2x² - 5x + 2 = 0

=> 2x² - 4x - x + 2 = 0

=> 2x(x - 2) - (x - 2) = 0

=> (2x - 1)(x - 2) = 0 

=> 2x - 1 = 0 => x = 1/2

hoặc x - 2 = 0 => x = 2

    Vậy x = 1/2 , x = 2

2x²-5x+2=0

=>2x²-4x-x+2=0

=>2x(x-2)-(x-2)=0

=>(2x-1)(x-2)=0

=>2x-1=0 hoặc x-2=0

+)Nếu 2x-1=0

=>2x=0+1=1

=>x=1:2=\(\frac{1}{2}\)

+)Nếu x-2=0

=>x=0+2=2

Vậy x=\(\frac{1}{2}\) hoặc x=2

Vì \(\frac{1}{3^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2002^2}< \frac{1}{2001.2002}\)

\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}\)

mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)\(=1-\frac{1}{2002}< 1\)

\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}< 1\)

\(\Rightarrow A=\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< 1\)(đpcm)

Nếu mà chỗ 3² ở phân số đầu tiên sửa thành 2² thì trông sẽ đẹp hơn nhé

x(x + 5) = 18

=> x² + 5x - 18 = 0

Có: \(\Delta=5^2-4.\left(-18\right)=97>0\Rightarrow\sqrt{\Delta}=\sqrt{97}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-5+\sqrt{97}}{2}\\x=\frac{-5-\sqrt{97}}{2}\end{cases}}\)

Vậy  pt có 2 nghiệm \(\orbr{\begin{cases}x=\frac{-5+\sqrt{97}}{2}\\x=\frac{-5-\sqrt{97}}{2}\end{cases}}\)

Ngọc Vĩ:chế làm thế lp 7 sao hỉu đc =.=

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+......+\frac{1}{50}\)  (đpcm)

Ta có:

1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50

= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/49 - 1/50

= (1 + 1/3 + 1/5 + ... + 1/49) - (1/2 + 1/4 + 1/6 + ... + 1/50)

= (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - 2.(1/2 + 1/4 + 1/6 + ... + 1/50)

= (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/49 + 1/50) - (1 + 1/2 + 1/3 + ... + 1/25)

= 1/26 + 1/27 + 1/28 + ... + 1/50

=> đpcm

\(\frac{3}{16}\)

2⁷ . 9³ / 6⁵ . 8²

= 2⁷ . (3²)³ / (2.3)⁵ . (2³)²

= 2⁷ . 3⁶ / 2⁵ . 3⁵ . 2⁶

= 2⁷ . 3⁶ / 2¹¹ . 3⁵

= 3/2⁴

= 3/16

a. \(1-2x< 7\)

mà: \(1-n\le1\)với mọi n

\(\Rightarrow2x=n\Rightarrow x=\frac{n}{2}\)với mọi n

b.để: (x-1).(x-2)>0

=> x-1>0hoặc x-2<0

=>x>1hoặc x<2

(mik chỉ làm 2 câu mẫu thôi, bạn cố gắng tự làm nha, rất vui được kết bạn với bạn)

AOx^ = AOC^ 

BOC^ = BOy^

AOB^ = COA^ + COB^ = 90o

Mà xOy^ = AOx^ + AOC^ + BOC^ + BOy^ = 2(AOC^ + COB^) = 2* 90o= 180o

=> Ox và Oy đối nhau

a) x - 3/5 <1/3

x < 14/15

-x +3/5 < 1/3

x> 3/5 - 1/3 

x > 4/15

vậy 4/5 < x < 14/15

b); c) tuong tu

d) x-5/1990 + x+5/1980 + x-25/1970=x-1990/5 + x-1980/15

 \(\Leftrightarrow\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)+\left(\frac{x-25}{1970}-1\right)=\left(\frac{x-1990}{5}-1\right)+\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1970}{25}-1\right)\)

\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\).

\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}=0\)

\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\right)=0\)

\(\Leftrightarrow x-1995=0\).Do \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\ne0\)

\(\Leftrightarrow x=1995\)