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c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
\(A=\dfrac{7^5}{7+7^2+7^3+7^4}=\dfrac{7^5}{\left(7+7^4\right)+\left(7^2+7^3\right)}=\dfrac{7^5}{7^5+7^5}=7^5\)
\(B=\dfrac{5^5}{5+5^2+5^3+5^4}=\dfrac{5^5}{\left(5+5^4\right)+\left(5^2+5^3\right)}=\dfrac{5^5}{5^5+5^5}=5^5\)
Vì 7 > 5 nên \(7^5>5^5\)
Vậy A > B
(Nhớ cho mik một tick nha cảm ơn bạn nhìu :3)
\(1-\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=1-\left(\frac{1}{90}+\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)
\(=1-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=1-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=1-\left(1-\frac{1}{10}\right)\)
\(=1-\frac{9}{10}\)
\(=\frac{1}{10}\)
a)×+1/ 53 + ×+2 /52 + ×+3/ 51+3 = 0
\(\Rightarrow\frac{x+1}{53}+1+\frac{x+2}{52}+1+\frac{x+3}{51}+1+\frac{3\left(x+54\right)}{\left(x+54\right)}=0\)
\(\Rightarrow\frac{x+54}{53}+\frac{x+54}{52}+\frac{x+54}{51}+\frac{x+54}{\frac{1}{3}\left(x+54\right)}=0\)
\(\Rightarrow\left(x+54\right)\left(\frac{1}{53}+\frac{1}{52}+\frac{1}{51}+\frac{1}{\frac{1}{3}\left(x+54\right)}\right)=0\)
\(\Rightarrow x+54=0\).Do \(\frac{1}{53}+\frac{1}{52}+\frac{1}{51}+\frac{1}{\frac{1}{3}\left(x+54\right)}\ne0\)
=>x=-54
b)×-2/ 72 + ×-3/ 71 + ×-4/ 70 -3 = 0
\(\Rightarrow\frac{x-2}{72}-1+\frac{x-3}{71}-1+\frac{x-4}{70}-1-\frac{3\left(x-74\right)}{x-74}=0\)
\(\Rightarrow\frac{x-74}{72}+\frac{x-74}{71}+\frac{x-74}{70}-\frac{x-74}{\frac{1}{3}\left(x-74\right)}=0\)
\(\Rightarrow\left(x-74\right)\left(\frac{1}{72}+\frac{1}{71}+\frac{1}{70}-\frac{1}{\frac{1}{3}\left(x-74\right)}\right)=0\)
\(\Rightarrow x-74=0\).Do \(\frac{1}{72}+\frac{1}{71}+\frac{1}{70}-\frac{1}{\frac{1}{3}\left(x-74\right)}\ne0\)
=>x=74
c)×+5/ 81 + ×+4/ 41 + ×-7/ 31 + 6 = 0
\(\Rightarrow\frac{x+5}{81}+1+\frac{x+4}{41}+2+\frac{x-7}{31}+3+\frac{6\left(x+86\right)}{x+86}=0\)
\(\Rightarrow\frac{x+86}{81}+\frac{x+86}{41}+\frac{x+86}{31}+\frac{x+86}{\frac{1}{6}\left(x+86\right)}=0\)
\(\Rightarrow\left(x+86\right)\left(\frac{1}{81}+\frac{1}{41}+\frac{1}{31}+\frac{1}{\frac{1}{6}\left(x+86\right)}\right)=0\)
\(\Rightarrow x+86=0\).Do \(\frac{1}{81}+\frac{1}{41}+\frac{1}{31}+\frac{1}{\frac{1}{6}\left(x+86\right)}\ne0\)
=>x=-86
d)tương tự nhé
neu bot mot canh hinnh vuong di 7 m va bot mot canh khac di 25 m thi duoc mot hinh chu nhat co chieu dai gap 3 lan chieu rong tinh chu vi va dien h hinh vuong
a)\(x=\dfrac{-14\cdot52}{72}\\ x=\dfrac{-91}{9}\)
b)\(x=\dfrac{120\cdot7.2}{70}\\ x=\dfrac{432}{35}\)
c)\(x=\dfrac{2\dfrac{2}{3}\cdot8.5}{5}\\ x=\dfrac{68}{15}\)
d)\(x=\dfrac{4\dfrac{2}{5}\cdot9.5}{8}\\ x=\dfrac{209}{40}\)
Hình như bạn chép sai đề, mình sửa nhé :
\(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{26.31}\\ =>\dfrac{S}{5}=\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{26.31}\\ =>\dfrac{S}{5}=\dfrac{6-1}{1.6}+\dfrac{11-6}{6.11}+\dfrac{16-11}{11.16}+...+\dfrac{31-26}{26.31}\\ =>\dfrac{S}{5}=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{26}-\dfrac{1}{31}=1-\dfrac{1}{31}=\dfrac{30}{31}\\ =>S=\dfrac{30}{31}.5=\dfrac{150}{31}\)
\(5^2\cdot7^2=25\cdot49=1225\)
\(\left(5^2\right).\left(7^2\right)\)
\(=\left(5.7\right)^2\)
\(=35^2\)
\(=1225\)