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\(\dfrac{\dfrac{7}{13}+\dfrac{7}{14}-\dfrac{7}{15}}{\dfrac{8}{13}+\dfrac{8}{14}-\dfrac{8}{15}}-\dfrac{\dfrac{5}{11}-\dfrac{5}{13}+\dfrac{5}{15}}{\dfrac{8}{11}-\dfrac{8}{13}+\dfrac{8}{15}}\)
\(=\dfrac{7\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}\right)}{8\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}\right)}-\dfrac{5\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{15}\right)}{8\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{15}\right)}\)
\(=\dfrac{7}{8}-\dfrac{5}{8}\)
\(=\dfrac{2}{8}=\dfrac{1}{4}\)
a. Theo t/c dãy tỉ số = nhau:
\(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{42}{7}=6\)
=>\(\frac{x}{2}=6\Rightarrow x=6.2=12\)
=>\(\frac{y}{5}=6\Rightarrow y=6.5=30\)
Vậy x=12; y=30.
b. \(\left|x-0,25\right|-\frac{5}{6}=1\frac{2}{3}\)
=> \(\left|x-0,25\right|=1\frac{2}{3}+\frac{5}{6}\)
=> \(\left|x-0,25\right|=\frac{5}{2}=2,5\)
+) x-0,25=2,5
=> x=2,5+0,25
=> x=2,75
+) x-0,25=-2,5
=> x=-2,5+0,25
=> x=-2,25
Vậy x \(\in\){-2,25; 2,75}.
c. y=kx
=> -17=k.8
=> k=-17/8
Vậy hệ số tỉ lệ là -17/8.
a) \(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{42}{7}=6\)
=> x=12 ; y = 30
b) \(\left|x-0,25\right|-\frac{5}{6}=1\frac{2}{3}=>\left|x-0,25\right|=\frac{5}{3}+\frac{5}{6}=\frac{5}{2}=2,5\)
=> x-0,25 = 2,5 hoac: -2,5
=> x = 2,75 hoac x= -2,25
Vay: x la { 2,75 ; -2,25 }
c) Ti le gi vay ban.
Neu thuan thi he so ti le la: \(-\frac{17}{8}\)
Neu nghich thi he so ti le la : -136
a) \(\sqrt{25}-\sqrt{\dfrac{49}{4}}+\sqrt{0,25}\)
\(=5-3,5+0,5\)
\(=2\)
b) \(\left(-\dfrac{25}{24}-\dfrac{31}{42}\right)-\left(\dfrac{-7}{27}-\dfrac{3}{42}\right)\)
\(=\dfrac{-25}{24}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}\)
\(=\left(\dfrac{-25}{24}+\dfrac{7}{27}\right)+\left(-\dfrac{31}{42}+\dfrac{3}{42}\right)\)
\(=-\dfrac{169}{216}+-\dfrac{2}{3}=-1\dfrac{97}{216}\)
c) \(\dfrac{103}{10}-\left(9,5-0,25.18\right):0,5\)
\(=10,3-\left(9,5-0,25.18\right):0,5\)
\(=10,3-5:0,5\)
\(=0,3\)
d) \(\left(10^3+15.10^2+5^3\right):7\)
\(=\left(10^3+10^2\right)\left(15+5^3\right):7\)
\(=\left(1100.140\right):7\)
\(=22000\)
\(M=\dfrac{2^2.3^2.4^2.....20^2}{1.3.2.4.3.5.4.6.5.7.6.8.7.9....19.21}=\)
\(=\dfrac{2^2.3^2.4^2....20^2}{1.2.3^2.4^2....19^2.20.21}=\dfrac{2.20}{21}=\dfrac{40}{21}\)
\(N=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{10}{11}=\dfrac{1}{11}\)
\(5^{200}=\left(5^2\right)^{100}=25^{100}\)
\(3< 25=>3^{100}< 25^{100}=>3^{100}< 5^{200}\)
\(\frac{75^{20}}{45^{10}.25^{15}}=\frac{25^{20}.3^{20}}{3^{10}.3^{10}.5^{10}.25^{15}}=\frac{25^{20}}{25^5.25^{15}}=1\)
\(=>75^{20}=45^{10}.25^{15}\left(dpcm\right)\)
P/S:nếu a=b=>a:b=1 mk làm theo cách đó cho nhanh mà bn ghi sai đề r
Ta có : \(A=\left(8.3^3.3^3\right).49.7^{13}\)
\(\Rightarrow A=\left(2^3.3^3.3^3\right).\left(7^2.7^{13}\right)\)
\(\Rightarrow A=\left(2.3.3\right)^3.7^{15}=6^3.3^3.7^{15}\)
\(\Rightarrow A=\left(6^3.7^3\right).3^3.7^{12}=\left(6.7\right)^3.27.7^{12}\)
\(\Rightarrow A=42^3.27.7^{12}\)
Vì trong tích A có chứa thừa số 42 => \(A⋮42\left(đpcm\right)\)
\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)
\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)
\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)
\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)