a)12a + 36b = 2(6a + 18b) chia hết cho 2

3211 không chia hết cho 2

=> không tìm được a,b thỏa mãn đề.

b)Đặt A=2a+7b

         B=4a+2b

xét hiệu:2A-B=2.(2a+7b)-(4a+2b)

=4a+14b-4a-2b

=12b

Vì A ⋮3 nên 2a⋮3;12b⋮3

⇒B⋮3 hay 4a+2b ⋮3(đpcm)

a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó (2a + 3c)(2b - 3d) 

= (2bk + 3dk)(2b - 3d)

= k(2b + 3d)(2b - 3d) (1)

(2a - 3c)(2b + 3d)

= (2bk - 2dk)(2b + 3d)

= k(2b - 3d)(2b + 3d) (2)

Từ (1)(2) => (2a + 3c)(2b - 3d) = (2a - 3c)(2b + 3d)

b) Sửa đề (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d) 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có (4a + 3b)(4c - 3d) = (4bk + 3b)(4dk - 3d) = bd(4k + 3)(4k - 3) (1)

Lại có (4a - 3b)(4c + 3d) = (4bk - 3b)(3dk + 3d) = bd(4k- 3)(4k + 3) (2)

Từ (1)(2) => (4a + 3b)(4c - 3d) = (4a - 3b)(4c + 3d) 

1, Ta có: \(\frac{a}{b}=\frac{c}{d}\)

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

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

        Vậy (2a + 3c).(2b - 3d) = (2a - 3c).(2b + 3d)

Câu 2 cũng tương tự nên tự làm đi

2a - b = 3 => b = 2a - 3.thay vào  4a + 2b = 34  ta đc

4a + 2.(2a - 3) = 34 => 4a + 4a - 6 = 34 => 8a = 40 =>a =  40 : 8 = 5  => b = 2.5 - 3 = 7

vậy a/b = 5/7

\(P=\dfrac{4a^2}{4b+2c}+\dfrac{4b^2}{4a+2c}+\dfrac{c^2}{4a+4b}\ge\dfrac{\left(2a+2b+c\right)^2}{8a+8b+4c}\)

\(=\dfrac{\left(2a+2b+c\right)^2}{4\left(2a+2b+c\right)}=\dfrac{1}{4}\left(2a+2b+c\right)\)

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r

a. Ta có: a > b

4a > 4b ( nhân cả 2 vế cho 4)

4a - 3 > 4b - 3 (cộng cả 2 vế cho -3)

b. Ta có: a > b

-2a < -2b ( nhân cả 2 vế cho -2)

1 - 2a < 1 - 2b (cộng cả 2 vế cho 1)

d. Ta có: a < b 

-2a > -2b ( nhân cả 2 vế cho -2)

5 - 2a > 5 - 2b (cộng cả 2 vế cho 5)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

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

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)