a) \(\dfrac{a}{b}=\dfrac{c}{d}\)suy ra\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

ta có \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

nên \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)

b)đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) suy ra a=bk;c=dk

ta có \(\dfrac{a}{b+a}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)(1)

\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)(2)

Từ (1);(2) suy ra \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

c)ta có \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

suy ra \(\dfrac{2a}{2c}=\dfrac{5b}{5d};\dfrac{3a}{3c}=\dfrac{4b}{4d}\)

suy ra \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3a-4d}\)

nên \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+3d}{3c-4d}\)

d)\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\left(\dfrac{a+c}{b+d}\right)^2\left(1\right)\)\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}\left(2\right)\)

từ (1);(2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

Cảm ơn bn.

Mk biết cách làm rồi!

Bài 2: 

a: \(\left|x\right|=-x\)

nên x<=0

b: \(\left|x\right|>x\)

=>x<0

a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (1)

\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)

Từ (1) và (2) suy ra: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

b.M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{50^2}\right)\)

= \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{2499}{2500}\)

= \(\dfrac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)

\(\dfrac{51}{2.50}=\dfrac{51}{100}\)

Lời giải:

a)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

\(\Rightarrow \left(\frac{a}{b}\right)^2=\left(\frac{b}{d}\right)^2=\frac{(a+c)^2}{(b+d)^2}(1)\)

Mặt khác, \(\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}(2)\) (áp dụng tính chất dãy tỉ số bằng nhau)

Từ \((1),(2)\Rightarrow \frac{(a+c)^2}{(b+d)^2}=\frac{a^2+c^2}{b^2+d^2}\)

b) Vì \(1-\frac{1}{2^2};1-\frac{1}{3^2};...;1-\frac{1}{50^2}<1\) nên:

\(\left\{\begin{matrix} \left \{ 1-\frac{1}{2^2} \right \}=1-\frac{1}{2^2}\\ \left \{ 1-\frac{1}{3^2} \right \}=1-\frac{1}{3^2}\\ ....\\ \left \{ 1-\frac{1}{50^2} \right \}=1-\frac{1}{50^2}\end{matrix}\right.\)

\(\Rightarrow M=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{50^2}\right)\)

\(\Leftrightarrow M=\frac{(2^2-1)(3^2-1)(4^2-1)....(50^2-1)}{(2.3....50)^2}\)

\(\Leftrightarrow M=\frac{[(2-1)(3-1)...(50-1)][(2+1)(3+1)...(50+1)]}{(2.3.4...50)^2}\)

\(\Leftrightarrow M=\frac{(2.3...49)(3.4.5...51)}{(2.3.4...50)^2}=\frac{(2.3.4...49)^2.50.51}{2.(2.3....49)^2.50^2}=\frac{50.51}{2.50^2}=\frac{51}{100}\)

Với \(a,b,c\ne0\) ta có:

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

Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=1\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)

Vì \(a=b=c\Rightarrow\dfrac{a^{49}\times b^{51}}{c^{100}}=\dfrac{a^{49}\times a^{51}}{a^{100}}=\dfrac{a^{100}}{a^{100}}=1\)

Chúc bn học tốt

(Bài này phải đc gọi là "Ác mộng dấu bằng")

xem lai de

để đúng bn ạ

mk giải ra rồi

2.

\(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\) . Ta có : +,ad < bc

\(\Rightarrow\)ad+ab < bc +ab (Cùng thêm ab vào 2 vế)

\(\Rightarrow\)a(b+d) < b(a+c)

\(\Rightarrow\)\(\dfrac{a}{b}\)< \(\dfrac{a+c}{b+d}\)

+, ad < bc

\(\Rightarrow\)ad + cd < bc + cd ( Cùng thêm cd vào 2 vế)

\(\Rightarrow\)d(a+c) < c(b+d)

\(\Rightarrow\)\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) Vậy \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

2.

ta có

\(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\Rightarrow ad< bc\)

xét

\(\dfrac{a}{b}=\dfrac{a\left(b+d\right)}{b\left(b+d\right)}=\dfrac{ab+ad}{b\left(b+d\right)}\)

\(\dfrac{a+c}{b+d}=\dfrac{b\left(a+c\right)}{b\left(b+d\right)}=\dfrac{ab+bc}{b\left(b+d\right)}\)

vì \(\dfrac{ab+ad}{b\left(b+d\right)}< \dfrac{ab+bc}{b\left(b+d\right)}\left(ad< bc\right)\)

\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)

xét

\(\dfrac{a+c}{b+d}=\dfrac{d\left(a+c\right)}{d\left(b+d\right)}=\dfrac{ad+cd}{d\left(b+d\right)}\)

\(\dfrac{c}{d}=\dfrac{c\left(b+d\right)}{d\left(b+d\right)}=\dfrac{bc+cd}{d\left(b+d\right)}\)

vì

\(\dfrac{ad+cd}{d\left(b+d\right)}< \dfrac{bc+cd}{d\left(b+d\right)}\left(ad< bc\right)\)

\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\)

từ (1) và (2) => ĐPCM

Để \(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) thì a(b+d) < b(a+c)

<=> ab + ad < ba + cb

<=> ad < cb

<=> \(\dfrac{a}{b}< \dfrac{c}{d}\)

Để \(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) thì (a+c)d < (b+d)c

<=> ad + cd < bc + dc

<=> ad < bc

<=> \(\dfrac{a}{b}< \dfrac{c}{d}\)

Chúc bạn học tốt!

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=t\) \(\Rightarrow a=bt\);\(c=dt\)

rồi bạn thế vào điều phải chứng minh là ra

Bn lm chi tiết từng bài giúp mk đc k

1.

- Theo đề bài ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}\left(a,b,c,d\ne0\right)\)

=> \(\dfrac{a}{c}=\dfrac{b}{d}\)

- Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{b}{d}\) =\(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)

=> \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)( đpcm).

2.

- Ta có:

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

=> ( a+b ).(c-a) = (a-b).(c+a)

=> ac - a² +bc-ba = ac +a² -bc -ba

=> ac - a² +bc-ba -(ac +a² -bc -ba) =0

=> ac - a² +bc-ba -ac -a² +bc +ba = 0

=>ac - aa +bc-ba -ac -aa +bc +ba = 0

=> ( ac-ac) +( -aa-aa) +( bc+bc) + ( -ba+ba) =0

=> -2aa +2bc = 0

=> 2bc = 2aa

=> bc = aa

=> bc = a²

- Vậy nếu bc = a² thì \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)( đpcm).