bài 3 : \(\left\{{}\begin{matrix}ab=2\\bc=3\\ca=54\end{matrix}\right.\)
hiển nhiên a;b;c =0 không phải nghiệm
\(\Leftrightarrow\left(abc\right)^2=2.3.54=18^2\)
\(\Leftrightarrow\left[{}\begin{matrix}abc=-18\\abc=18\end{matrix}\right.\)
abc=-18 => c=-9; a=-6; b=-1/3
abc=18 => c=9; a=6; b=1/3
1) Tính
\(A=\dfrac{1}{13}+\dfrac{3}{13.23}+\dfrac{3}{23.33}+...+\dfrac{3}{2003.2013}\)
\(B=\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{2018}-1\right)\)
2) Tìm x biết:
a) \(x^2-2x-15=0\)
b) \(\dfrac{3}{\left(x+2\right).\left(x+5\right)}+\dfrac{5}{\left(x+5\right).\left(x+10\right)}+\dfrac{7}{\left(x+10\right).\left(x+17\right)}=\dfrac{x+1}{\left(x+2\right).\left(x+17\right)}\)
3) Cho \(\dfrac{a}{b}=\dfrac{d}{c}\) . Chứng minh: \(\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4) Cho \(f\left(x\right)=x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
Tính giá trị của hiệu \(f\left(x\right)-g\left(x\right)\) tại x=0,1
5) Cho tam giác ABC có \(\widehat{A}=\ge90\) ; \(M\in AB,N\in AC\)
Chứng minh: BC > MN
6) Cho tam giác ABC, M là trung điểm BC, biết \(\widehat{BAM}>\widehat{CAM}\) . So sánh B và C
1)Thực hiện phép tính
a) \(\dfrac{9^8\cdot5^6}{3^7\cdot27^3\cdot25^4}\) b) \(\dfrac{5^{20}\cdot6^{18}}{15^{20}\cdot4^{10}}\) c) \(\left(\dfrac{1}{3}\right)^{50}\cdot9^{25}-\left(\dfrac{2}{3}\right)^{100}:\left(\dfrac{8}{27}\right)^{33}\) d) \(\left(\dfrac{3}{8}\right)^{60}\cdot\left(\dfrac{2}{3}\right)^{60}:\left(\dfrac{1}{2}\right)^{119}\cdot\left(\dfrac{1}{2}\right)\)
2) Tìm x, biết:
a) \(\dfrac{x}{12}=\dfrac{5}{x}\) b)\(8^x=2^{2x+3}\) c)\(\dfrac{1}{2}\sqrt{\dfrac{1}{2}x-2}-\dfrac{2}{3}=\dfrac{1}{3}\)
a,\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+......+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\)
b,A=\(1+5+5^2+5^3+5^4+.....+5^{49}+5^{50}\)
c,A=\(\left(\dfrac{1}{2^2}-1\right).\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)
d,A=\(2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
Ai giúp mình thực hiện phép tính này với ạ?? Cảm ơn nhiều!!
Bài 1. a, Cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right).....\left(\dfrac{1}{10}-1\right)\)
So sánh A với \(\dfrac{-1}{9}\)
Bài 2. Cho A = \(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)....\left(\dfrac{1}{2008}-1\right)\left(\dfrac{1}{2009}-1\right)\)
B = \(\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)....\left(-1\dfrac{1}{2007}\right)\left(-1\dfrac{1}{2008}\right)\)
Tính A . B ?
Tìm x biết :
a) \(\dfrac{1}{4}x-1+\dfrac{1}{3}\left(\dfrac{5}{2}x-7\right)-\left(\dfrac{5}{8}x-2\right)=\dfrac{7}{2}\)
b) \(\left|2-\dfrac{3}{2}x\right|-4=x+2\)
c) \(-3\left(\dfrac{2}{5}x-\dfrac{1}{5}\right)-x\left(x-\dfrac{1}{2}\right)=\dfrac{1}{6}-x^2\)
tính :
a, \(\left[6.\left(-\dfrac{1}{3}\right)^2-3.\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
b, \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
1.thực hiện phép tính
a.\(\dfrac{-5}{21}-\dfrac{1}{3}+3\dfrac{1}{2}.\left(\dfrac{-2}{3}\right)^3\)
b)\(-2\dfrac{1}{3}+\left(\dfrac{3}{8}-\dfrac{3}{4^{^{ }}}\right)^3:\dfrac{5}{9}-\dfrac{1}{2}\)
c)\(\left(6-2\dfrac{4}{5}\right).3\dfrac{1}{8}-1\dfrac{3}{5}:\left(-\dfrac{1}{2}\right)^2\)
d)\(-\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{7}.\dfrac{-1}{3}-\left(0,5\right)^3\)
e)\(\dfrac{15}{5}-\dfrac{2}{17}+\dfrac{15}{23}-\dfrac{15}{17}+\dfrac{4}{19}\)
Giúp mik nha mai mik phải nộp bài cho thầy rùi nì
Bài 1: Thu gọn đơn thức ( a là hằng só )
a) \(1\dfrac{1}{4}x^2y\left(\dfrac{-5}{6}xy\right)^0.\left(-2\dfrac{1}{3}xy\right)\)
b) \(\dfrac{1}{2}x.\dfrac{1}{4}x^2.\dfrac{x^3}{8}.2y.4y^2.x^3\)
c) \(\left(\dfrac{-9}{2}\right)^3.3xy\left(4a^2x^3\right)\left(4\dfrac{1}{3}ay^2\right)\)
d) \(\left(\dfrac{1}{3}xy^2\right)^3\left(\dfrac{-3}{7}x^4y\right)^2\)
Tìm x biết
a) \(\dfrac{-2}{5}+\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=\dfrac{-7}{6}\)
b) \(\left(\dfrac{1}{4}x-1\right)+\left(\dfrac{2}{3}x-2\right)-\left(\dfrac{5}{8}x+1\right)=5\)
Ai nhanh 3 tick nha
