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BT1: 20152014 có tận cùng là 5
20142015=2014.(20142)1007=2014.40561961007=2014.(...6) => Có tận cùng là ...4
=> 20152014-20142015 có tận cùng là ...5-...4=...1
BT2: f(1)=a.1+b=1 (1)
f(2)=a.2+b=4 (2)
Trừ (2) cho (1) => a=3
Thay a=3 vào (1) => b=-2
ĐS: a=3; b=-2
\(\frac{x+199}{200}+\frac{x+198}{201}+\frac{x+197}{202}=-3\)
\(\frac{x+199}{200}+1+\frac{x+198}{201}+1+\frac{x+197}{202}+1=0\)
\(\frac{x+399}{200}+\frac{x+399}{201}+\frac{x+399}{202}=0\)
\(\left(x+399\right)\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)=0\)
Mà \(\left(\frac{1}{200}+\frac{1}{201}+\frac{1}{202}\right)\ne0\)
=> x + 399 = 0
=> x = -399
\(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}+5=0\)
\(\Leftrightarrow\frac{x+1}{203}+1+\frac{x+2}{202}+1+\frac{x+3}{201}+1+\frac{x+4}{200}+1+\frac{x+5}{199}+1=0\)
\(\Leftrightarrow\frac{x+204}{203}+\frac{x+204}{202}+\frac{x+204}{201}+\frac{x+204}{200}+\frac{x+204}{199}=0\)
\(\Leftrightarrow\left(x+204\right)\left(\frac{1}{203}+\frac{1}{203}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)=0\)
\(\Leftrightarrow x+204=0\).Do \(\frac{1}{203}+\frac{1}{203}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\ne0\)
\(\Leftrightarrow x=-204\)
Ta có :
\(\frac{x+1}{203}+\frac{x+2}{202}+\frac{x+3}{201}+\frac{x+4}{200}+\frac{x+5}{199}+5=0\)
\(\Leftrightarrow\left(\frac{x+1}{203}+1\right)+\left(\frac{x+2}{202}+1\right)+\left(\frac{x+3}{201}+1\right)+\left(\frac{x+4}{200}+1\right)+\left(\frac{x+5}{199}+1\right)=0\)
\(\Leftrightarrow\left(\frac{x+204}{203}\right)+\left(\frac{x+4}{202}\right)+\left(\frac{x+4}{201}\right)+\left(\frac{x+204}{200}\right)+\left(\frac{x+204}{199}\right)=0\)
\(\Leftrightarrow\left(x+204\right)\left(\frac{1}{203}+\frac{1}{202}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)=0\)
Dễ thấy \(\left(\frac{1}{203}+\frac{1}{202}+\frac{1}{201}+\frac{1}{200}+\frac{1}{199}\right)\ne0\)
=> x + 204 = 0
<=> x = - 204
Vậy pt có nghiệm x = - 204
Ta có : \(\frac{x+4}{200}+\frac{x+3}{201}=\frac{x+2}{202}+\frac{x+1}{203}\)
=> \(\frac{x+4}{200}+\frac{x+3}{201}-\frac{x+2}{202}-\frac{x+1}{203}=0\)
=> \(\frac{x+4}{200}+1+\frac{x+3}{201}+1-\frac{x+2}{202}-1-\frac{x+1}{203}-1=0\)
=> \(\frac{x+204}{200}+\frac{x+204}{201}-\frac{x+204}{202}-\frac{x+204}{203}=0\)
=> \(\left(x+204\right)\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
=> \(x+204=0\)
=> \(x=-204\)
Vậy phương trình có tập nghiệm là S = { -204 }
Đặt \(\frac{x}{4}=\frac{y}{3}=\frac{z}{5}=kak\left(kak\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=4kak\\y=3kak\\z=5kak\end{cases}}\)
Mà \(x^2+y^2+z^2=200\)
\(\Leftrightarrow\left(4kak\right)^2+\left(3kak\right)^2+\left(5kak\right)^2=200\)
\(\Leftrightarrow16.kak^2+9.kak^2+25.kak^2=200\)
\(\Leftrightarrow kak^2.\left(16+9+25\right)=200\)
\(\Leftrightarrow kak^2.50=200\)
\(\Leftrightarrow kak^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}kak=2\\kak=-2\end{cases}}\)
+) Với \(kak=2\)thì \(\hept{\begin{cases}x=4kak=8\\y=3kak=6\\z=5kak=10\end{cases}}\)
+) Với \(kak=-2\)thì \(\hept{\begin{cases}x=4kak=-8\\y=3kak=-6\\z=5kak=-10\end{cases}}\)
Vậy ...
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Ta có : \(xyz=-30\)
\(\Leftrightarrow2k\times3k\times5k=-30\)
\(\Leftrightarrow30k^3=-30\)
\(\Leftrightarrow k^3=-1\)
\(\Leftrightarrow k=-1\)
Thay vào ta được :
\(\hept{\begin{cases}x=2k=-2\\y=3k=-3\\z=5k=-5\end{cases}}\)
Vậy ...
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
=> \(\left(1+\frac{x+5}{200}\right)+\left(1+\frac{x+4}{201}\right)=\left(1+\frac{x+3}{202}\right)+\left(1+\frac{x+2}{203}\right)\)
=> \(\frac{x+205}{200}+\frac{x+205}{201}=\frac{x+205}{202}+\frac{x+205}{203}\)
=> \(\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)
=> \(\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
Do \(\frac{1}{200}>\frac{1}{202};\frac{1}{201}>1-\frac{1}{203}\)
=> \(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)
=> \(x+205=0\)
=> \(x=-205\)
\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)
\(=>\frac{x+5+200}{200}+\frac{x+4+201}{201}-\frac{x+3+202}{202}-\frac{x+2+203}{203}=0\)
\(=>\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)
\(=>\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)
\(Do:\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)
\(=>x+205=0\)
\(=>x=-205\)