Vòng 3 - Chung kết

Câu 1:

a)\(14,58.460+7,29.540.2=14,58.460+14,58.540=14,58\left(460+540\right)=14,58.1000=14580\)

b)\(200-198+196-194+...+8-6+4-2=\left(200-198\right)+\left(198-196\right)+...+\left(8-6\right)+\left(4-2\right)\)

=2+2+...+2+2 (50 số hạng)

=2.50

=100

c)\(\frac{2010.2009-1}{2008.2010+2009}:\frac{1}{1999}=\frac{2010.\left(2008+1\right)-1}{2008.2010+2009}.1999=\frac{2010.2008+2010-1}{2008.2010+2009}.1999=\frac{2010.2008+2009}{2008.2010+2009}.1999=1.1999=1999\)

Câu 2:

a) Ta có: \(1000^{1000}< M< 1000+1000^2+1000^3+...+1000^{999}+1000^{1000}\)

<=>1000...000 (1000 chữ số 0)<M<1001001...1000 (3001 chữ số)

=>3 chữ số bên trái đầu tiên của M là 100

b)Chứng minh bài toán phụ: Với \(\frac{a}{b}< 1\) (\(a;b\in N,a\ne0\)) thì \(\frac{a}{b}< \frac{a+m}{b+m}\)(m\(\in\)N^*)

\(\frac{a}{b}< 1\Leftrightarrow a< b\)<=> am<bm <=> am+ab < bm+ab <=>a(m+b)<b(m+a)<=>\(\frac{a}{b}< \frac{a+m}{b+m}\)

Áp dụng vào bài toán ta có: \(B=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}=A\)

Vậy B<A

Câu 3:

a)\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=2013\)

<=>\(\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(b-a\right)-\left(b-c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{\left(c-b\right)-\left(c-a\right)}{\left(c-a\right)\left(c-b\right)}=2013\)

<=>\(\frac{a-c}{\left(a-b\right)\left(a-c\right)}-\frac{a-b}{\left(a-b\right)\left(a-c\right)}+\frac{b-a}{\left(b-a\right)\left(b-c\right)}-\frac{b-c}{\left(b-a\right)\left(b-c\right)}+\frac{c-b}{\left(c-a\right)\left(c-b\right)}-\frac{c-a}{\left(c-a\right)\left(c-b\right)}=2013\)

<=>\(\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}=2013\)

<=>\(\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}+\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}=2013\)

<=>\(2.\frac{1}{a-b}+2.\frac{1}{c-a}+2.\frac{1}{b-c}=2013\)

<=>\(2\left(\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\right)=2013\)

<=>\(\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}=\frac{2013}{2}\)

b)\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{4\left(3x-2y\right)}{16}=\frac{3\left(2z-4x\right)}{9}=\frac{2\left(4y-3z\right)}{4}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}=\frac{12x-8y}{16}=\frac{6z-12x}{9}=\frac{8y-6z}{4}=\frac{12x-8y+6z-12x+8y-6z}{16+9+4}=\frac{0}{29}=0\)

hay: 3x-2y=0 <=> 3x=2y <=> \(\frac{x}{2}=\frac{y}{3}\) (1)

4y-3z=0 <=> 4y=3z <=> \(\frac{y}{3}=\frac{z}{4}\) (2)

Từ (1) và (2) => \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\left(đpcm\right)\)

Câu 4:

a) Vì \(\left|x+\frac{1}{1.3}\right|\ge0;\left|x+\frac{1}{3.5}\right|\ge0;...;\left|x+\frac{1}{97.99}\right|\ge0\)

=>\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|\ge0\Leftrightarrow51x\ge0\Leftrightarrow x\ge0\)

\(x\ge0\Rightarrow\left|x+\frac{1}{1.3}\right|=x+\frac{1}{1.3};\left|x+\frac{1}{3.5}\right|=x+\frac{1}{3.5};...;\left|x+\frac{1}{97.99}\right|=x+\frac{1}{97.99}\)

<=>\(x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=51x\)

<=>\(49x+\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}=51x\)

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\) <=> \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\Leftrightarrow A=\frac{98}{99}:2=\frac{49}{99}\)

<=>\(49x+\frac{49}{99}=51x\Leftrightarrow\frac{49}{99}=2x\Leftrightarrow x=\frac{49}{99}:2=\frac{49}{198}\)

Vậy x=49/198

b)Gọi 3 số cần tìm lần lượt là a;b;c

Theo đề bài ta có: \(15a=10b=6c\Leftrightarrow\frac{a}{60}=\frac{b}{90}=\frac{c}{150}\Leftrightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{5}\)

Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)=> \(\begin{cases}a=2k\\b=3k\\c=5k\end{cases}\)

=>ƯCLN(a;b;c)=ƯCLN(2k;3k;5k)=30k=1680

=>k=56

=>\(\begin{cases}a=2.56=112\\b=3.56=168\\c=5.56=280\end{cases}\)

Vậy a=112, b=168, c=280

Câu 6:

Kẻ DH//AC và H\(\in BC\)

a)Xét \(\Delta AKI\) và \(\Delta DBH\) có:

\(\widehat{KAI}=\widehat{BDH}\) (vì DH//AC nên 2 góc đồng vị bằng nhau)

AK=DB (giả thiết)

\(\widehat{AKI}=\widehat{DBH}\) (vì KI//BC nên 2 góc đồng vị bằng nhau)

=>\(\Delta AKI\)=\(\Delta DBH\)(g.c.g) =>AI=DH (1)

Xét \(\Delta DEH\) và \(\Delta CHE\) có:

\(\widehat{DEH}=\widehat{EHC}\) (DE//BC nên 2 góc so le trong bằng nhau)

EH là cạnh chung

\(\widehat{DHE}=\widehat{HEC}\) (DH//AC nên 2 góc so le trong bằng nhau)

=>\(\Delta DEH\)=\(\Delta CHE\) (g.c.g) => DH=EC (2)

Từ (1) và (2) => AI=EC (đpcm)

b) Theo chứng minh phần a có:

\(\Delta AKI=\Delta DBH\Rightarrow KI=BH\)
\(\Delta DEH=\Delta CHE\Rightarrow DE=HC\)

=>DE+KI=BH+HC=5cm