Không có điều kiện gì à ( Kiểu \(\dfrac{a}{b}=\dfrac{c}{d}\) ấy )

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

\(BDT\Leftrightarrow2a^4b+2b^4c+2c^4a+3ab^4+3bc^4+3ca^4\ge5a^2b^2c+5a^2bc^2+5ab^2c^2\)

Ta chứng minh được \(ab^4+bc^4+ca^4\ge a^2b^2c+a^2bc^2+ab^2c^2\)

\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)

\(VT=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=VP\)

Vậy ta cần chứng minh \(2a^4b+2b^4c+2c^4a+2ab^4+2bc^4+2ca^4\ge4a^2b^2c+4a^2bc^2+4ab^2c^2\)

\(\Leftrightarrow\sum_{cyc}\left(2c^3+bc^2-b^2c+ac^2-a^2c+3ab^2+3a^2b\right)\left(a-b\right)^2\ge0\)

Dấu "=" xảy ra khi \(a=b=c\)

Em có cách này tuy nhiên không chắc,do em mới học sos thôi,mong mọi người giúp đỡ ạ!

BĐT \(\Leftrightarrow\Sigma_{cyc}\left(\frac{7b^3+3ab^2-7a^2b-3a^3}{2a+3b}\right)\ge0\)\(\Leftrightarrow\Sigma_{cyc}\left(\frac{7b\left(b^2-a^2\right)+3a\left(b^2-a^2\right)}{2a+3b}\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\left(\frac{\left(b^2-a^2\right)\left(7b+3a\right)}{2a+3b}-2\left(b^2-a^2\right)\right)\ge0\) (ta không cần cộng thêm \(\Sigma_{cyc}2\left(b^2-a^2\right)\) vì \(\Sigma_{cyc}2\left(b^2-a^2\right)=\Sigma_{cyc}2\left(b^2-a^2+c^2-b^2+a^2-c^2\right)=0\))

\(\Leftrightarrow\Sigma_{cyc}\left(b^2-a^2\right)\left(\frac{7b+3a-4a-6b}{2a+3b}\right)\ge0\)\(\Leftrightarrow\Sigma_{cyc}\frac{\left(a+b\right)\left(a-b\right)^2}{2a+3b}\ge0\)

P/s: Hình như có gì đó sai sai ạ,mong mọi người check hộ em!Em cảm ơn nhiều ạ!

sos helps :3

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)

\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)

*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)

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

Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )

=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)

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

Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

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

b: Đặt a/b=c/d=k

=>a=bk; c=dk

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

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

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

c: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)

\(\dfrac{a^3-b^3}{c^3-d^3}=\dfrac{b^3k^3-b^3}{d^3k^3-d^3}=\dfrac{b^3}{d^3}\)

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

d: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{b^{2018}k^{2018}-b^{2018}}{b^{2018}k^{2018}+b^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

\(\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

Do đó: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}\)

Bài 1:

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

Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)

\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)

Ta có đpcm.

Bài 2:

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

Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)

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

Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.

Bài 3:

a) Sửa điều kiện: \(\frac{a}{b}=\frac{c}{d}\neq -1\)

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

Theo đkđb thì \(k\neq -1\) nên \(k^3+1\neq 0\); \(k+1\neq 0\)

Ta có: \(\frac{a^3+b^3}{c^3+d^3}=\frac{(bk)^3+b^3}{(dk)^3+d^3}=\frac{b^3(k^3+1)}{d^3(k^3+1)}=\frac{b^3}{d^3}\)

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

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

b)

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

Ta cần cm \(k=t\)

Khi đó:

\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b(2k+13)}{b(3k-7)}=\frac{2k+13}{3k-7}\)

\(\frac{2c+13d}{3c-7d}=\frac{2dt+13d}{3dt-7d}=\frac{d(2t+13)}{d(3t-7)}=\frac{2t+13}{3t-7}\)

Vì \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\Rightarrow \frac{2k+13}{3k-7}=\frac{2t+13}{3t-7}\)

\(\Rightarrow (2k+13)(3t-7)=(2t+13)(3k-7)\)

\(-14k+39t=-14t+39k\Rightarrow k=t\)

Ta có đpcm.

Bài 1:

a) Có: 4a = 3b => \(\dfrac{a}{3}=\dfrac{b}{4}\) => \(\dfrac{a}{15}=\dfrac{b}{20}\)

7b = 5c => \(\dfrac{b}{5}=\dfrac{c}{7}\) => \(\dfrac{b}{20}=\dfrac{c}{28}\)

=> \(\dfrac{a}{15}=\dfrac{b}{20}=\dfrac{c}{28}\)

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

\(\dfrac{a}{15}=\dfrac{b}{20}=\dfrac{c}{28}=\dfrac{2a+3b-c}{30+60-28}=\dfrac{186}{62}=3\)

=> \(\left\{{}\begin{matrix}a=45\\b=60\\c=84\end{matrix}\right.\)

b) Tương tự câu a

c) Đặt \(\dfrac{a-1}{2}=\dfrac{b-2}{3}=\dfrac{c-3}{4}=k\)

=> \(\left\{{}\begin{matrix}a=2k+1\\b=3k+2\\c=4k+3\end{matrix}\right.\)

Mà a - 2b + 3c = 14 => 2k + 1 - 6k - 4 + 12k + 9 = 8k + 6 = 14 => k = 1

=> \(\left\{{}\begin{matrix}a=3\\b=5\\c=7\end{matrix}\right.\)

d) Từ a:b:c = 3:4:5 => \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}\)

Đặt \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\)

=> \(\left\{{}\begin{matrix}a=3k\\b=4k\\c=5k\end{matrix}\right.\)

Mà 2a² + 2b² - 3c² = -100 => 18k² + 32k² - 75k² = -100 => k² = 4 => k = \(\pm\)2

Với k = 2 => \(\left\{{}\begin{matrix}a=6\\b=8\\c=10\end{matrix}\right.\)

Với k = -2 => \(\left\{{}\begin{matrix}a=-6\\b=-8\\c=-10\end{matrix}\right.\)

Bài 2:

Nửa chu vi hình chữ nhật là: 90:2 = 45 (m)

Tỉ số giữa chiều dài và chiều rộng = \(\dfrac{2}{3}\)=> chiều rộng = \(\dfrac{2}{5}\) nửa chu vi

=> chiều rộng = 18(m) => chiều dài = 27(m)

thánh nhân xuất hiện đê

Ta có:

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

<=>\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{3a}{3c}=\dfrac{2b}{2d}\)

<=>\(\dfrac{5a-3b}{5c-3d}=\dfrac{3a-2b}{3c-2d}\)(đpcm)

Các câu sau tương tự

Nguyễn Thị Hồng Nhung chị làm bài f đc ko ạ ???

c/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
d/ tương tự câu b/
e/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2} \)(1)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)(đpcm)
f/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(1\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2-c^2}{b^2-d^2}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(a+c\right)^2}{a^2-c^2}=\dfrac{\left(b+d\right)^2}{b^2-d^2}\)

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