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a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)
<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
<=> x = 2010
cau a dau nhi cuoi cung k phai j dau nha ! mk an lom !
\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)
ta có |x+5| \(\ge\)0 \(\forall x\)
Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn
b,
\(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)
Bài 1:
a) \(\frac{1}{5}x^4y^3-3x^4y^3\)
= \(\left(\frac{1}{5}-3\right)x^4y^3\)
= \(-\frac{14}{5}x^4y^3.\)
b) \(5x^2y^5-\frac{1}{4}x^2y^5\)
= \(\left(5-\frac{1}{4}\right)x^2y^5\)
= \(\frac{19}{4}x^2y^5.\)
Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.
Chúc bạn học tốt!
\(VP=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(a+5\right)}=\frac{2\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{2\left(a+3\right)\left(a+5\right)}\)
\(=\frac{2\left(a+5\right)}{4\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+5\right)+2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left[\left(a+3\right)+\left(a+5\right)\right]}{4\left(a+3\right)\left(a+5\right)}=\frac{\left(a+3\right)+\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}\)
\(=\frac{\left(a+a\right)+\left(3+5\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{2a+8}{2\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+4\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)
\(VT=\frac{x-2}{\left(a+3\right)\left(a-5\right)}\)
\(\Rightarrow\frac{x-2}{\left(a+3\right)\left(a-5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)
\(\Rightarrow\frac{x-2}{a+4}=\frac{\left(a+3\right)\left(a-5\right)}{\left(a+3\right)\left(a+5\right)}\Rightarrow\frac{x-2}{a+4}=\frac{a-5}{a+5}\Rightarrow\left(x-2\right)\left(a+5\right)=\left(a-5\right)\left(a+4\right)\)
chịu