Bài làm:

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

=> \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=0\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=0\) (1)

Mà \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\), cách CM như sau:

\(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Tương tự: \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\) ; \(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\)

Cộng vế 3 BĐT trên lại ta sẽ được: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

Thay vào (1) ta được:

\(0=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\ge3\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\le0\)

Dấu "=" xảy ra khi: \(a=b=c\)

a) Từ x - y = 2(x + y) = x : y 

x - y = 2(x + y)

=> x - y = 2x + 2y 

=> x = -3y

=> => x : y = - 3

Khi đó 2(x + y) = - 3

=> x + y = -1,5 (1)

=> x - y = -3 (2)

Từ (1) (2) => x = [(-1,5) + (-3)] : 2 = -2,25

=> y = -1,5 - (-2,25) = 0,75

Vậy x =  -2,25 ; y = 0,75

b) Từ x + y = x.y = x : y (1)

=> xy = x : y

=> \(xy=\frac{x}{y}\Rightarrow y=\frac{x}{y}:x\Rightarrow y=\frac{1}{y}\Rightarrow y^2=1\Rightarrow y=\pm1\)

Từ (1) => x + y = xy

TH1 : Nếu y = 1 

=> x + 1 = x

=> 0x = 1 (loại) 

TH2 : Nếu y = -1

=> x - 1 = -x

=> 2x = 1 

=> x = 0,5 (tm)

Vây y = - 1 ; x = 0,5

a>b>0

\(\Rightarrow\frac{a}{b}>\frac{b}{b}=1\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)

\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}=\frac{49}{50}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{49}{50}\)

\(\Rightarrow1-\frac{1}{n+1}=\frac{49}{50}\)

\(\Rightarrow\frac{1}{n+1}=\frac{1}{50}\)

\(\Rightarrow n+1=50\)

\(\Rightarrow n=49\)

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)

\(\Rightarrow\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{50}{51}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{50}{51}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2n+1}=\frac{50}{51}\)

\(\Rightarrow\frac{1}{2n+1}=\frac{1}{51}\)

\(\Rightarrow2n+1=51\)

\(\Rightarrow2n=50\)

\(\Rightarrow n=25\)

\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)

\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\cdot\frac{98}{303}=\frac{49}{303}\)

\(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{2550}\)

\(=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{50\cdot51}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{50}-\frac{1}{51}\)

\(=\frac{1}{3}-\frac{1}{51}\)

\(=\frac{16}{51}\)

nhanh nhé mik tích cho 8 cái lun

Vì AB // DM :

⇒DMAˆ=BAMˆ⇒DMA^=BAM^(2 góc so le trong)

⇒CAMˆ=EMAˆ⇒CAM^=EMA^(2 góc so le trong)

⇒DMAˆ+EMAˆ=CAMˆ+BAMˆ⇔DMEˆ=CABˆ⇒DMA^+EMA^=CAM^+BAM^⇔DME^=CAB^(1)

Vì EM // AC

⇒MECˆ=ACEˆ⇒MEC^=ACE^(2 góc so le trong)

⇒DECˆ=ECMˆ⇒DEC^=ECM^(2 góc so le trong)

⇒MECˆ+DECˆ=ACEˆ+ECMˆ⇔MEDˆ=ACMˆ⇒MEC^+DEC^=ACE^+ECM^⇔MED^=ACM^(2)

nhanh để mik tích

Đặt ab + 4 = m22 (m ∈ N)

 ⇒ab = m22− 4 = (m − 2) (m + 2)

 ⇒b =(m−2).(m+2)a(m−2).(m+2)a

Ta có:m=a+2⇒⇒ m-2=a

⇒⇒b=a(a+4)aa(a+4)a=a+4

Vậy với mọi số tự nhiên a luôn tồn tại b = a + 4 để ab + 4 là số chính phương.