở phân thức thứ tư thì ta sẽ được các tử bằng nhau rồi dùng phân phối sẽ ra.
giải phương trình
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
(x+2/2008+1)+(x+3/2007+1)+(x+4/2006+1)+(x+2028/6-3)=0
=x+2010/2008+ x+2010/2007+ x+2010/2006+ x+2010/6=0
=(x+2010)(1/2008+1/2007+1/2006+1/6)=
VÌ 1/2008 +1/2007 +1/2006+1/6 khác 0
=>x+2010=0=>x=-2010
giải phương trình hộ nhé
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
3 hạng tử đầu , mỗi hạng tử cùng cộng 1
Hạng tử cuối trừ 3
Nhân tử chung : x + 2010
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\)
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
Vậy....
Giải các phương trình sau :
a) ( 3x - 2 )( 4x + 3 ) = ( 2 - 3x )( x - 1)
b) x2 + ( x + 3 )( 5x - 7 ) = 9
c) 2x2 + 5x + 3 = 0
d) \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}=\frac{3-2x}{2009}+\frac{3-2x}{2010}\)
Giúp mik vs !!!
a, (3x - 2)(4x + 3) = (2 - 3x)(x - 1)
\(\Leftrightarrow\) (3x - 2)(4x + 3) - (2 - 3x)(x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(4x + 3) + (3x - 2)(x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(4x + 3 + x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(5x + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-2}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{2}{3}\); \(\frac{-2}{5}\)}
b, x2 + (x + 3)(5x - 7) = 9
\(\Leftrightarrow\) x2 - 9 + (x + 3)(5x - 7) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + (x + 3)(5x - 7) = 0
\(\Leftrightarrow\) (x + 3)(x - 3 + 5x - 7) = 0
\(\Leftrightarrow\) (x + 3)(6x - 10) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy S = {-3; \(\frac{5}{3}\)}
c, 2x2 + 5x + 3 = 0
\(\Leftrightarrow\) 2x2 + 2x + 3x + 3 = 0
\(\Leftrightarrow\) 2x(x + 1) + 3(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(2x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy S = {-1; \(\frac{3}{2}\)}
d, \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}=\frac{3-2x}{2009}+\frac{3-2x}{2010}\)
\(\Leftrightarrow\) \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}-\frac{3-2x}{2009}-\frac{3-2x}{2010}=0\)
\(\Leftrightarrow\) (3 - 2x)\(\left(\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)\) = 0
\(\Leftrightarrow\) 3 - 2x = 0
\(\Leftrightarrow\) x = \(\frac{3}{2}\)
Vậy S = {\(\frac{3}{2}\)}
Chúc bn học tốt!!
giải phương trình
a \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
b \(\frac{x^2+2x+1}{x^2+2x+2}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)
c \(x^2+\frac{1}{x^2}-\frac{9}{2}\left(x+\frac{1}{x}\right)+7=0\)
Hướng dẫn:
a) Đặt : \(x^2-2x+1=t\)Ta có:
\(\frac{1}{t+1}+\frac{2}{t+2}=\frac{6}{t+3}\)
b) Đặt : \(x^2+2x+1=t\)
Ta có pt: \(\frac{t}{t+1}+\frac{t+1}{t+2}=\frac{7}{6}\)
c)ĐK: x khác 0
Đặt: \(x+\frac{1}{x}=t\)
KHi đó: \(x^2+\frac{1}{x^2}=t^2-2\)
Ta có pt: \(t^2-2-\frac{9}{2}t+7=0\)
a) Đặt \(x^2-2x+3=v\)
Phương trình trở thành \(\frac{1}{v-1}+\frac{2}{v}=\frac{6}{v+1}\)
\(\Rightarrow\frac{v\left(v+1\right)+2\left(v+1\right)\left(v-1\right)}{v\left(v+1\right)\left(v-1\right)}=\frac{6v\left(v-1\right)}{v\left(v+1\right)\left(v-1\right)}\)
\(\Rightarrow v\left(v+1\right)+2\left(v+1\right)\left(v-1\right)=6v\left(v-1\right)\)
\(\Rightarrow v^2+v+2v^2-2=6v^2-6v\)
\(\Rightarrow3v^2-7v+2=0\)
Ta có \(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)
\(\Rightarrow\orbr{\begin{cases}v=\frac{7+5}{6}=2\\v=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2-2x+3=2\\x^2-2x+3=\frac{1}{3}\end{cases}}\)
+) \(x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
+)\(x^2-2x+3=\frac{1}{3}\)
\(\Rightarrow x^2-2x+\frac{8}{3}=0\)
Ta có \(\Delta=2^2-4.\frac{8}{3}=\frac{-20}{3}< 0\)
Vậy phương trình có 1 nghiệm là x = 1
c) Đặt \(\left(x+\frac{1}{x}\right)=a\) Khi đó pt có dạng :
\(a^2-\frac{9}{2}a+7-2=0\)
\(\Leftrightarrow2a^2-9a+10=0\)
\(\Leftrightarrow2a^2-4a-5a+10=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=2\\a=\frac{5}{2}\end{cases}}\)
+) Với \(a=\frac{5}{2}\Rightarrow x+\frac{1}{x}=\frac{5}{2}\)
\(\Rightarrow x^2+1=\frac{5x}{2}\)
\(\Rightarrow2x^2+2-5x=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\) ( thỏa mãn)
+) Với \(a=2\Rightarrow x+\frac{1}{x}=2\)
\(\Rightarrow x^2-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\) ( thỏa mãn )
Vậy pt đã cho có tập nghiệm \(S=\left\{1,\frac{1}{2},2\right\}\)
Giải PT: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\\ \Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\\ \Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\\ \Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\\ \Leftrightarrow x+2010=0\\ \Leftrightarrow x=-2010\)
Vậy pt có tập nghiệm \(S=\left\{-2010\right\}\)
Bài 15: Giải phương trình sau:
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
Bài 17: Giải phương trình sau:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
Bài 15:
Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2+2008}{2008}+\frac{x+3+2007}{2007}+\frac{x+4+2006}{2006}+\frac{x+2028-18}{6}=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\)
nên x+2010=0
hay x=-2010
Vậy: x=-2010
Bài 17:
Ta có: \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+3+63}{63}=\frac{x+5+61}{61}+\frac{x+7+59}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Vì \(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
nên x+66=0
hay x=-66
Vậy: x=-66
GIẢI PHƯƠNG TRÌNH SAU
A) \(\frac{X^2+2X+1}{X^2+2X+2}+\frac{X^2+2X+2}{X^2+2X+3}=\frac{7}{6}\)
B) \(\frac{\left(X^2-3X-4\right)^4}{\left(X-3\right)^5\left(X+2\right)^3}+\frac{\left(X^2+4X+3\right)^6}{\left(X-3\right)^3\left(X+2\right)^5}=0\)