Bài 1:

Theo đề ta có : A₁=G₁; T₁=3A₁; X₁ = 2T₁

=>\(\left\{{}\begin{matrix}X_1=6A_1\\T_1=3A_1\\G_1=A_1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}A=4A_1\\G=7A_1\end{matrix}\right.\)

Lại có: 2A + 3G= 5800

=> 29A₁ = 5800 => A₁=200

=> G = 1400

6:

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

mà 8<9

nên \(2^{225}< 3^{150}\)

4: \(\left|5x+3\right|>=0\forall x\)

=>\(-\left|5x+3\right|< =0\forall x\)

=>\(-\left|5x+3\right|+5< =5\forall x\)

Dấu = xảy ra khi 5x+3=0

=>x=-3/5

1:

\(\left(2x+1\right)^4>=0\)

=>\(\left(2x+1\right)^4+2>=2\)

=>\(M=\dfrac{3}{\left(2x+1\right)^4+2}< =\dfrac{3}{2}\)

Dấu = xảy ra khi 2x+1=0

=>x=-1/2

Câu 7:

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)

c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)

d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)

Câu 8:

a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)

b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)

c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

Bài 6: 

a, \(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)

b, \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaSO_3}=n_{SO_2}=0,05\left(mol\right)\)

\(\Rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{0,05.74}{100}.100\%=3,7\%\)

c, \(m_{CaSO_3}=0,05.120=6\left(g\right)\)

Câu 1: 

Chương trình bảng tính là phần mềm được thiết kế giúp ghi lại và trình bày thông tin dưới dạng bảng, thực hiện các tính toán cũng như xây dựng các biểu đồ một cách trực quan các số liệu có trong bảng

Mary if she could speak some foreign languages
Lan if she was going to visit her aunt the day after
what I was doing
how she was feeling then
what I usually did in my free time
why he why he didn't come there to meet her
why I was so lazy and naughty
like playing soccer, don't you?
goes to school late, doesn't he?
can swim very well, can't you?
is going to the party, isn't she?
was published in Germany in 1550, wasn't it?
are sold all over the world, aren't they?
have been built this year, haven't they?
was given a book, wasn't he?
was bought by Mrs Brown yesterday, wasn't she?
is used every day, isn't it?
be beautiful sights in this village when I lived here

tách nhỏ câu hỏi ra

\(a,x+\dfrac{1}{2}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{2}\\ x=\dfrac{1}{2}\\ b,-\dfrac{2}{3}-x=1\\x=-\dfrac{2}{3}-1\\ x=-\dfrac{5}{3}\\ d,\dfrac{1}{4}+\dfrac{3}{4}:x=\dfrac{5}{2}\\ \dfrac{3}{4}:x=\dfrac{5}{2}-\dfrac{1}{4}\\ \dfrac{3}{4}:x=\dfrac{9}{4}\\ x=\dfrac{3}{4}:\dfrac{9}{4}\\ x=\dfrac{1}{3}\\ e,\left(x+\dfrac{1}{4}\right)\cdot\dfrac{3}{4}=-\dfrac{5}{8}\\ x+\dfrac{1}{4}=-\dfrac{5}{8}:\dfrac{3}{4}\\ x+\dfrac{1}{4}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{1}{4}\\ x=\dfrac{7}{12}\)

\(g,\dfrac{x-3}{15}=\dfrac{-2}{5}\\ 5\left(x-3\right)=-30\\ x-3=-6\\ x=-6+3\\ x=-3\\ h,\dfrac{x}{-2}=\dfrac{-8}{x}\\ x^2=16\\ x=\pm\sqrt{16}\\ x=\pm4\\ k,\dfrac{x+2}{3}=\dfrac{x-4}{5}\\ 5\left(x+2\right)=3\left(x-4\right)\\ 5x+10=3x-12\\ 5x-3x=-12-10\\ 2x=-22\\ x=-11\)

\(m,\left(2x-1\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Đánh giá giúp nhá cảm ơn