mình nghĩ đề thế này, do bạn ko viết a+1,b+1,c+1 dưới mẫu

Cho abc = 1 . CMR : \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=1\)

                                             GIẢI

Ta có : \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{a^2bc+abc+ab}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\)

\(=\frac{ab+a+1}{ab+a+1}=1\)

Em kiểm tra lại đề bài nhé !

Chắc bạn viết nhầm đề, cho \(a=b=c=1\) đâu có đúng

Sửa lại đề: cho \(abc=1\) chứng minh \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\)

Ta có

\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{c}{ac+c+abc}\)

\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)

\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{ab+a+1}+\dfrac{1}{ab+a+1}\)

\(=\dfrac{a+ab+1}{ab+a+1}=1\) (đpcm)

Đề bạn Lâm đúng đấy!

Lời giải:
Do $0\leq a,b,c\le1 1$ nên: \(\text{VT}\leq \frac{a+b+c}{1+abc}\)

Giờ ta cần cm: $a+b+c\leq 2(1+abc)(*)$

Thật vậy:
$c(a-1)(b-1)\geq 0$

$\Leftrightarrow c(ab-a-b+1)\geq 0$

$\Leftrightarrow abc\geq ac+bc-c$

$\Leftrightarrow 2(abc+1)\geq ac+bc-c+abc+2$

Mà:

$ac+bc-c+abc+2-(a+b+c)=abc+(a+b)(c-1)-2(c-1)$

$=abc+(a+b-2)(c-1)\geq 0$ với mọi $0\leq a,b,c\leq 1$

$\Rightarrow ac+bc-c+abc+2\geq a+b+c$

$\Rightarrow 2(abc+1)\geq a+b+c$

Do đó BĐT $(*)$ đúng nên ta có đpcm.

ac+c+1

\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac}{abc+ac+1}+\dfrac{ab}{abc+ab+1}+\dfrac{bc}{abc+bc+1}\)

\(=\dfrac{ac}{ac+2}+\dfrac{ab}{ab+2}+\dfrac{bc}{bc+2}\)

\(=abc\left(\dfrac{b}{abc+2}+\dfrac{c}{abc+2}+\dfrac{a}{abc+2}\right)\)

\(=1.1=1\)(đpcm).

Vậy \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\).

Hình gửi kèm

\(2\cdot\hat{B}=4\cdot\hat{A}\)

=>\(\hat{B}=2\cdot\hat{A}\)

Đặt \(\hat{A}=x\)

=>\(\hat{B}=2x;\hat{C}=4x\)

Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)

=>x+2x+4x=180 độ

=>3x+4x=180 độ

=>sin3x=sin4x

Xét ΔABC có \(\frac{BC}{\sin A}=\frac{AC}{\sin B}=\frac{AB}{\sin C}=2R\)

=>\(BC=2R\cdot\sin A=2R\cdot\sin x\)

\(AC=2R\cdot\sin B=2R\cdot\sin2x\)

\(AB=2R\cdot\sin C=2R\cdot\sin4x\)

\(\frac{1}{AB}+\frac{1}{AC}=\frac{1}{2R\cdot\sin2x}+\frac{1}{2R\cdot\sin4x}\)

\(=\frac{1}{2R}\left(\frac{1}{\sin2x}+\frac{1}{\sin4x}\right)=\frac{1}{2R}\cdot\frac{\sin4x+\sin2x}{\sin2x\cdot\sin4x}\)

\(=\frac{1}{2R}\cdot\frac{2\sin3x \cos x}{\sin2x (2\sin2x \cos2x)}=\frac{1}{2R}\cdot\frac{\sin3x \cos x}{\sin^2 2x \cos2x}\)

\(=\frac{1}{2R}\cdot\frac{(2\sin2x \cos2x) \cos x}{\sin^2 2x \cos2x}=\frac{1}{2R}\cdot\frac{2\cos x}{\sin2x}=\frac{1}{2R}\cdot\frac{2\cos x}{2\sin x \cos x}=\frac{1}{2R\cdot\sin x}\)

=1/BC

\(\dfrac{1}{1+a+ab}+\dfrac{a}{a+ab+abc}+\dfrac{1}{abc+ac+c}\)

\(=\dfrac{1}{1+a+ab}+\dfrac{a}{1+a+ab}+\dfrac{1}{c\left(1+a+ab\right)}\)

\(=\dfrac{ac+c+1}{c\left(1+a+ab\right)}=\dfrac{c\left(a+1+ab\right)}{c\left(1+a+ab\right)}=1\)