Ta có: \(a^{2000}+b^{2000}=a^{2001}+b^{2001}=a^{2002}+b^{2002}\)

\(\Rightarrow\hept{\begin{cases}a^{2000}\left(a-1\right)+b^{2000}\left(b-1\right)=0\\a^{2001}\left(a-1\right)+b^{2001}\left(b-1\right)=0\end{cases}}\)

\(\Leftrightarrow a^{2000}\left(a-1\right)\left(a-1\right)+b^{2000}\left(b-1\right)\left(b-1\right)=0\)

\(\Leftrightarrow a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2=0\)

Ta có: \(\hept{\begin{cases}a^{2000}\left(a-1\right)^2\ge0\forall a>0\\b^{2000}\left(b-1\right)^2\ge0\forall b>0\end{cases}}\)\(\Leftrightarrow a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2\ge0\)

Mà \(a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}a^{2000}\left(a-1\right)^2=0\\b^{2000}\left(b-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a-1=0\left(a>0\right)\\b-1=0\left(b>0\right)\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}\)

\(M=a^{2017}+b^{2017}=1+1=2\)

Vậy \(M=2\)

không biết cách này đúng không nữa 

\(a^{2000}+b^{2000}=a^{2001}+b^{2001}\Rightarrow a^{2001}+b^{2001}-a^{2000}-b^{2000}=0\)

\(\Rightarrow a^{2000}.\left(a-1\right)+b^{2000}.\left(b-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}a-1=0\\b-1=0\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}}\)(1)

\(a^{2002}+b^{2002}=a^{2001}+b^{2001}\Rightarrow a^{2002}+b^{2002}-a^{2001}-b^{2001}=0\)

\(\Rightarrow a^{2001}.\left(a-1\right)+b^{2001}.\left(b-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}a-1=0\\b-1=0\end{cases}\left(\text{vì a,b dương nên }a^{2001}\text{và }b^{2001}\text{ lớn hơn 0}\right)\Rightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}}\)(2)

từ (1) và (2) => a=b=1=> M=2

p/s: trình độ thấp, sai bỏ qua

sửa đề: a,b,c là 3 số nguyên dương

\(\text{vì }a,b,c\text{ là 3 số nguyên dương}\)

\(\text{Có: }\hept{\begin{cases}\frac{a}{a+b+c}< \frac{a}{b+c}\\\frac{b}{a+b+c}< \frac{b}{c+a}\\\frac{c}{a+b+c}< \frac{c}{a+b}\end{cases}}\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}>1 \)

\(y\left(x-1\right)=x^2+2\Leftrightarrow y\left(x-1\right)-x^2+1=3\Leftrightarrow y\left(x-1\right)-\left(x-1\right)\left(x+1\right)=3\)

\(\Leftrightarrow\left(x-1\right)\left(y-x-1\right)=3=1.3=3.1=\left(-1\right)\left(-3\right)=\left(-3\right)\left(-1\right)\)

Phương pháp: Phân tích đa thức thành nhân tử, VP ko nhất thiết phải bằng 0 (vì đây là PT nghiệm nguyên)

Bạn tự xét tiếp từng trường hợp nhé.

Chúc bạn học tốt, năm mới vui vẻ!

\(A=n^3+2n^2-3=n^3-n^2+3n^2-3=n^2\left(n-1\right)+3\left(n-1\right)\left(n+1\right)\)

\(A=\left(n-1\right)\left(n^2+3n+3\right)\)

Vì A là hợp số nên \(A>0\)lại có \(n^2+3n+3\ge3>0\)nên \(n-1>0\Leftrightarrow n>1\)

Xét TH \(n=2\Rightarrow A=n^2+3n+3=13\)là SNT.

Với \(n>2\), A luôn có ít nhất 3 ước là \(1;n-1;A\)nên nó là hợp số.

Vậy để A là hợp số thì \(n>2\)

Gọi độ dài quãng đường AB là: \(x\left(km\right)\left(x>0\right)\)

Thời gian xe máy dự định đi là: \(\frac{x}{40}\left(h\right)\)

Sau 30 phút, xe máy đã đi được quãng đường là: \(40.\frac{1}{2}=20\left(km\right)\)

Thời gian xe máy đi hết quãng đường AB theo thực tế là: \(\frac{1}{2}+\frac{x-20}{45}\left(h\right)\)

Theo bài ra: \(\frac{1}{2}+\frac{x-20}{45}=\frac{x}{40}\)

\(\Leftrightarrow\frac{180+8\left(x-20\right)}{360}=\frac{9x}{360}\)

\(\Leftrightarrow180+8\left(x-20\right)=9x\Leftrightarrow8x+20=9x\Leftrightarrow x=20\) 

Quãng đường AB dài 20 km

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