TA có 13-13*1+10-10= 13- 13+10-10=0+0=0

(10*11+12*13+14*15+...+22*23+24*25)*0=0

* là dấu x nhe bạn 

thick minh nha

( 10 x 11 + 12 x 13 + 14 x 15 + ... + 22 x 23 + 24 x 25) x ( 13 - 13 x 1 + 10 -10)

= ( 10 x 11 + 12 x 13 + 14 x 15 + ... + 22 x 23 + 24 x 25) x 0

= 0

 k cho anh nha !!!!!!!!!!!!!!!! : học lớp 5 rùi

  (10 x 11 + 12 x 13 + 14 x 15 + ... + 22 x 23 + 24 x 25) x ( 13 - 13 x 1 + 10 - 10)

=( 10 x 11 + 12 x 13 + 14 x 15 + ... + 22 x 23 + 24 x 25) x 0

= 0 

tk nha

\(\frac{x+22}{11}+\frac{x+23}{12}=\frac{x+24}{13}+\frac{x+25}{14}\)

\(\Leftrightarrow\left(\frac{x+22}{11}+1\right)+\left(\frac{x+23}{12}+1\right)=\left(\frac{x+24}{13}+1\right)+\left(\frac{x+25}{14}+1\right)\)

\(\Leftrightarrow\frac{x+33}{11}+\frac{x+35}{12}=\frac{x+37}{13}+\frac{x+39}{14}\)

\(\Leftrightarrow\frac{x+33}{11}+\frac{x+35}{12}-\frac{x+37}{13}-\frac{x+39}{14}=0\)

\(\Leftrightarrow\frac{2184\cdot\left(x+33\right)+2002\cdot\left(x+35\right)-1848\cdot\left(x+37\right)-1716\cdot\left(x+39\right)}{24024}=0\)

\(\Leftrightarrow2184x+72072+2002x+70070-1848x-68376-1716x-66924=0\)

\(\Leftrightarrow622x+6842=0\)

\(\Leftrightarrow x=-11\)

\(x+\left(x+1\right)+\left(x+2\right)+....+13+14=14\Leftrightarrow x+\left(x+1\right)+....+13\Leftrightarrow x=-13\)

\(25+24+23+....+x+\left(x-1\right)+\left(x-2\right)=25\Leftrightarrow24+....+\left(x-2\right)=0\Leftrightarrow x-2=-24\)

\(\Leftrightarrow x=-22\)

bạn làm rõ hơn đi

\(\dfrac{x^2-26}{10}+\dfrac{x^2-25}{11}\ge\dfrac{x^2-24}{12}+\dfrac{x^2-23}{13}\)

\(\Leftrightarrow\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)

\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)

\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)

\(\Leftrightarrow\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)

Vì \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\Rightarrow x^2-36\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x\ge6\end{matrix}\right.\)

Bất phương trình đó tương đương với:

 \(\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)

⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)

⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)

⇔ \(\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)

+)Vì \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\) 

⇔ \(x^2-36\ge0\)

⇔ \(x^2\ge36\)

⇔ \(\sqrt{x^2}\ge6\)

⇔ \(\left|x\right|\ge6\)

⇔ \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)

➤ Vậy \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)

a: =>14x+2*(14*13/2)=14

=>14x+14*13=14

=>14x=-12*14

=>x=-12

b: =>3x+(0-1-2-3+1+2+...+25)=25

=>3x+4+...+25=25

=>3x+4+...+24=0

=>3x+(24-4+1)*28/2=0

=>3x+21*28/2=0

=>3x=-21*14

=>x=-7*14=-98

Hình như đề sai nếu viết thế này:x + (x+1) + (x+2) + (x+3) + .... + 13 + 14 = 14

Thì biết chỗ nào ko có x chỗ nào có x

Chúc bn học tốt