ơ kìa đăng 2 hôm r ko ai giúp

a/ Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=k^3\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)

Áp dụng tính chất của tỉ lệ thức ta có:\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)

Mặt khác: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\)

\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(=k^3\right)\)

giup minh nha: Tinh nhanh lop 4

42 x 43 - 12 x 9 - 42 x 3

Đặt 

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

a)

Sửa đề nhá : 

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

\(\frac{2a+5c}{2b+5d}=\frac{2bk+5dk}{2b+5d}=\frac{k\left(2b+5d\right)}{2b+5d}=k\)(2)

Từ (1) và (2) 

=> \(\frac{3a+2c}{3b+2d}=\frac{2a+5c}{2b+5d}\)

b)

\(\frac{a^3}{b^3}=\frac{b^3k^3}{b^3}=k^3\)(3)

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

Từ (3) và (4) 

=> \(\frac{a^3}{b^3}=\frac{\left(a+c\right)^3}{\left(b+d\right)^3}\)

nhân chéo r rút gọn

Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\Leftrightarrow\frac{4a}{4c}=\frac{3b}{3d}\)

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

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

Vậy \(\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\left(ĐPCM\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\left(k\ne0\right)\)

a) \(\frac{4a-3b}{4c-3d}=\frac{4bk-3b}{4dk-3d}=\frac{b.\left(4k-3\right)}{d.\left(4k-3\right)}=\frac{b}{d}\)

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

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

b) \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3k^3+b^3}{d^3k^3+d^3}=\frac{b^3.\left(k^3+1\right)}{d^3.\left(k^3+1\right)}=\frac{b^3}{d^3}\)

     \(\frac{a^3-b^3}{c^3-d^3}=\frac{\left(bk\right)^3-b^3}{\left(dk\right)^3-d^3}=\frac{b^3k^3-b^3}{d^3k^3-d^3}=\frac{b^3.\left(k^3-1\right)}{d^3.\left(k^3-1\right)}=\frac{b^3}{d^3}\)

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

Gọi \(\frac{a}{b}=\frac{c}{d}=x\Rightarrow a=bx;c=dx\)

Thay vào vế trái ta được 

\(\frac{3a-5c}{4a+7c}=\frac{3.bx-5.dx}{4.bx+7.dx}=\frac{x\left(3b-5d\right)}{x\left(4b+7d\right)}=\frac{3b-5d}{4b+7d}\)

Vậy vế trái bằng vế phải

Ta có:\(\frac{a}{b}=\frac{c}{d}=\frac{3a-5c}{3b-5d}\left(1\right)\)

Ta lại có:\(\frac{a}{b}=\frac{c}{d}=>\frac{4a+7c}{4b+7d}\left(2\right)\)

Từ (1) và (2),suy ra : \(\frac{3a-5c}{4a+7c}=\frac{3b-5d}{4b+7d}\)

Cách của mình cũng đúng nhưng khác cách làm của thang Tam thôi

Đặt: \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=k\)

=> 2a = k .3b; 3b = k. 4c; 4c = k. 5d; 5d = k.2a

Mà  \(1=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=\frac{k.3b+k4c+k.5d+k.2a}{3b+4c+5d+2a}=\frac{k.\left(3b+4c+5d+2a\right)}{3b+4c+5d+2a}=k\)

=> C = 1+1+1+1 = 4

Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:

\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)

\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)

\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)

Cộng (1),(2) và (3) có:

\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)

\(\Rightarrow2VP\ge2VT\)

\(\RightarrowĐPCM\)