Đặt a/b=c/d=k

=>a=bk; c=dk

1: \(\dfrac{2a+15b}{5a-7b}=\dfrac{2\cdot bk+15b}{5\cdot bk-7b}=\dfrac{2k+15}{5k-7}\)

\(\dfrac{2c+15d}{5c-7d}=\dfrac{2dk+15d}{5dk-7d}=\dfrac{2k+15}{5k-7}\)

Do đó: \(\dfrac{2a+15b}{5a-7b}=\dfrac{2c+15d}{5c-7d}\)

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

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

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

hay (a+2c)(b+d)=(a+c)(b+2d)

a: \(\dfrac{2a+15b}{5a-7b}=\dfrac{2c+15d}{5c-7d}\)

\(\Leftrightarrow\left(2a+15b\right)\left(5c-7d\right)=\left(5a-7b\right)\left(2c+15d\right)\)

\(\Leftrightarrow10ac-14ad+75bc-105bd=10ac+75ad-14bc-105bd\)

\(\Leftrightarrow-14ad+75bc=-14bc+75ad\)

=>ad=bc

hay a/b=c/d

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\)

\(\dfrac{2c^2-ac}{2d^2-bd}=\dfrac{2\cdot d^2k^2-bk\cdot dk}{2\cdot d^2-bd}=k^2\)

Do đó; \(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)

Vì \(\frac{2a+15b}{5a-7b}=\frac{2c+15d}{5c-7d}\)

\(\Rightarrow\frac{2a+15b}{2c+15d}=\frac{5a-7b}{5c-7d}=\frac{2c}{2c}=\frac{15b}{15b}=\frac{5a}{5c}=\frac{7b}{7d}\)( áp dụng tc của dãy tỉ số = nhau )

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

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Thay a = bk; c = dk vào đẳng thức \(\frac{2a+15b}{5a-7b}=\frac{2a+15d}{5c-7d}\). Ta được: 

+, \(\frac{2bk+15b}{5bk-7b}=\frac{b\left(2k+15\right)}{b\left(5k-7\right)}=\frac{2k+15}{5k-7}\)(1)

+, \(\frac{2dk+15d}{5dk-7d}=\frac{d\left(2k+15\right)}{d\left(5k-7\right)}=\frac{2k+15}{5k-7}\)(2)

Từ (1) và (2) 

\(\Rightarrow\frac{2bk+15b}{5bk-7b}=\frac{2dk+15d}{5dk-7d}\)

Hay \(\frac{2a+15b}{5a-7b}=\frac{2c+15d}{5c-7d}\)<đpcm>

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

Khi đó : \(\frac{2a+15b}{5a-7b}=\frac{2bk+15b}{5bk-7b}=\frac{b\left(2k+15\right)}{b\left(5k-7\right)}=\frac{2k+15}{5k-7}\left(1\right)\)

\(\frac{2c+15d}{5c-7d}=\frac{2dk+15d}{5dk-7d}=\frac{d\left(2k+15\right)}{d\left(5k-7\right)}=\frac{2k+15}{5k-7}\left(2\right)\)

Từ (1) và (2) 

=> \(\frac{2a+15b}{5a-7b}=\frac{2c+15d}{5c-7d}\left(\text{đpcm}\right)\)

1, Ta có:\(\frac{2a+15b}{5a-7b}=\frac{2c+15d}{5c-7d}\)\(\Rightarrow\frac{2a+15b}{2c+15d}=\frac{5a-7b}{5c-7d}\)

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

\(\frac{2a+15b}{2c+15d}=\frac{5a-7b}{5c-7d}=\frac{2a+15b+5a-7b}{2c+15d+5c-7d}=\frac{7a-8b}{7c-8d}\)

\(\Rightarrow\frac{7a-8b}{7c-8d}=\frac{7a}{7c}=\frac{8b}{8d}\)\(\Rightarrow\frac{7a}{7c}=\frac{8b}{8d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)(đpcm)

2, Ta có: \(4^{30}=2^{30}.2^{30}=2^{30}.\left(2^2\right)^{15}=2^{30}.4^{15}\)

Lại có: \(3.24^{10}=3.3^{10}.8^{10}=3^{11}.\left(2^3\right)^{10}=3^{11}.2^{30}\)

Vì \(4^{15}>3^{11}\)\(\Rightarrow2^{30}.4^{15}>2^{30}.3^{11}\)\(\Rightarrow4^{30}>3.24^{10}\)\(\Rightarrow2^{30}+3^{30}+4^{30}>3.24^{10}\)

Sửa lại câu 1.

Với đk: \(5a\ne7b;5c\ne7d\);  \(b;d\ne0\).

\(\frac{2a+15b}{5a-7b}=\frac{2c+15d}{5c-7d}\)

TH1: \(2c+15d=0\)=> \(2a+15b=0\)=> \(\frac{a}{b}=\frac{c}{d}\)

TH2: \(2c+15d\ne0\)

=> \(\frac{2a+15b}{2c+15d}=\frac{5a-7b}{5c-7d}\)

=> \(\frac{5\left(2a+15b\right)}{5\left(2c+15d\right)}=\frac{2\left(5a-7b\right)}{2\left(5c-7d\right)}\)

=> \(\frac{10a+75b}{10c+75d}=\frac{10a-14b}{10c-14d}\)

Áp dụng dãy tỉ số bằng nhau ta có: 

\(\frac{10a+75b}{10c+75d}=\frac{10a-14b}{10c-14d}=\frac{10a+75b-10a+14b}{10c+75d-10c+14d}=\frac{89b}{89d}=\frac{b}{d}\)

=> \(\frac{10a+75b}{10c+75d}=\frac{b}{d}=\frac{75b}{75d}=\frac{10a+75b-75b}{10c+75d-75d}=\frac{10a}{10c}=\frac{a}{c}\)

=> \(\frac{b}{d}=\frac{a}{c}\)

=> \(\frac{a}{b}=\frac{c}{d}\).

Bạn thiếu điều kiện cô Linh Chi đã bổ sung thêm rồi còn mình chỉ làm bài thôi

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

\(\frac{2a+15b}{5a-7b}=\frac{2bk+15b}{5bk-7b}=\frac{b\left(2k+15\right)}{b\left(5k-7\right)}=\frac{2k+15}{5k-7}\left(1\right)\)

\(\frac{2c+15d}{5c-7d}=\frac{2dk+15d}{5dk-7d}=\frac{d\left(2k+15\right)}{d\left(5k-7\right)}=\frac{2k+15}{5k-7}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{c}{d}\)

Chúc bạn học tốt !!!