Bài 1 :

a) \(C=\frac{-4}{\left(2x-3\right)^2+5}\)

Vì \(\left(2x-3\right)^2\ge0\forall x\)

\(\Rightarrow C\ge\frac{-4}{5}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)

Vậy....

b) \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\)

\(\Rightarrow\left(a+b\right)\left(c-a\right)=\left(c+a\right)\left(a-b\right)\)

\(\Leftrightarrow ac-a^2+bc-ab=ac-bc+a^2-ab\)

\(\Leftrightarrow ac-a^2-ab-ac+ab-a^2=-bc-bc\)

\(\Leftrightarrow-2a^2=-2bc\)

\(\Leftrightarrow a^2=bc\left(đpcm\right)\)

b) a+b/a-b = c+a/c-a

=> (a+b).(c-a) = (a-b).(c+a)

<=> (a+b).c - (a+b).a = (a-b).c + (a-b).a

<=> ac+bc - a^2-ba = ac-bc + a^2 - ba

<=> ac -ac + bc + bc -ba +ba = a^2 +a^2

<=> 2bc = 2a^2

<=> bc = a^2 (đccm)

Chúc bạn hc tốt 

hình nào ?

Hình đâu bạn ei

\(\left|x-2007\right|+\left|x-2010\right|+\left|x-2008\right|+\left|y-2009\right|\)

\(\ge\left|x-2007+2010-x\right|+\left|x-2008\right|+\left|y-2009\right|=3+0+0=3\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-2007\right)\left(2010-x\right)\ge0\\\left|x-2008\right|=0\\\left|y-2009\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=2008\\y=2009\end{cases}}\)

Vậy x = 2008 và y = 2009

\(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\)

\(\Rightarrow\left|x-2017\right|+\left|x-2018\right|+\left|2010-x\right|+\left|y-2009\right|=3\)

Ta có :+) \(\left|x-2007\right|+\left|2010-x\right|\ge\left|x-2007+2010-x\right|=3\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-2007\right)\left(2010-x\right)\ge0\Leftrightarrow2007\le x\le2010\)

+) \(\left|x-2008\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow x-2008=0\Leftrightarrow x=2008\)

+)\(\left|y-2009\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow y-2009=0\Leftrightarrow y=2009\)

\(\Rightarrow\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|\ge3\)

\(\Rightarrow\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\)

\(\Leftrightarrow\hept{\begin{cases}2007\le x\le2010\\x=2008\\y=2009\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2008\\y=2009\end{cases}}\)

Vậy................................

có ai giải đc bài này hông, mình cần gấp

Áp dụng ta đc:

\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{5a+5b+5c}{a+b+c}=5\left(\text{vì: a,b,c khác 0}\right)\)

\(\Rightarrow\hept{\begin{cases}b+c=2a\\c+a=2b\\a+b=2c\end{cases}}\Rightarrow a=b=c\)

\(\Rightarrow P=6\)

\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)

\(\Rightarrow\frac{3a+b+c}{a}-2=\frac{a+3b+c}{b}-2=\frac{a+b+3c}{c}-2\)

\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)

Xét \(a+b+c\ne0\)

\(\Rightarrow a=b=c\)

Thay vào P ta được P=6

Xét \(a+b+c=0\)

\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)

Thay vào P ta được P= -3

Vậy P có 2 gtri là ...........

D = 10 + 12 + 14 + ... + 994 + 996 + 998

D = 10 + (12 + 998) + (14 +996) + (16 + 994) + ... + (504 + 506)

D = 10 + 1010 + 1010 + 1010 + ... + 1010

D = 10 + 1010.494

D = 10 + 498940

D = 498950