a) ĐKXĐ : \(x\ne\pm1\)

+ \(B=\left(\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x^2-4x-1\right)}{x^2-1}\right)\cdot\frac{x-2014}{x-1}\)

\(B=\frac{4x+x^2-4x-1}{x^2-1}\cdot\frac{x-2014}{x+1}\)

\(B=\frac{x^2-1}{x^2-1}\cdot\frac{x-2014}{x+1}=\frac{x-2014}{x+1}\)\

b) B có giá trị nguyên

\(\Leftrightarrow x-2014⋮x+1\)

\(\Leftrightarrow x+1-2015⋮x+1\)

\(\Leftrightarrow2015⋮x+1\)

ĐKXĐ : \(x\ne\pm1\)

a) \(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)

\(A=\left[\frac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}-\frac{x\left(1-x\right)}{1-x}\right]:\frac{\left(1-x\right)\left(x+1\right)}{\left(1-x\right)-x^2\left(1-x\right)}\)

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

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

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

\(A=\left(1-x\right)\left(x^2+1\right)\)

b) Để A < 0 thì \(1-x\)và \(x^2+1\)trái dấu

Mà \(x^2+1>0\forall x\)

Vậy để A < 0 thì \(1-x< 0\Leftrightarrow x>1\)

Vậy....

a) ta có: A=\(\frac{\left(1-x\right)+x^2\left(1-x\right)}{1-x}:\frac{1-x^2}{\left(1-x\right)\left(1-x^2\right)}=\left(x^2+1\right)\cdot\left(1-x\right)=1-x^3\)

b) Để A<0 <=> x^3>1 <=>x>1

\(\frac{2-x}{2012}-2=\frac{1-x}{2013}-\frac{x}{1007}\)

\(\Leftrightarrow\frac{2-x}{2012}-\frac{1-x}{2013}+\frac{x}{1007}-2=0\)

\(\Leftrightarrow\left(\frac{2-x}{2012}+1\right)-\left(\frac{1-x}{2013}+1\right)+\left(\frac{x}{1007}-2\right)=0\)

\(\Leftrightarrow\frac{2014-x}{2012}-\frac{2014-x}{2013}+\frac{x-2014}{1007}=0\)

\(\Leftrightarrow\left(2014-x\right)\left(\frac{1}{2012}-\frac{1}{2013}-\frac{1}{1007}\right)=0\Leftrightarrow2014-x=0\Leftrightarrow x=2014\)

1.This flower is watered (by my father) every morning.

2.Fiona was invented to John’s birthday party last month.

3.The dinner is being prepared (by her mother) in the kitchen.

4.Our teeth should be cleaned twice a day.

5.The English grammar has been explained (by our teacher).

Passive Voice 

1. My father waters this flower every morning.

-->..........this flower is watered by my father every morning....................................

2. John invited Fiona to his birthday party last night.

-->....................Fiona was invited to john's birthday party by him last night. ...................................................

3. Her mother is preparing the dinner in the kitchen.

-->.............................. the dinner is being prepared in the kitchen by her mother. ........................................

4. We should clean our teeth twice a day.

-->.............our teeth should be cleaned by us twice a day...........................................................

5. Our teachers have explained the English grammar.

-->.......the English grammar has been explained by our teachers........................................

bn bị rảnh ak ?

ko trả lời thì đừng có viết linh tinh

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+2045}{10}=0\)

\(\Leftrightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1+\frac{x+2045}{10}-3=0\)

\(\Leftrightarrow\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}+\frac{x+3+2012}{2012}+\frac{x+2045-3.10}{10}=0\)

\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{10}=0\)

\(\Leftrightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{10}\ne0\)

Nên x + 2015 = 0 <=> x = -2015

Vậy x = -2015

22+1+2005=2028

nhé

22 + 1 + 2005

= 23 + 2005

= 2028

Ta có: CA=3,6cm =>CB=AB-CA=6-3,6=2,4cm

=> CA/CB=3,6/2,4=3/2

=> DA/DB=3/2

Mà ta có DA-DB=AB=6 cm ( Do điểm B nằm giữa A và D)

Áp dụng hiệu tỉ ta có:

DA=6:(3-2).3=18(cm)

2/\(\frac{AB}{CD}=\frac{2}{3}\Rightarrow\frac{AB}{2}=\frac{CD}{3}\Rightarrow\frac{AB}{4}=\frac{CD}{6}\)

\(\frac{CD}{EF}=\frac{4}{6}=\frac{2}{3}\Rightarrow\frac{CD}{2}=\frac{EF}{3}\Rightarrow\frac{CD}{6}=\frac{EF}{9}\)

=>\(\frac{AB}{4}=\frac{CD}{6}=\frac{EF}{9}=\frac{AB+CD+EF}{4+6+9}=\frac{70}{19}\)

=>AB=280/19 cm

CD=420/19 cm

EF=630/19 cm

Chúc e hc tốt :)

Cô giáo của mk kết quả lại ra AB=16cm ,CD=24cm ,EF=30cm. mk ko hiểu là sai ở đâu ạ

xy+yz+xz=3xyz

<=> xy+yz+xz/xyz = 3

<=> 1/x + 1/y + 1/z = 3

Do vai trò x ; y ; z như nhau , ko mất tính tổng quát , giả sử 

\(x\ge y\ge z\) . Khi đó , ta có : 

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3.\frac{1}{x}\)

\(\Rightarrow3\le3.\frac{1}{x}\)

\(\Rightarrow1\le\frac{1}{x}\)

\(\Rightarrow x\le1\)

Mà x nguyên dương nên x = 1

Làm tương tự như vậy , ta có : y = 1 ; z = 1

Vậy .... 

Sai rồi bạn , nếu làm như bạn , phải giả sử 

z \(\ge y\ge x\)chứ 

:v 

ban tham khao link nay co bai tuong tu chi khac moi bc thoi

https://baitapsgk.com/lop-9/sbt-toan-lop-9/cau-18-trang-102-sach-bai-ta%CC%A3p-sbt-toan-9-ta%CC%A3p-2-chung-minh-rang-h-ma-mb-khong-doi.html

bạn phải nói là lấy M sao cho M cách B hoặc C gì đó nữa thì tụi mình mới giúp bạn giải được chứ. Nếu không thì bài toán này có vô số đáp án đấy chứ. Bạn ghi thiếu đề hẳ?