bn ơi đề bài k cs d mà bắt Cm cs cả d là sao

nhầm đầu bn nha bạn phải là \(\frac{a}{b}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\) 

Thay:

\(\frac{2a^2-3ab+4b^2}{5b^2+6ab}=\frac{2k^2b^2-3kb^2+4b^2}{5b^2+6kb^2}=\frac{b^2\left(2k^2-3k+4\right)}{b^2\left(5+6k\right)}=\frac{2k^2-3k+4}{5+6k}\)

\(\frac{2c^2-3cd+4d^2}{5d^2+6cd}=\frac{d^2\left(2k^2-3k+4\right)}{d^2\left(5+6k\right)}=\frac{2k^2-3k+4}{5+6k}\) 

=>đpcm

Bạn tham Khảo: https://hoc24.vn/hoi-dap/question/230602.html

Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)

\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2\left(bk^2\right)-3bkb+5b^2}{2b^2+3bkb}=\frac{2b^2.k^2-3kb^2+5b^2}{2b^2+3b^2.k}\)\(=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+5}{2+3k}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)\(=\frac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3dkd}\)

Tương tự nhóm tiếp là ra

=>bằng nhau

Đặt a/b=c/d=k rồi thay vào nha bạn

GỢI Ý:

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

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

\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+5}{3k+2}\)

\(\frac{2c^2-3cd+5d^2}{2d^2+3cd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\frac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\frac{2k^2-3k+5}{3k+2}\)

nên 2 phân số trên bằng nhau (đpcm)

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

Ta có : \(\frac{2a^2-3ab+5b^2}{2b^2+3ab}\)

<=> \(\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}\)

<=> \(\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}\)

<=> \(\frac{2k^2-3k+5}{2+3k}\left(1\right)\)

Ta có: \(\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)

<=> \(\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}\)

<=> \(\frac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}\)

<=> \(\frac{2k^2-3k+5}{2+3k}\left(2\right)\)

Từ 1 và 2 => đpcm

Mình hướng dẫn thôi. Chứ giờ đang bận.

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\).Rồi thay a = kb; c=kd vào từng vế. Thấy hai vế bằng nhau => đpcm

\(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=>\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{2a^2}{2c^2}=\frac{5b^2}{5d^2}=\frac{3ab}{3ab}=\frac{3cd}{3cd}=\frac{2a^2-3ab+5b^2}{2b^2-3cd+5d^2}=\frac{2b^2+3ab}{2d^2+3cd}\)

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

Đặt\(\frac{a}{b}=\frac{c}{d}=k\),ta có:

\(a=bk\)\(c=dk\)\(\Rightarrow\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2\left(bk\right)^2-3bkb+5b^2}{2b^2+3bkb}=\frac{2b^2k^2-3b^2k+5b^2}{2b^2+3b^2k}=\frac{b^2.\left(2k^2+3k+5\right)}{b^2.\left(2+3k\right)}\)\(=\frac{2k^2+3k+5}{2+3k}\left(1\right)\)

\(\Rightarrow\frac{2c^2-3cd+5d^2}{2d^2+3cd}=\frac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\frac{d^2.\left(2k^2+3k+5\right)}{d^2.\left(2+3k\right)}\)

\(=\frac{2k^2+3k+5}{2+3k}\)(2)

Từ (1) và (2) suy ra:

\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)(đpcm)

Điều kiện mẫu thức xác định là sao?