làm tự luận giúp e ạ
K
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2022
5.
\(y=\dfrac{2x-1}{1-x}\Rightarrow y'=\dfrac{\left(2x-1\right)'\left(1-x\right)-\left(1-x\right)'\left(2x-1\right)}{\left(1-x\right)^2}\)
\(=\dfrac{2\left(1-x\right)+\left(2x-1\right)}{\left(1-x\right)^2}=\dfrac{1}{\left(1-x\right)^2}=\dfrac{1}{\left(x-1\right)^2}\)
9.
\(\lim\limits\dfrac{2n^2+4}{3-n^2}=\lim\dfrac{2+\dfrac{4}{n^2}}{\dfrac{3}{n^2}-1}=\dfrac{2+0}{0-1}=-2\)
giúp em trắc nghiệm thui ạ tự luận em tự làm cảm ơn ạ
KV
18 tháng 10 2023
Dạng 1
Đại số
Câu 1. B
Câu 2. B
Câu 3. D
Câu 4. B
Câu 5. B
Câu 6. C
12 tháng 9 2021
\(n_{HCl}=\dfrac{3,65\%.25}{36,5}=0,025\left(mol\right)\\ Đặt:A\left(I\right)\\ 2A+2HCl\rightarrow2ACl+H_2\\ n_A=n_{HCl}=0,025\left(mol\right)\\ M_A=\dfrac{0,575}{0,025}=23\left(\dfrac{g}{mol}\right)\\ \Rightarrow A\left(I\right):Natri\left(Na=23\right)\)
=> Chọn B
AH
Akai Haruma
Giáo viên
16 tháng 8 2021
Bài 1:
a.
\(=(6\sqrt{5}-3\sqrt{5}+3\sqrt{5}-6\sqrt{5}):\sqrt{5}=0:\sqrt{5}=0\)
b.
\(=3\sqrt{a}-\frac{1}{2a}\sqrt{(3a)^2.a}+\sqrt{a^2}.\sqrt{4^2}.\sqrt{\frac{1}{a}}-\frac{2}{a^2}.\sqrt{(6a^2)^2.a}\)
\(=3\sqrt{a}-\frac{1}{2a}.3a\sqrt{a}+4\sqrt{a^2.\frac{1}{a}}-\frac{2}{a^2}.6a^2\sqrt{a}\)
\(=3\sqrt{a}-1,5\sqrt{a}+4\sqrt{a}-12\sqrt{a}=-6,5\sqrt{a}\)
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Cau 23: B
Câu 24: A