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Ta có: \(-3x=5y\Rightarrow\dfrac{-3x}{15}=\dfrac{5y}{15}\Rightarrow\dfrac{x}{-5}=\dfrac{y}{3}\Rightarrow\dfrac{x}{35}=\dfrac{y}{-21}\)(1)
\(-2y=7z\Rightarrow\dfrac{-2y}{14}=\dfrac{7z}{14}\Rightarrow\dfrac{y}{-7}=\dfrac{z}{2}\Rightarrow\dfrac{y}{-21}=\dfrac{z}{6}\) (2)
Từ (1) và (2) ta có: \(\dfrac{x}{35}=\dfrac{y}{-21}=\dfrac{z}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{35}=\dfrac{y}{-21}=\dfrac{z}{6}=\dfrac{2x-3y+z}{2\cdot35-3\cdot-21+6}=\dfrac{42}{139}\)
\(\Rightarrow\dfrac{x}{35}=\dfrac{42}{139}\Rightarrow x=\dfrac{1470}{139}\)
\(\Rightarrow\dfrac{y}{-21}=\dfrac{42}{139}\Rightarrow y=-\dfrac{882}{139}\)
\(\Rightarrow\dfrac{z}{6}=\dfrac{42}{139}\Rightarrow z=\dfrac{252}{139}\)
AC = \(\sqrt{BC^2-AB^2}\) = 12 (cm)
AD là trung tuyến, Ta có: \(AD^2=\dfrac{1}{2}\left(AB^2+AC^2\right)-\dfrac{BC^2}{4}\)
=> AD = \(\sqrt{\dfrac{1}{2}\left(AB^2+AC^2\right)-\dfrac{BC^2}{4}}\)
=> AD = 6.5
Tương tự với BE, CF ta đc:
BE = \(\sqrt{61}\)
CF = \(\dfrac{\sqrt{601}}{2}\)
Đây chắc ko phải bài lớp 7
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}\)
\(\dfrac{x^2}{8:x}=\dfrac{y^2}{64:y}=\dfrac{x^2+y^2}{8:x+64:y}=\dfrac{20}{8:x+64:y}\)
\(8:x+64:y\in\left\{-1;1;-2;2;-4;4;-5;5;-10;10;-20;20\right\}\)
Bạn kẻ bảng và xét từng trường hợp một, hoặc dùng phương pháp loại trừ để giảm số lượng thử trường hợp, khi đó ta tìm được.
\(\left\{{}\begin{matrix}x=-2\\y=-4\end{matrix}\right.or\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
I. Supply the correct tenses or forms in brackets (2pt ).
1. She’d like ( buy ) some bananas.
..................to buy...............................................................................
2. I enjoy ( listen ) in my free time.
.................listening................................................................................
3. Lan ( be ) very tired when she ( catch ) a bad cold two days ago.
................was-caught.........................................................................
4. Everybody ( wait ) for the president in the hall now.
....................is waiting.............................................................................
5. You should ( finish ) your homework before watching TV.
...................finish..............................................................................
6. Lan ( be )fourteen years old next week?
....................Will-be.............................................................................
7. She never ( allow ) her children to stay up late.
....................allows...........................................................................
II. Match the answers in column B to the questions in column A.
A
1. Would you like to play chess?
2. What is his height?
3. What do you do in your free time ?
4. What sport do you like best ?
B
a.1,5 meters
b.I take part in outdoor activities with friends.
c.Badminton.
d.I’d like to but I can’t. I have to go now.
1 - d, 2 - a, 3 - b, 4 - c
Read the passage carefully. Then do the exercises below. (2pts )
Well, I get up early. I always get up at 6.15 and I have a small breakfast. Then, my dad takes me to the swimming pool. I practice every day. I usually arrive at the pool at 7.20 and leave the pool at 8.20. After that, I go to school. Our lessons start at 9.10 and I don’t go home for lunch. I have my lunch at school.
We finish school at 3.45 and I go to the swimming pool again. I swim from 4.30 to 6.00. I usually have supper at 7.00, then I do my homework or watch TV. I sometimes write letters to my pen pals in Spain and Greece, but I always go to bed at 9.45. My friends go to bed at eleven o’clock or midnight! But I want to be a champion swimmer, so I go to bed early.
v True ( T) or False ( F ) (1pt)
1. The writer’s school day lasts six hours and thrity – five minutes. T
2. The writer always writes letters to his pen pals in Spain and Greece. F
v Answer the question. (1pt).
1. What does the write do after supper ?
The writer does his homework or watch T, and sometimes writes letter to pen pals
2. How often does the writer go to the swimming pool?
The writer goes to the swimming pool everyday
1. Mr Robinson speaks Vietnamese very ________.
A. fluent B. fluently C. influent D. fluency
2. A lot of people love football. ________.
A. So do I B. So am I C. Neither do I D. Neither am I
3. Don't let the children play with knives. They are ________.
A. danger B. dangers C. dangerous D. dangerously
4. Nam isn't going to the movie theatre. Trung ________.
A. isn't, too B. isn't, either C. isn't, neither D. is, too
5. I like________ cartoons on television.
A. watch B. to watch C. watched D. watching
6. My mother cycles________.
A. slowly B. slow C. lately D. usually
7. He was absent ________ school yesterday.
A. at B. away C. for D. from
8. He didn't do ________ yesterday.
A. morning exercise B. morning exercises
C. exercise morning D. exercises morning
Ta có \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{a+c}{b}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, thu được:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{a+c}{b}=\dfrac{a+b+b+c+a+c}{c+a+b}\) \(=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
Trừ theo vế 2 hệ thức đầu tiên, ta có
\(\left(a+b\right)-\left(b+c\right)=2c-2a\)
\(\Leftrightarrow a-c=2\left(c-a\right)\)
\(\Leftrightarrow3\left(c-a\right)=0\)
\(\Leftrightarrow c=a\)
Hoàn toàn tương tự, ta có \(a=b=c\)
Do đó \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=1\)
Vậy \(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
\(=8\)
a; Xét tam giác ABD và tam giác EBD có:
\(\widehat{ABD}\) = \(\widehat{EBD}\) (vì BD là phân giác góc ABC)
AB = BE (gt)
Cạnh BD chung
⇒ \(\Delta\)ABD = \(\Delta\)EBD (c-g-c)
⇒\(\widehat{BED}\) = \(\widehat{BAD}\) = 900
⇒DE \(\perp\) BC
b; Xét tam giác BEF và tam giác ABC có:
\(\widehat{BAC}\) = \(\widehat{BEF}\) = 900
AB = BE (gt)
\(\widehat{ABE}\) chung
⇒ \(\Delta\)FBE = \(\Delta\)CBA (g-c-g)
⇒ BC = BF
BC = BE + EC = AB + AF
⇒ AF = EC
c; Xét tam giác BCF có BC = BF (cmt)
⇒ \(\Delta\)BCF cân tại B
BD là phân giác của góc B ⇒ BD là trung tuyến tam giác BCF
⇒BD \(\equiv\) BI ⇒ B;D;I thẳng hằng (vì qua một đỉnh chỉ kẻ được một đường trung tuyến của tam giác)
d; \(\widehat{AEB}\) = \(\widehat{EAC}\) + \(\widehat{ECA}\) (góc ngoài của tam giác bằng hai góc trong không kề với nó)
Xét tam giác ABE có: AB = BE (gt)
⇒ \(\Delta\)ABE cân tại B
⇒ \(\widehat{BAE}\) = \(\widehat{AEB}\) = \(\widehat{EAC}\) + \(\widehat{ECA}\) (đpcm)