a) Để A có nghĩa <=> \(\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne-5\\x\ne0\\x\ne0;x\ne-5\end{cases}}\) <=> \(\hept{\begin{cases}x\ne-5\\x\ne0\end{cases}}\)

b) A = \(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

A = \(\frac{x\left(x^2+2x\right)+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)

A = \(\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)

A = \(\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

A = \(\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

A = \(\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)

A = \(\frac{x^2+5x-x-5}{2\left(x+5\right)}\)

A = \(\frac{\left(x-1\right)\left(x+5\right)}{2\left(x+5\right)}=\frac{x-1}{2}\)

a) a² - b² + 2a - 2b = (a + b)(a - b) + 2(a - b) = (a - b)(a + b + 2)

b) x² + x - 12 = x² + 4x - 3x - 12 = x(x + 4) - 3(x + 4) = (x - 3)(x + 4)

c) x⁴ + 4 = (x⁴ + 4x² + 4) - 4x² = (x² + 2)² - 4x² = (x² - 2x + 2)(x² + 2x + 2)

                                               Bài làm :

a) a² - b² + 2a - 2b = (a + b)(a - b) + 2(a - b) = (a - b)(a + b + 2)

b) x² + x - 12 = x² + 4x - 3x - 12 = x(x + 4) - 3(x + 4) = (x - 3)(x + 4)

c) x⁴ + 4 = (x⁴ + 4x² + 4) - 4x² = (x² + 2)² - 4x² = (x² - 2x + 2)(x² + 2x + 2)

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

a) \(\left(x-2\right)\left(2x+1\right)\)

\(=2x^2-3x-2\)

b) \(\left(x-2y\right)\left(x+y\right)\)

\(=x^2-xy-2y^2\)

a) \(P=\frac{3\left(x+\sqrt{x}-3\right)}{x+\sqrt{x}-2}+\frac{\sqrt{x}+3}{\sqrt{x}+2}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\) \(\left(x\ge0;x\ne1\right)\)

\(P=\frac{3x+3\sqrt{x}-9}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{3\sqrt{x}+8}{\sqrt{x}+2}\)

b) \(P=\frac{7}{2}\)

\(\Leftrightarrow\frac{3\sqrt{x}+8}{\sqrt{x}+2}=\frac{7}{2}\)

\(\Rightarrow6\sqrt{x}+16=7\sqrt{x}+14\)

\(\Leftrightarrow\sqrt{x}=2\Rightarrow x=4\)

[\(\frac{-75}{59}\).\(\frac{-107}{93}\)]\(\frac{31}{50}\)=\(\frac{2675}{1829}\).\(\frac{31}{50}\)=\(\frac{107}{118}\)

\(\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{1}{6}\cdot\frac{20}{3}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{6-15+\frac{2}{3}}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)

\(=\left[\left(-\frac{150}{107}\right)\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}=\frac{50}{31}\cdot\frac{31}{50}=1\)

\(\frac{\frac{25}{108}.\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\frac{-41}{21}}\)=\(\frac{\frac{5755}{108}+\frac{187}{4}}{\frac{-973}{410}}\)=\(\frac{\frac{8531}{84}}{\frac{-973}{410}}\)=-241,0180

\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230\frac{1}{5}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

\(=\frac{\left(\frac{53}{4}-\frac{59}{27}-\frac{65}{6}\right)\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}:\left(-\frac{41}{21}\right)}\)

\(=\frac{\frac{25}{108}\cdot\frac{1151}{5}+\frac{187}{4}}{\frac{139}{30}\cdot\left(-\frac{21}{41}\right)}=\frac{\frac{2701}{27}}{-\frac{973}{410}}\)

Tính nốt vì số dữ quá , lần sau để số ít thôi

a) Xét tam giác AEQ và tam giác BEC có
EQ=EC
AEQ=BEC đối đỉnh
EA=EB
=> tam giác AEQ = tam giác BEC(c.g.g).
=> AQ=BC(cạnh tuognư ứng). (1)
Xét Tam giác AFP và tam giác CFB có
AF=CF
AFP=CFB đối đỉnh
FB=FP
=> tam giác AFB = tam giác CFB(c.g.c)
=> AP = BC (2)
từ (1) và (2) suy ra AP=AQ

c)
xét tam giác BEQ và tam giác AEC có
EQ=EC
BEQ=AEC đối đỉnh
EB=EA
=> tam giác BEQ = tam giác AEC(c.g.c)
=> BQE=AEC (góc tương ứng) 
mà chúng ở vị trí so le trong nên BQ//AC.
xét tam giác PFC và BFA có:
FA=FC
AFB=CFP
BF=PF
=> tam giác PFC = BFA (c.g.c)
=> FAB = FCB(góc tương ứng)
mà chúng ở vị trí số le trong nên
CP//AB

Bài tập 1

1,will he?

2,aren't you?

3,have you?

4,don't you?

5,are they?

6, does she?

7,don't you?

8, doesn't it?

9, can't you?

10, aren't I

Chú ý: phần đằng sau dấu phẩy của câu hỏi đuôi (Question tag) sẽ ở khẳng định (nếu câu trước dấu phẩy là câu phủ định) và ngược lại

Bài tập 2

câu 1;2;4;6 chọn fewer. Các câu còn lại chọn less

Chú ý:  Nếu danh từ đi với nó đếm được, chọn less. Còn không chọn fewer

Put a question tag on the end of these sentences

1,Tom wont be late, .will he.....? No ,heis never late.

2,You are tired,...aren't you...? Yes,a little.

3,You haven t seen mary today,..have you...? No,I am afraid not.

4,You go a camera,....don't you...Yes why.

5,These arent many people here yet ,....are they.......No,but it still quite early.

6,Sue doesn t know ann,...does she... No ,they are never met.

7,You cme in yours sports car,....don't you.......No,i came in a taxi.

8,These cakes look good,...don't you ?.....  I can t wait to try them.

9,You can speak German,....can't you.........Yes ,but not very fluently.

10,I am to impatient,....amnot I..... Yes, you are sometimes.

Fill in each blank with fewer or less  

1,They work..fewer..hour than us a day .

2,Susan s her new life in Ha Noi now . She writes.fewer.letters home.

3,You should spend ....less.....time watching tv.

4,He has .....fewer........day off than anyone in the factory .

5,Lan drinks.........less....... water than hoa a day .

6,There are .........fewer............trees along the strees in the now town.

7,I get ....less.........money than him.

8,Lan spends ......less.....money on books than Kien.

9,They are very tired .They can do ...............exercises this evening.

10,They are very tired ,too .They will do .................work tonight.

Ta có : \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

\(=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}\right).\frac{2005}{1890}+115\)

\(=\left(\frac{3}{5}-\frac{3}{5}\right).\frac{2005}{1890}+115=0+115=115\)

= ( \(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)+ \(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)) x \(\frac{2005}{1890}\)+ 115

= ( \(\frac{3(\frac{1}{2}+\frac{1}{3}-\frac{1}{4})}{5(\frac{1}{1}+\frac{1}{3}-\frac{1}{4})}\)+ \(\frac{3(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}{-5(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}\)) x \(\frac{2005}{1890}\)+ 115

=( \(\frac{3}{5}\)+\(\frac{3}{-5}\)) x \(\frac{2005}{1890}\)+115 = 0 +115 = 115

chắc D bạn à :D

B nha bạn