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 \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\ne0\)

\(\Leftrightarrow x=1995\)

Ta có: x - y - z = 0 suy ra x = y + z (1)

Thay (1) vào biểu thức A ta được:

A = \(\left(x+y\right)\left(zy-y^2-z^2\right)\)

   = \(y^2z-y^3-yz^2+yz^2-y^2z-z^3\)

  = \(-\left(y^3+z^3\right)\)

  = - B 

Vậy A và B là hai đa thức đối nhau.

sai đề rồi