Giúp e vs ạ! Em cảm ơn!
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Where's John? He is listening to a new CD in his room. He speaks German so well because he comes from Germany. She is holding some roses. They smell lovely. Where is Molly? She is feeding her cat downstairs. Does she need to go and see a doctor? Is your sister wearing sunglasses? He frequently does yoga. We are moving to Canada in August. I don’t like to take selfies. Megan is going on holiday to Cornwall this summer. When does the film start?
Câu a:
125\(^5\) + 4.5\(^{12}\)
= 125\(^5\) + 4.(5\(^3\))\(^4\)
= 125\(^5\) + 4.125\(^4\)
= 125\(^4\).(125 + 4)
= 125\(^4\).129 ⋮ 129 (đpcm)
a: \(125^5+4\cdot5^{12}\)
\(=\left(5^3\right)^5+4\cdot5^{12}\)
\(=5^{15}+4\cdot5^{12}=5^{12}\left(5^3+4\right)=5^{12}\cdot129\) ⋮129
b: \(1+7+7^2+\cdots+7^{101}\)
\(=\left(1+7\right)+\left(7^2+7^3\right)+\cdots+\left(7^{100}+7^{101}\right)\)
\(=\left(1+7\right)+7^2\left(1+7\right)+\cdots+7^{100}\left(1+7\right)\)
\(=8\left(1+7^2+\cdots+7^{100}\right)\) ⋮8
c: \(2+2^2+2^3+\cdots+2^{100}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\cdots+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{97}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+\cdots+2^{97}\right)\) ⋮5
\(2+2^2+2^3+\cdots+2^{100}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+\cdots+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+\cdots+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(=31\cdot\left(2+2^6+\cdots+2^{96}\right)\) ⋮31
Olm chào em, em chọn học bài, chọn lớp, chọn môn tiếng anh, như vậy em có thể học các bài giảng tiếng anh của Olm, em nhé. Cảm ơn em đã đồng hành cùng Olm. Chúc em luôn học tập hiệu quả và có những giây phút giao lưu thú vị cùng cộng đồng tri thức Olm.
a: \(\left(-\frac54x+3,25\right)\left\lbrack\frac35-\left(-\frac52x\right)\right\rbrack=0\)
=>\(\left(\frac54x-\frac{13}{4}\right)\left(\frac52x+\frac35\right)=0\)
=>\(\left[\begin{array}{l}\frac54x-\frac{13}{4}=0\\ \frac52x+\frac35=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x=\frac{13}{4}\\ \frac52x=-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{13}{4}:\frac54=\frac{13}{5}\\ x=-\frac35:\frac52=-\frac{6}{25}\end{array}\right.\)
b: \(\left(-\frac72x+1,75\right)\left\lbrack\frac45-\left(-\frac53x\right)\right\rbrack=0\)
=>\(\left[\begin{array}{l}-\frac72x+1,75=0\\ \frac45-\left(-\frac53x\right)=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}-\frac72x=-1,75=-\frac74\\ \frac53x=-\frac45\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{-7}{4}:\frac{-7}{2}=\frac24=\frac12\\ x=-\frac45:\frac53=-\frac45\cdot\frac35=-\frac{12}{25}\end{array}\right.\)
c: \(\left(x^2-4\right)\left(x+\frac27\right)=0\)
=>\(\left[\begin{array}{l}x^2-4=0\\ x+\frac27=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=4\\ x=-\frac27\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-2\\ x=-\frac27\end{array}\right.\)
d: \(\left(25-x^2\right)\left(5x-\frac59\right)=0\)
=>\(\left[\begin{array}{l}25-x^2=0\\ 5x-\frac59=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=25\\ 5x=\frac59\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-5\\ x=\frac19\end{array}\right.\)