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\(\left(125^3\cdot7^5-175^5:5\right):2001^{2002}\)
\(\left(5^9\cdot7^5-7^5\cdot5^{10}:5\right):2001^{2002}\)
\(\left(5^9\cdot\left(7^5-7^5\right)\right):2001^{2002}\)
\(\left(5^9\cdot0\right):2001^{2002}\)
\(0:2001^{2002}\)
= \(0\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy \(x=97\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy x=97
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
\(0,25x+175\%x=x+\dfrac{9}{7}\\ x\left(0,25+1,75\right)=x+\dfrac{9}{7}\\ 2x=x+\dfrac{9}{7}\\ \dfrac{9}{7}=2x-x=x\\ x=\dfrac{9}{7}\)
a) \(0,25x+175\%x=x+\frac{9}{7}\)
\(0,25x+175\%x+x=\frac{9}{7}\)
\(0,25x+\frac{7}{4}x+x=\frac{9}{7}\)
\(\left(0,25+\frac{7}{4}+1\right)x=\frac{9}{7}\)
\(3x=\frac{9}{7}\)
\(x=\frac{9}{7}\div3\)
\(x=\frac{9}{7}\times\frac{1}{3}\)
\(x=\frac{3}{7}\)
vậy \(x=\frac{3}{7}\)
b)\(2x-\frac{3}{2}-x=1\frac{1}{5}\)
\(2x-x=1\frac{1}{5}+\frac{3}{2}\)
\(x=\frac{27}{10}\)
vậy \(x=\frac{27}{10}\)
Đặt
\(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(S=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(S=\dfrac{49}{99}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{9999}=\dfrac{49}{99}\)
\(\Rightarrow\dfrac{1}{x}=\dfrac{50}{101}\Leftrightarrow50x=101\Leftrightarrow x=\dfrac{101}{50}\)
Đặt \(M=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(M=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(M=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)\)
\(M=\dfrac{1}{2}\cdot\dfrac{98}{99}\)
\(M=\dfrac{49}{99}\)
A và B dễ
Bài 2:
sai đề bài vì ngay từ cái phép tính đầu đã ko theo quy luật rồi
\(A=\frac{-3}{5}-\frac{2}{5}+2\)
\(A=-1+2=1\)
\(B=\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}=\frac{1}{4}\)
nÀ NÍ sao lại = đây là dấu trừ hay cộng 1/4
1/2.x-3/5=-4/5
1/2.x=-4/5+3/5
1/2.x=-1/5
x=-1/5:1/2
x=-2/5
kl:.....
câu đầu mik tính ra sốn to lắm
câu cuối mik tính ko chia hết nên chỉ làm đc câu giữa
Mk sửa đề nha :
20202020 x ( 710 : 78 - 3 x 24 - 22020 : 22020 )
= 20202020 x ( 72 - 48 - 20 )
= 20202020 x ( 49 - 48 - 1 )
= 20202020 x 0
= 0
Study well ! >_<
Bài làm
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\frac{1}{x+2}=\frac{1}{2016}\)
\(\Rightarrow x+2=2016\)
\(x=2014\)
5x+2.7y=175x
=>5x.25.7y=175x
=>25.7y=35x
=>52.7y=35x
=>352.7y-2=35x
=>7y-2=35x-2
=>x-2=y-2=0
=>x=y=2
vậy x=y=2