a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\frac{4a-3b}{4a+3b}=\frac{4c-3d}{4c+3d}\Rightarrow\frac{4a-3d}{4c-3d}=\frac{4a+3b}{4c+3d}\)

b) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

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

a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)

Áp dụng tỉ lệ thức ta có :

\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\)\(\frac{4a}{4c}=\frac{3b}{3d}\Rightarrow\frac{4a+3b}{4c+3d}=\frac{4c-3d}{4c-3d}\)

b) Có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)

Áp dụng tỉ lệ thức ta có "

\(\frac{2a}{3b}=\frac{2c}{3d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\Rightarrow\frac{2a-3b}{2c-3d}=\frac{2a3b}{2c+3d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)

Các câu còn lại bạn làm tương tự

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\). Khi đó ta có:

a)

\((a+c)(b-d)=(bk+dk)(b-d)=k(b+d)(b-d)\)

\((a-c)(b+d)=(bk-dk)(b+d)=k(b-d)(b+d)=k(b+d)(b-d)\)

\(\Rightarrow (a+c)(b-d)=(a-c)(b+d)\) (đpcm)

b)

\((a+c)b=(bk+dk)b=k(b+d).b=bk(b+d)\)

\((b+d).a=(b+d).bk=bk(b+d)\)

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

c)

\(a(b-d)=bk(b-d)\)

\(b(a-c)=b(bk-dk)=bk(b-d)\)

\(\Rightarrow a(b-d)=b(a-c)\)

d)

\((b+d).c=(b+d).dk=dk(b+d)\)

\((a+c)d=(bk+dk)d=k(b+d)d=dk(b+d)\)

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

e)

\((b-d).c=(b-d).dk=dk(b-d)\)

\((a-c)d=(bk-dk)d=k(b-d)d=dk(b-d)\)

\(\Rightarrow (b-d)c=(a-c)d\)

f)

\((a+b)(c-d)=(bk+b)(dk-d)=b(k+1)d(k-1)=bd(k-1)(k+1)\)

\((a-b)(c+d)=(bk-b)(dk+d)=b(k-1)d(k+1)=bd(k-1)(k+1)\)

\(\Rightarrow (a+b)(c-d)=(a-b)(c+d)\)

g)

\((2a+3c)(2b-3d)=(2bk+3dk)(2b-3d)=k(2b+3d)(2b-3d)\)

\((2a-3c)(2b+3d)=(2bk-3dk)(2b+3d)=k(2b-3d)(2b+3d)\)

\(\Rightarrow (2a+3c)(2b-3d)=(2a-3c)(2b+3d)\)

h)

\((4a+3b)(4c-3d)=(4bk+3b)(4dk-3d)=b(4k+3)d(4k-3)=bd(4k+3)(4k-3)\)

\((4a-3b)(4c+3d)=(4bk-3b)(4dk+3d)=b(4k-3)d(4k+3)=bd(4k+3)(4k-3)\)

\(\Rightarrow (4a+3b)(4c-3d)=(4a-3b)(4c+3d)\)

i,k: Hoàn toàn tương tự.

Gọi \(\frac{a}{b}=\frac{c}{d}=k\left(k\in R\right)\)thì a = bk ; c = dk . Ta có :

\(\frac{1111c-99d}{9999c-11d}=\frac{1111dk-99d}{9999dk-11d}=\frac{d\left(1111k-99\right)}{d\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\)(1)

\(\frac{1111a-99b}{9999a-11b}=\frac{1111bk-99b}{9999bk-11b}=\frac{b\left(1111k-99\right)}{b\left(9999k-11\right)}=\frac{1111k-99}{9999k-11}\)(2)

Từ (1) và (2) , ta có \(\frac{1111c-99d}{9999c-11d}=\frac{1111a-99b}{9999a-11b}\)

day la cac tinh chat ma

a/

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}\)

Áp dụng tỉ lệ thức ta có:

\(\frac{a^3}{b^3}=\frac{c^3}{d^3}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)

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

\(\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)

Vậy \(\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)

a) Ta có:

+) a/2=b/3

=>a=2b/3

+) b/5=c/4

=>c=4b/5

Lại có:

a-b+c=49

=> 2b/3 -b + 4b/5 =49

=> 7b/15==49

=> b= 105

Khi đó:

+) a=2b/3=2.105/3=70

+)c=4b/5=4.105/5=84

Vậy a=70; b=105; c=84...

chúc bạn học tốt

b: 3a=2b nen a/2=b/3

5b=7c nên b/7=c/5

=>a/14=b/21=c/15

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{14}=\dfrac{b}{21}=\dfrac{c}{15}=\dfrac{3a+5b-7c}{3\cdot14+5\cdot21-7\cdot15}=\dfrac{60}{42}=\dfrac{10}{7}\)

=>a=20; b=30; c=150/7

c: a/b=8/5 nên a/8=b/5

b/3=c/5 nên b/15=c/25

=>a/24=b/15=c/25=k

=>a=24k; b=15k; c=25k

a+b+c=61

=>24k+15k+25k=61

=>k=61/64

=>a=183/8; b=915/64; c=1525/64

e: a/2=b/3 nên a/8=b/12

b/4=c/5 nên b/12=c/15

=>a/8=b/12=c/15=k

=>a=8k; b=12k; c=15k

a^2-b^2=-16

=>64k^2-144k^2=-16

=>80k^2=16

=>k^2=1/5

TH1: k=1/căn 5

=>\(a=\dfrac{8}{\sqrt{5}};b=\dfrac{12}{\sqrt{5}};c=3\sqrt{5}\)

TH2: k=-1/căn 5

=>\(a=\dfrac{-8}{\sqrt{5}};b=\dfrac{-12}{\sqrt{5}};c=-3\sqrt{5}\)

a)Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

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

=>\(\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)

=>\(\frac{ac}{bd}=\frac{a^2+b^2}{c^2+d^2}\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}=\frac{a^2-c^2}{b^2-d^2}\)

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

\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}\Rightarrow\frac{a^2+b^2}{a^2-b^2}=\frac{c^2+d^2}{c^2-d^2}\)

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

a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{2a+5b}{2c+5d}=\dfrac{2bk+5b}{2dk+5d}=\dfrac{b}{d}\)

\(\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5k}=\dfrac{b}{d}\)

Do đó: \(\dfrac{2a+5b}{2c+5d}=\dfrac{2a-5b}{2c-5d}\)

b: \(\dfrac{a^2-b^2}{a^2+b^2}=\dfrac{b^2k^2-b^2}{b^2k^2+b^2}=\dfrac{k^2-1}{k^2+1}\)

\(\dfrac{c^2-d^2}{c^2+d^2}=\dfrac{d^2k^2-d^2}{d^2k^2+d^2}=\dfrac{k^2-1}{k^2+1}\)

Do đó: \(\dfrac{a^2-b^2}{a^2+b^2}=\dfrac{c^2-d^2}{c^2+d^2}\)