\(∘backwin\)

\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)

\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)

\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)

\(100 x + 5050 = 5750\)

\( 100 x = 5750 − 5050\)

\(100 x = 700\)

\(x = 700 : 100\)

\(x = 7\)

\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)

\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)

\(B<1-\)\(\dfrac{1}{2021}\)

\(B<\)\(\dfrac{2020}{2021}\)

\(\dfrac{2020}{2021}< 1\)

\(B<1\)

a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7

b: \(\Leftrightarrow6\sqrt{x-2}-15\cdot\dfrac{\sqrt{x-2}}{5}=20+4\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{x-2}\cdot6-3\sqrt{x-2}-4\sqrt{x-2}=20\)

\(\Leftrightarrow-\sqrt{x-2}=20\)(vô lý)

\(4\left(x-3\right)-8x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(4-8x\right)=0\\ \Leftrightarrow2\left(1-2x\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\\ 5x\left(x-7\right)-10\left(7-x\right)=0\\ \Leftrightarrow\left(x-7\right)\left(5x+10\right)=0\\ \Leftrightarrow5\left(x+2\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\\ 2x-8=3x\left(x-4\right)\\ \Leftrightarrow2\left(x-4\right)-3x\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(2-3x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\\ 3x\left(x-5\right)=10-2x\\ \Leftrightarrow3x\left(x-5\right)+2\left(x-5\right)=0\\ \Leftrightarrow\left(3x+2\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=5\end{matrix}\right.\\ 6x\left(x-3\right)-3\left(3-x\right)=0\\ \Leftrightarrow\left(6x+3\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

\(x^2\left(x+4\right)+9\left(-x-4\right)=0\\ \Leftrightarrow\left(x^2-9\right)\left(x+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-4\end{matrix}\right.\)

\(\left(4-8x\right)\left(x-3\right)=0\)

\(\left[{}\begin{matrix}4-8x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\3\end{matrix}\right.\)

\(2\left(x-4\right)-3x\left(x-4\right)=0\)

\(\left(2-3x\right)\left(x-4\right)=0\)

\(\left[{}\begin{matrix}2-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)

Có A = 1998 x 2002

         = (2000 - 2) x (2000 + 2)

         = 2000² - 2²

         = 2000² - 4

Mà B = 2000 x 2000 = 2000² ; 2000² - 4 < 2000²

=> A < B

Có B = 35 x 53 - 18

         = (34 + 1) x 53 - 18

         = 34 x 53 + 53 - 18

         = 34 x 53 + 35

=> B = D

9/13 x 7/12 + 9/13 x 5/12 - 9/13

= 9/13 x (7/12 + 5/12 - 1)

= 9/13 x 0

= 0

4/13 x 5/12 + 4/13 x 7/12 - 4/3

= 4/13 x (5/12 + 7/12) - 4/3

= 4/13 x 1 - 4/3

= 4/13 - 4/3

= -40/39

a) 9/13 x 7/12 + 9/13 x 5/12 - 9/13

= 9/13 x ( 5/12 + 7/12 - 1 )

= 9/13 x 0

= 0

b) 4/13 x 5/12 + 4/13 x 7/ 12 - 4/3

= 4/13 x ( 5/12 + 7/12 - 13/3 )

= 4/13 x ( - 10/3 )

= -40/39 

10 - x - |x - 5| = 0

=> x - |x - 5|   = 10 - 0

=> x - |x - 5|   = 10

=>      |x - 5|   = 10 - x

Điều kiện : 10 - x \(\ge\)0

Khi đó : |x - 5| = 10 - x

       => x - 5   = 10 - x

       => x + x  = 10 + 5

       => 2x      = 15

       => x        = \(\frac{15}{2}\)( x \(\notin\)Z )

Vậy không tồn tại giá trị x.