\(\text{a) }\dfrac{2x^2-x-1}{2}-3x^2+x+4=\left(5-x\right)\left(2x+4\right)\\ \Leftrightarrow\left(\dfrac{2x^2-x-1}{2}-3x^2+x+4\right)2=\left(5-x\right)\left(2x+4\right)2\\ \Leftrightarrow2x^2-x-1-6x^2+2x+8=\left(5-x\right)\left(4x+8\right)\\ \Leftrightarrow-4x^2+x+7=20x+40-4x^2-8x\\ \Leftrightarrow-4x^2+x+4x^2-12x=40-7\\ \Leftrightarrow-11x=33\\ \Leftrightarrow x=-3\\ \text{Vậy }S=\left\{-3\right\}\)

\(\text{b) }\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\dfrac{\left(x-1\right)\left(2x+4\right)}{2}+1\\ \Leftrightarrow\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1=\left(x-1\right)\left(x+2\right)+1\\ \Leftrightarrow\left(\dfrac{\left(2x-5\right)\left(3x+7\right)}{4}+2x-1\right)4=\left(x^2-x+2x-2+1\right)4\\ \Leftrightarrow\left(2x-5\right)\left(3x+7\right)+8x-4=\left(x^2+x-1\right)4\\ \Leftrightarrow6x^2-15x+14x-35+8x-4=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-39=4x^2+4x-4\\ \Leftrightarrow6x^2+7x-4x^2-4x-39+4=0\\ \Leftrightarrow2x^2+3x-35=0\\ \Leftrightarrow2x^2+10x-7x-35=0\\ \Leftrightarrow\left(2x^2+10x\right)-\left(7x+35\right)=0\\ \Leftrightarrow2x\left(x+5\right)-7\left(x+5\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-5\end{matrix}\right.\\ \\ \text{Vậy }S=\left\{\dfrac{7}{2};-5\right\}\)

\(\text{c) }8x^3-1=8x^2+4x+2\\ \Leftrightarrow\left(2x-1\right)\left(4x^2+2x+1\right)=2\left(4x^2+2x+1\right)\\ \Leftrightarrow2x-1=2\\ \Leftrightarrow2x=3\\ \Leftrightarrow x=\dfrac{3}{2}\\ \text{Vậy }S=\left\{\dfrac{3}{2}\right\}\)

\(\text{d) }\left(x^2+x+1\right)\left(x^2-x+1\right)=x^6-1\\ \Leftrightarrow\left(x^3+1\right)\left(x^3-1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x^2-x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\\ \Leftrightarrow\left(x+1\right)\left(x-1\right)=1\\ \Leftrightarrow x^2-1=1\\ \Leftrightarrow x^2=2\\ \Leftrightarrow x=\sqrt{2}\\ \text{Vậy }S=\left\{\sqrt{2}\right\}\)

\(\text{e) }\left(x^3+2x\right)\left(x^2+4\right)=\left(x^2+6x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+2x^2+4x^2+8\right)\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[\left(x^2+2x^2\right)+\left(4x^2+8\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left[x^2\left(x^2+2\right)+4\left(x^2+2\right)\right]\left(3-2x\right)\\ \Leftrightarrow x\left(x^2+2\right)\left(x^2+4\right)=\left(x^2+4\right)\left(x^2+2\right)\left(3-2x\right)\\ \Leftrightarrow x=3-2x\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\\ \text{Vậy }S=\left\{1\right\}\)

f) Kiểm tra lại hạng tử thứ 2 ở vế phải.

2

TICK VOI NHA

\(\text{a) }ĐKXĐ:x\ne\pm5\)

Với \(x\ne\pm5\), ta có:

\(A=\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\\=\dfrac{2x}{\left(x+5\right)\left(x-5\right)}-\dfrac{5}{x-5}-\dfrac{1}{x+5}\\ =\dfrac{2x}{\left(x+5\right)\left(x-5\right)}-\dfrac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\\ =\dfrac{2x-5x-25-x+5}{\left(x+5\right)\left(x-5\right)}\\ =\dfrac{-4x-20}{\left(x+5\right)\left(x-5\right)}\\ =\dfrac{-4\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{4}{5-x}\)

Vậy \(A=\dfrac{4}{5-x}\) với \(x\ne\pm5\)

b) Với \(x\ne\pm5\)

Để A nhận giá trị nguyên

thì \(\Rightarrow\dfrac{4}{5-x}\in Z\)

\(\Rightarrow4\text{ }⋮\text{ }5-x\\ \Rightarrow5-x\inƯ_{\left(4\right)}\)

Mà \(Ư_{\left(4\right)}=\left\{\pm1;\pm2;\pm4\right\}\)

Lập bảng giá trí:

Vậy với \(x=\left\{9;7;6;4;3;1\right\}\)

thì A nhận giá trị nguyên.

My mother said that she would rather ____ to Hoi An than Nha Trang.

A. to travel

B. travelling

C. not to travel

D. travel

5421 nha

nhớ ko tick

\(\text{a) }pthh:CH_4+2O_2\text{ }\underrightarrow{t^0}\text{ }CO_2+2H_2O\left(1\right)\)

\(\text{b) }n_{CH_4}=\dfrac{m}{M}=\dfrac{6,4}{16}=0,4\left(mol\right)\)

Theo \(pthh\) \(\left(1\right)\), ta có :

\(n_{O_2}=2n_{CH_4}=2\cdot0,4=0,8\left(mol\right)\\ \Rightarrow V_{O_2}=n\cdot22,4=0,8\cdot22,4=17,92\left(l\right)\)

Theo \(pthh\) \(\left(1\right)\), ta có :

\(n_{H_2O}=2\cdot n_{CH_4}=2\cdot0,4=0,8\left(mol\right)\\ \Rightarrow m_{H_2O}=n\cdot M=0,8\cdot18=14,4\left(g\right)\)

\(\text{Ta có : }\%S=50\%\\ \Rightarrow\%O=100-50=50\%\)

\(m_S=\dfrac{64\cdot50}{100}=32\left(g\right)\\m_O=\dfrac{64\cdot50}{100}=32\left(g\right)\\ \Rightarrow n_S=\dfrac{m}{M}=\dfrac{32}{32}=1\left(mol\right)\\ n_O=\dfrac{m}{M}=\dfrac{32}{16}=2\left(mol\right)\)

\(\Rightarrow\) Trong một phân tử có: 1 nguyên tử S; 2 nguyên tử O

\(\Rightarrow CTHH:SO_2\)

Vậy \(CTHH\) cần lập là: \(SO_2\)

\(16x-5x^2-3\\ =15x+x-5x^2-3\\ =\left(15x-5x^2\right)-\left(3-x\right)\\ =5x\left(3-x\right)-\left(3-x\right)\\ =\left(5x-1\right)\left(3-x\right)\)

Số lít nước mắm cửa hàng mua về là:

78 : 60 x 100 = 130 (l nước mắm)

Số lít nước mắm cửa hàng còn lại sau hai ngày bán là:

130 - 78 - 42 = 10 (l nước mắm)

Đáp số: 10 l nước mắm