\(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)

=>\(\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)=\left(\frac{x+8}{11}+1\right)+\left(\frac{x+10}{9}+1\right)+\left(\frac{x+12}{7}+1\right)\)

=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}=\frac{x+19}{11}+\frac{x+19}{9}+\frac{x+19}{7}\)

=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)

\(\Rightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)

=> x + 19 = 0 Vì \(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\ne0\)

=> x = - 19

Bài làm:

Ta có: \(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)

\(\Leftrightarrow\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)-\left(\frac{x+8}{11}+1\right)-\left(\frac{x+10}{9}+1\right)-\left(\frac{x+12}{7}+1\right)=0\)

\(\Leftrightarrow\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)

\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}\right)=0\)

Mà \(\hept{\begin{cases}\frac{1}{15}< \frac{1}{11}\\\frac{1}{13}< \frac{1}{9}\end{cases}\Rightarrow}\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}< 0\)

\(\Rightarrow x+19=0\)

\(\Rightarrow x=-19\)

Vì n lẻ => n = 2k + 1 (k \(\inℕ^∗\))

=>  A = 1 + 3 + 5 + 7 + ... + (2k + 1)

         = [(2k + 1 - 1) : 2 + 1] . (2k + 1 + 1) : 2

         =  (k + 1).2(k + 1): 2

         = (k + 1)²

=> A là số chính phương

n lẻ => n có dạng 2k + 1 ( \(k\inℕ^∗\))

=> A = 1 + 3 + 5 + 7 + ... + n

         = 1 + 3 + 5 + 7 + ... + ( 2k + 1 )

         = \(\frac{\left[\left(2k+1\right)+1\right]\left[\frac{\left(2k+1\right)-1}{2}+1\right]}{2}\)

         = \(\frac{\left(2k+2\right)\left(k+1\right)}{2}\)

         = \(\frac{2\left(k+1\right)\left(k+1\right)}{2}\)

         = \(\left(k+1\right)\left(k+1\right)\)

         = \(\left(k+1\right)^2\)

=> A là số chính phương ( đpcm ) 

( x + 5 )( x² + 9 ) < 0

Xét 2 TH :

TH1: x + 5 < 0 và x² + 9 > 0

* x + 5 < 0 => x < -5

* x² + 9 > 0 ( đúng ∀ x )

=> Nhận

TH2: x + 5 > 0 và x² + 9 < 0

* x + 5 > 0 => x > 5

* x² + 9 < 0 ( vô lí ) 

=> Loại 

Vậy nghiệm của bpt là x < -5

Sợ trình bày như này bạn không hiểu :(

Do tích của chung <0 nên 2 thừa số ở vế trái trái dấu nhau.Mà x^2+9>0 với mọi x thuộc R 

\(\Rightarrow x+5< 0< x^2+9\Rightarrow x< -5\)

Vậy x<-5

\(x^2-3y^2-8z^2+2xy-10yz+2xz\)

\(=x^2-3y^2-8z^2+3xy-xy-4yz-6yz+4xz-2xz\)

\(=\left(x^2+3xy+4xz\right)+\left(-xy-3y^2-4yz\right)+\left(-2xz-6yz-8z^2\right)\)

\(=x\left(x+3y+4z\right)-y\left(x+3y+4z\right)-2z\left(x+3y+4z\right)\)

\(=\left(x+3y+4z\right)\left(x-y-2z\right)\)

a) |3 - 8x| < 19

=> \(-19< 3-8x< 19\)

=> -22 < 8x < 16

=> -22 : 8 < x < 16 : 8

=> - 2,75  < x < 2

b) |x - 4| > 3

=> \(\orbr{\begin{cases}x-4>3\\-\left(x-4\right)< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>7\\-x+4< 3\end{cases}\Rightarrow\orbr{\begin{cases}x>7\\x< 1\end{cases}}}}\)

(1/3)^2 - 1/3y^4 - 1/3y^4 - (y^4)^2

1/3(1/3 - y^4) -  y^4(1/3 - y^4)

(1/3 - y^4)(1/3-y^4)

còn lại bạn làm.

\(2ab-a-b=2\)

\(\Leftrightarrow2a\left(b-\frac{1}{2}\right)-\left(b-\frac{1}{2}\right)=\frac{3}{2}\)

\(\Leftrightarrow\left(b-\frac{1}{2}\right)\left(2a-1\right)=\frac{3}{2}\)

\(\Leftrightarrow\left(2a-1\right)\left(2b-1\right)=3\)

Xét ước nhé bạn

TH1: Nếu \(x< -3\)\(\Rightarrow\hept{\begin{cases}\left|x+3\right|=-\left(x+3\right)\\\left|x+1\right|=-\left(x+1\right)\end{cases}}\)

\(\Rightarrow-\left(x+3\right)-\left(x+1\right)=3x\)

\(\Leftrightarrow-x-3-x-1=3x\)

\(\Leftrightarrow-2x-4=3x\)\(\Leftrightarrow-5x=4\)\(\Leftrightarrow x=\frac{-4}{5}\)( không thỏa mãn )

TH2: Nếu \(-3\le x< -1\)\(\Rightarrow\hept{\begin{cases}\left|x+3\right|=x+3\\\left|x+1\right|=-\left(x+1\right)\end{cases}}\)

\(\Rightarrow\left(x+3\right)-\left(x+1\right)=3x\)

\(\Leftrightarrow x+3-x-1=3x\)\(\Leftrightarrow3x=2\)\(\Leftrightarrow x=\frac{2}{3}\)( không thỏa mãn )

TH3: Nếu \(x\ge-1\)\(\Rightarrow\hept{\begin{cases}\left|x+3\right|=x+3\\\left|x+1\right|=x+1\end{cases}}\)

\(\Rightarrow x+3+x+1=3x\)

\(\Leftrightarrow2x+4=3x\)\(\Leftrightarrow x=4\)( thỏa mãn )

Vậy \(x=4\)

\(|x+3|+|x+1|=3x\\ \)

\(\Leftrightarrow2x+4=3x\\ \)

\(\Rightarrow x=4\)

\(A=2-4\sqrt{x-3}\)

Điều kiện để A xác định: \(x\ge3\)

Vì \(\sqrt{x-3}\ge0\)\(\Rightarrow4\sqrt{x-3}\ge0\)

\(\Rightarrow2-4\sqrt{x-3}\le2\)

Dấu " = " xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)( thỏa mãn )

Vậy \(maxA=2\)\(\Leftrightarrow x=3\)