giải giúp vs ạ
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Part 1:
1. with
2. does
3. after
4. if
5. where
6. I won't
Part 2:
1B 2F 3A 4E 5C 6D
Part 3:
1. invited
2. will play
3. have just won
4. be cleaned
5. to take
6. beautifully
7. impression
Part 4:
1. read -> reading
2. disappointing -> disappointed
3. environment -> environmental
4. who -> that
Part 5:
1. the bad weather
2. are made to study
3. were good at learning
4. going to the English
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\(1,\\ a,=6x^4y^4-x^3y^3+\dfrac{1}{2}x^4y^2\\ b,=4x^3+5x^2-8x^2-10x+12x+15\\ =4x^3-3x^2+2x+15\\ 2,\\ a,=7\left(x^2-6x+9\right)=7\left(x-3\right)^2\\ b,=\left(x-y\right)^2-36=\left(x-y-6\right)\left(x-y+6\right)\\ 3,\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow x\left(x-0,6\right)\left(x+0,6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,6\\x=-0,6\end{matrix}\right.\)
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5: Để A nguyên thì \(x^2-4+6⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{-1;-3;0;-4;1;-5;4;-8\right\}\)
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I
1 A phát âm là /t/ còn lại phát âm là /d/
2 B phát âm là /ai/ còn lại phát âm là /i/
3 B phát âm là /t∫/ còn lại phát âm là /∫/
II
1 C âm 1 còn lại âm 2
2 C âm 2 còn lại âm 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2-x-2\right)\sqrt{x-1}=0\left(đk:x\ge1\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\sqrt{x-1}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\) (do x+1>0)
Ý B.
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\(3x\left(x-y\right)+x-y\)
\(=3x\left(x-y\right)+1\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+1\right)\)
a/ Kẽ \(AK\perp AB\)tại A
\(\Rightarrow\widehat{CAK}=\widehat{ABC}\left(1\right)\)
Ta lại có \(CH\perp AB\)
\(\Rightarrow CH//AK\)
\(\Rightarrow\widehat{ACH}=\widehat{CAK}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\widehat{CBA}=\widehat{ACH}\)
b/ Ta có:
\(\widehat{ACH}=\widehat{CBH}\)
\(\widehat{CHA}=\widehat{BHC}=90^o\)
\(\Rightarrow\Delta ACH\)~ \(\Delta CBH\)
\(\Rightarrow\frac{CH}{BH}=\frac{AH}{CH}\)
\(\Rightarrow CH^2=AH.BH\)