a) Vì \(\frac{a}{b}>1\Rightarrow a>b\Rightarrow a-b>0\)

Xét hiệu : \(\frac{a}{b}-\frac{a+c}{b+c}=\frac{a\left(b+c\right)-b\left(a+c\right)}{b\left(b+c\right)}=\frac{ab+ac-ba-bc}{b\left(b+c\right)}=\frac{ac-bc}{b\left(b+c\right)}=\frac{c\left(a-b\right)}{b\left(b+c\right)}\)

Mà a-b>0 (cmt) suy ra :\(\frac{a}{b}-\frac{a+c}{b+c}>0\Leftrightarrow\frac{a}{b}>\frac{a+c}{b+c}\left(đpcm\right)\)

b) Chứng minh tương tự

2/Cho b,d>0

Chứng minh \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

1, 

\(\frac{a+2}{a-2}=\frac{b+3}{b-3}\)

<=> (a - 2)(b + 3) = (a + 2)(b - 3)

<=> ab + 3a - 2b - 6 = ab - 3a + 2b - 6

<=> 3a - 2b = -3a + 2b

<=> 6a = 4b

<=> 3a = 2b 

<=> \(\frac{a}{2}=\frac{b}{3}\)(Đpcm)

2,

Có:

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(=\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}\)

\(=\frac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}=0\)

=> bz - cy = 0

=> bz = cy

=> \(\frac{b}{y}=\frac{c}{z}\)(1)

=> cx - az = 0

=> cx = az

=> \(\frac{c}{z}=\frac{a}{x}\)(2)

Từ (1) và (2)

=> \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)(Đpcm)

do abc=1 nên \(\frac{a}{ab+a+1}\)=\(\frac{a}{ab+a+abc}\)=\(\frac{a}{a\left(bc+b+1\right)}\)=\(\frac{1}{bc+b+1}\)

\(\frac{c}{ac+c+1}\)=\(\frac{bc}{abc+bc+b}\)(nhân cả 2 vế cho b)=\(\frac{bc}{bc+b+1}\)

=>\(\frac{a}{ab+a+1}\)+\(\frac{b}{bc+b+1}\)+\(\frac{c}{ac+c+1}\)=\(\frac{bc+b+1}{bc+b+1}\)=1

Có: 1+x = \(\frac{a+b+a-b}{a+b}\) = \(\frac{2a}{a+b}\)

Tương tự, 1 + y = \(\frac{2b}{b+c}\)

1 + z = \(\frac{2c}{c+a}\)

1 - x = \(\frac{q+b-a+b}{a+b}\) = \(\frac{2a}{a+b}\)

Tương tự như thế rồi nhân (1+x), (1+y), (1+z) với nhau; (1-z), (1-y), (1-z) với nhau

bạn nhân ra hết cho mk

thanks bạn nhiều nha

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Đề có sai không cậu ơi??

hơi sai đó bn

\(\frac{a+c}{b+c}>\frac{a}{b}\)

\(\Leftrightarrow b\left(a+c\right)>a\left(b+c\right)\)

\(\Leftrightarrow ab+bc>ab+ac\)

\(\Leftrightarrow bc>ac\)

\(\Leftrightarrow b>a\) 

\(\Rightarrow\frac{a}{b}< 1\) (luôn đúng)