Đề bài này chỉ đúng khi a;b;c là độ dài 3 cạnh của 1 tam giác

Nếu đề đúng như thế thì chứng minh như sau:

\(VT=\frac{1}{a+b-c}+\frac{1}{a+c-b}+\frac{1}{b+c-a}\)

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

\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{2}{b}\) ; \(\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{2}{c}\)

Cộng vế với vế:

\(2VT\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow VT\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Dấu "=" xảy ra khi \(a=b=c\)

a) \(\frac{1}{a}-\frac{1}{a+1}=\frac{\left(a+1\right)-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)

b) \(\frac{1}{b}-\frac{1}{b+m}=\frac{\left(b+m\right)-b}{b\left(b+m\right)}=\frac{m}{b\left(b+m\right)}\)

a) \(\frac{1}{a}-\frac{1}{a+1}=\frac{\left(a+1\right)-a}{a\cdot\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)(đpcm)

b) \(\frac{1}{b}-\frac{1}{b+m}=\frac{\left(b+m\right)-b}{b\left(b+m\right)}=\frac{m}{b\left(b+m\right)}\)(đpcm)

Ta có:

1+\(\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a}\)

Thay a=1

=>\(1+\dfrac{1}{b}=b+\dfrac{1}{c}=c+1\)

*Lấy \(1+\dfrac{1}{b}=c+1\Rightarrow\dfrac{1}{b}=c\Rightarrow b=\dfrac{1}{c}\)

=>\(1+\dfrac{1}{b}=\dfrac{2}{c}=c+1\)

*Lấy \(\dfrac{2}{c}=\dfrac{c+1}{1}\)

=> 2=c(c+1)

<=> 2=c²+c

=>c=-2

*Lấy \(1+\dfrac{1}{b}=\dfrac{2}{c}\)

Thay c=-2 và quy đồng

=>\(\dfrac{b+1}{b}=-1\)

=>b+1=-b

=> b+b=-1

=>2b=-1

=> b=-1/2

Vậy b=\(-\dfrac{1}{2};c=-2\)

Lời giải:

Với $a,b,c$ nguyên dương thì:

$1=\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+b+c}< \frac{3}{a}$

$\Rightarrow a< 3$

$a$ là số nguyên dương nên $a=1,2$

Nếu $a=1$ thì $\frac{1}{b+a}+\frac{1}{a+b+c}=1-\frac{1}{a}=0$ (vô lý - loại)

$\Rightarrow a=2$

Khi đó:

$\frac{1}{b+2}+\frac{1}{b+c+2}=\frac{1}{2}$

Mà $\frac{1}{b+2}+\frac{1}{b+c+2}< \frac{2}{b+2}$

$\Rightarrow \frac{1}{2}< \frac{2}{b+2}$

$\Rightarrow b+2<4\Rightarrow b<2\Rightarrow b=1$

Khi đó: $\frac{1}{3}+\frac{1}{c+3}=\frac{1}{2}$

$\frac{1}{c+3}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$

$\Rightarrow c+3=6\Rightarrow c=3$

a) ĐK: a>0,a\(\ne1\)

b) \(M=\left(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\right)\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{a+\sqrt{a}-2}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}-\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}=\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{2}{a-1}\)

c) Để M là số nguyên thì \(\dfrac{2}{a-1}\in Z\Rightarrow a-1\inƯ\left(2\right)\in\left(\pm1,\pm2\right)\)\(\Rightarrow\left[{}\begin{matrix}a-1=1\\a-1=-1\\a-1=2\\a-1=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=2\left(tm\right)\\a=0\left(ktm\right)\\a=3\left(tm\right)\\a=-1\left(ktm\right)\end{matrix}\right.\)

Vậy để M là số nguyên thì a=2 hoặc a=3

Bài 1:

a) Ta có: (a-b)+(c-d)-(a+c)

=a-b+c-d-a-c

=-b-d(1)

Ta lại có: -(b+d)=-b-d(2)

Từ (1) và (2) suy ra (a-b)+(c-d)-(a+c)=-(b+d)

b) Ta có: (a-b)-(c-d)+(b+c)

=a-b-c+d+b+c

=a+d(đpcm)

c) Ta có: a(b-c)-b(a-c)

=ab-ac-ab+cb

=cb-ca

=c(b-a)(đpcm)

d) Ta có: b(c-a)+a(b-c)

=bc-ba+ab-ac

=bc-ac

=c(b-a)(đpcm)

e) Ta có: -c(-a+b)+b(c-a)

=ca-cb+bc-ba

=ca-ba

=a(c-b)(đpcm)

g) Ta có: a(c-b)-b(-a-c)

=ac-ab+ba+bc

=ac+bc

=c(a+b)(đpcm)