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

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

a) Ta có:

+) a/2=b/3

=>a=2b/3

+) b/5=c/4

=>c=4b/5

Lại có:

a-b+c=49

=> 2b/3 -b + 4b/5 =49

=> 7b/15==49

=> b= 105

Khi đó:

+) a=2b/3=2.105/3=70

+)c=4b/5=4.105/5=84

Vậy a=70; b=105; c=84...

chúc bạn học tốt

b: 3a=2b nen a/2=b/3

5b=7c nên b/7=c/5

=>a/14=b/21=c/15

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

\(\dfrac{a}{14}=\dfrac{b}{21}=\dfrac{c}{15}=\dfrac{3a+5b-7c}{3\cdot14+5\cdot21-7\cdot15}=\dfrac{60}{42}=\dfrac{10}{7}\)

=>a=20; b=30; c=150/7

c: a/b=8/5 nên a/8=b/5

b/3=c/5 nên b/15=c/25

=>a/24=b/15=c/25=k

=>a=24k; b=15k; c=25k

a+b+c=61

=>24k+15k+25k=61

=>k=61/64

=>a=183/8; b=915/64; c=1525/64

e: a/2=b/3 nên a/8=b/12

b/4=c/5 nên b/12=c/15

=>a/8=b/12=c/15=k

=>a=8k; b=12k; c=15k

a^2-b^2=-16

=>64k^2-144k^2=-16

=>80k^2=16

=>k^2=1/5

TH1: k=1/căn 5

=>\(a=\dfrac{8}{\sqrt{5}};b=\dfrac{12}{\sqrt{5}};c=3\sqrt{5}\)

TH2: k=-1/căn 5

=>\(a=\dfrac{-8}{\sqrt{5}};b=\dfrac{-12}{\sqrt{5}};c=-3\sqrt{5}\)

b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)

Vậy (a,b,c) = (18,16,15) 

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

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

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

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

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

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