Câu hỏi của Phạm Ngọc Ánh

Question

Câu 1: Cho $abc=2010$. Tính giá trị biểu thức

$\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}$

<...

Nguyễn Huy Tú · Accepted Answer

Câu 1:

Ta có: $\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}$

$=\frac{abca}{ab+abca+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}$

$=\frac{abca}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}$

$=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}$

$=\frac{ac+c+1}{ac+c+1}$

$=1$

Vậy $\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}$

Câu trả lời:

Câu 1:

Ta có: $\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}$

$=\frac{abca}{ab+abca+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}$

$=\frac{abca}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}$

$=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}$

$=\frac{ac+c+1}{ac+c+1}$

$=1$

Vậy $\frac{2010a}{ab+2010a+2010}+\frac{b}{bc+b+2010}+\frac{c}{ac+c+1}$

Nguyễn Huy Tú · Answer

Câu 2:

Đặt $B=4^{2009}+4^{2008}+...+4^2+5$

$\Rightarrow B=1+4+4^2+...+4^{2009}$

$\Rightarrow4B=4+4^2+4^3+...+4^{2010}$

$\Rightarrow4B-B=4^{2010}-1$

$\Rightarrow3B=4^{2010}-1$

$\Rightarrow B=\frac{4^{2010}-1}{3}$

Thay vào A ta có:

$A=75.\frac{4^{2010}-1}{3}+25$

$\Rightarrow A=25.\left(4^{2010}-1\right)+25$

$\Rightarrow A=25\left(4^{2010}-1+1\right)$

$\Rightarrow A=25.4^{2010}$

Vậy $A=25.4^{2010}$