Question

[galaxyLâm]

Bài 4: Tìm a,b,c biết a+1/2 = b-1/3 = c+2/4 = a+b+c+2/2a+5

Bài 5: Cho a/2002 = b/2003 = c/2004 . Chứng minh rằng 4(a-b)(b-c) = (c-a)²

Để xem ai IQ 10 mik tik cho nhé !

Answer 1

[l҉o҉n҉g҉ d҉z҉]

Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)

\(\Rightarrow\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)

\(VT=4\left(a-b\right)\left(b-c\right)=4\left(2002k-2003k\right)\left(2003k-2004k\right)=4\left(-1k\right)\left(-1k\right)=4k^2\)

\(VP=\left(c-a\right)^2=\left(2004k-2002k\right)^2=\left(2k\right)^2=4k^2\)

\(\Rightarrow VT=VP\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\left(đpcm\right)\)
 

Answer 2

[Xyz OLM]

4) Ta có :\(\frac{a+1}{2}=\frac{b-1}{3}=\frac{c+2}{4}=\frac{a+b+c+2}{2a+5}=\frac{a+b+c+1-1+2}{2+3+4}=\frac{a+b+c+2}{9}\)(1)

=> 2a + 5 = 9

=> 2a = 4

=> a = 2

Thay a vào (1) ta có : 

\(\frac{b-1}{3}=\frac{c+2}{4}=\frac{3}{2}\)

=> \(\hept{\begin{cases}\frac{b-1}{3}=\frac{3}{2}\\\frac{c+2}{4}=\frac{3}{2}\end{cases}}\Rightarrow\hept{\begin{cases}2\left(b-1\right)=9\\2\left(c+2\right)=12\end{cases}}\Rightarrow\hept{\begin{cases}2b-2=9\\2c+4=12\end{cases}}\Rightarrow\hept{\begin{cases}2b=11\\2c=8\end{cases}\Rightarrow\hept{\begin{cases}b=5,5\\c=4\end{cases}}}\)

Vậy a = 2 ; b = 5,5 ; c = 4

5) Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)

=> \(\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)

4(a - b)(b - c) = (c - a)²

=> 4(2002k - 2003k)(2003k - 2004k) = (2002k - 2004k)²

=> 4(-k)(-k) = (-2k)²

=> (-2)²(-k)² = (-2k)²

=> 2²k² = (2k)²

=> (2k)² = (2k)²

=> 4(a - b)(b - c) = (c - a)² (đpcm)