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\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)
\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(cộng 2 vế cho 3)
\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
x+2010=0
x=-2010
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)
\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)
\(=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
a.
\(\frac{2^5\times4^5\times5^{43}}{125^{44}}=\frac{2^5\times\left(2^2\right)^5\times5^{43}}{\left(5^3\right)^{44}}=\frac{2^5\times2^{10}\times5^{43}}{5^{132}}=\frac{2^{15}}{5^{89}}\)
b.
\(\frac{9^{24}}{27^{18}}=\frac{\left(3^2\right)^{24}}{\left(3^3\right)^{18}}=\frac{3^{48}}{3^{54}}=\frac{1}{3^6}=\frac{1}{729}\)
c.
\(\left(-\frac{2}{3}\right)^2+\left|-\frac{7}{8}\right|-\frac{11}{12}=\frac{4}{9}+\frac{7}{8}-\frac{11}{12}=\frac{29}{72}\)
Chúc bạn học tốt ^^
a) \(2^5.4^5.5^{43}:125^{44}=\frac{2^5.2^{10}.5^{43}}{5^{132}}=\frac{2^{15}}{5^{89}}\)
b) \(\frac{9^{24}}{27^{13}}=\frac{3^{42}}{3^{39}}=3^3=27\)
c) \(\left(\frac{-2}{3}\right)^2+\left|\frac{-7}{8}\right|-\frac{11}{12}=\frac{4}{9}+\frac{7}{8}-\frac{11}{12}\)
Sau đó quy đồng lên đươc kết quả là \(\frac{29}{72}\)
Chúc bạn làm bài tốt
Bn tự vẽ hình nha
Xét tg AHB và tg AHC có
AB=AC; góc AHB = góc AHC =90 độ;
Ah cạnh chung
=> tg AHB = tg AHC (ch cgv)
=> BH = HC
=> H là trung điểm BC
Xét tg BKC có
H là trung điểm BC (cmt)
DH//KC ( gt)
=> D là trung điểm BK
( đpcm )
Ầy mk chỉ biết câu a thui mà đằng nào chúng ta mới 2k5 thui biết vận dụng cả lớp 8 là tốt lắm rùi ....!
( 3x - 1/2 ) + ( 1/2y + 3/5 ) = 0
=> ( 3 x - 1/2 ) = 0
3x = 0+1/2
3x = 1/2
x = 1/2 : 3
x = 1/6
=> ( 1/2 y + 3/5 ) = 0
1/2y = 0 - 3/5
1/2 y = -3/5
y = -3/5 : 1/2
y = -6/5
\(B=\frac{3^{12}.13+3^{12}.3}{3^{11}.2^{24}}\)
\(B=\frac{3^{12}.\left(13+3\right)}{3^{11}.2^{24}}\)
\(B=\frac{3^{12}.16}{3^{11}.2^{24}}\)
\(B=\frac{3^{12}.2^4}{3^{11}.2^{24}}\)
\(B=\frac{3}{2^{20}}\)
B = 3/2 mũ 20