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\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)
\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))
\(⇔x=23\)
\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)
\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)
\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))
\(⇔x=-100\)
\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)
\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)
\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)
\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))
\(⇔x=-2013\)
\(\frac{201-x}{99}+\frac{203}{97}=\frac{205}{95}+3\)
\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
\(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
Giúp mk với ạ
a. \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x=23\) (Vì \(\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)\ne0\) )
b. \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=-100\) (Vì \(\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)\ne0\) )
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
câu a tự quy đồng cùng mẫu rồi làm thôi :"))
b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)
Đặt \(x^2-x=k\), ta có:
\(k.\left(k-2\right)=24\)
\(\Leftrightarrow k^2-2k+1=25\)
\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)
\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)
c)\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)
\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)
p/s: bn tự kết luận nha :))
Đặt \(a=24-x,b=x-25\)
Khi đó pt ban đầu trở thành :
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow30a^2+68ab+30b^2=0\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3a=-5b\\5a=-3b\end{cases}}\)
Đến đây bạn thay vào là dễ rồi nhé ! Chúc bạn học tốt !