giải phương trình sau:
2x^2-5x+2=0
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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 \(\orbr{\begin{cases}x=\frac{-5+\sqrt{97}}{2}\\x=\frac{-5-\sqrt{97}}{2}\end{cases}}\)
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+......+\frac{1}{50}\) (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
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)
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 \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\ne0\)
\(\Leftrightarrow x=1995\)
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