mn ơi giúp mik câu này vs ạ
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\(P=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{3-x}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)
\(=\left(\dfrac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10}{x+3}-\dfrac{x+3}{x+3}\right)\)
\(=\left(\dfrac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)
\(=\left(\dfrac{-2x-14}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x+7}{x+3}\right)\)
\(=\dfrac{-2\left(x+7\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+7}\)
\(=\dfrac{-2}{x-3}\)
đk : x khác -3 ; 3 ; -7
\(P=\left(\dfrac{x^2+1+x\left(x-3\right)+5x+15}{x^2-9}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)
\(=\dfrac{2x^2+1+2x+15}{x^2-9}:\dfrac{x+7}{x+3}=\dfrac{2x^2+2x+16}{\left(x-3\right)\left(x+7\right)}\)
a)\(2Ca+O_2\underrightarrow{t^o}2CaO\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)
\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
2: Để (d)//y=(m2+1)x-4 thì \(\left\{{}\begin{matrix}m^2=1\\m-5\ne-4\end{matrix}\right.\Leftrightarrow m=1\)
1
Với \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\)
\(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\left(\dfrac{x^2+2x+1}{4x^4-4x^2+1}\right)\\ =\left(\dfrac{\left(x-1\right)\left(x+1\right)}{\left(2-x\right)\left(x+1\right)}+\dfrac{x^2}{\left(x+1\right)\left(2-x\right)}\right)\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{x^2-1+x^2}{\left(x+1\right)\left(2-x\right)}\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{\left(2x^2-1\right)\left(x+1\right)^2}{\left(x+1\right)\left(2-x\right)\left(2x^2-1\right)^2}\\ =\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}\)
2
Để M = 0 thì \(\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}=0\Rightarrow x+1=0\Rightarrow x=-1\) (loại)
Vậy không có giá trị x thỏa mãn M = 0
1) \(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\cdot\dfrac{x^2+2x+1}{4x^4-4x^2+1}\) (ĐK: \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\))
\(M=\left(\dfrac{-\left(x-1\right)}{x-2}-\dfrac{x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-\left(x^2-1\right)-x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-x^2+1-x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\dfrac{-2x^2+1}{\left(x-2\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\dfrac{-\left(2x^2-1\right)\left(x+1\right)^2}{\left(x-2\right)\left(x+1\right)\left(2x^2-1\right)^2}\)
\(M=\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}\)
2) Ta có: \(M=0\)
\(\Rightarrow\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}=0\)
\(\Leftrightarrow-\left(x+1\right)=0\)
\(\Leftrightarrow-x=1\)
\(\Leftrightarrow x=-1\left(ktm\right)\)
1. My brother (listen)listened to music last night.
2. When I came to see Peter, he (do)was doinghis homework.
3. Mr Manh wishes he (be)were in Paris now.
4. Up to present, Mr John (write)has written more than one hundred novels.
5. Tim (always blame)is always blaming his faults on the others.
6. Mary (write)writes her pen pal once a week.
7. When the milkman came, my family (have)was having dinner.
8. Lan and Maryam (not respond)haven't responded each other for along time.
9. If she (study)studies harder, she (pass)will pass the examination.
10. It's important that you arm (operate)be operated now
e tk:
https://vietjack.com/van-mau-lop-7/images/so-do-tu-duy-duc-tinh-gian-di-cua-bac-ho-a01.PNG