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

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: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)

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\(\dfrac{3a-2b}{5}\)=\(\dfrac{2c-5a}{3}\)=\(\dfrac{5b-3c}{2}\)=\(\dfrac{15a-10b}{5}\)=\(\dfrac{6c-15a}{9}\)=\(\dfrac{10b-6c}{2}\)

Suy ra: \(\dfrac{15a-10b+6c-15a+10b-6c}{25+9+4}\)=\(\dfrac{0}{38}\)=0

Suy ra: 3a=2b\(\Leftrightarrow\)\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)(1)

2c=5a\(\Leftrightarrow\)\(\dfrac{c}{5}\)=\(\dfrac{a}{2}\)(2)

Từ (1) và (2) suy ra: \(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{5}\)

Theo tính chất của dãy tỉ số bằng nhau ta được:

\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{5}\)=\(\dfrac{a+b+c}{2+3+5}\)=\(\dfrac{-50}{10}\)=-5

Tự làm nốt nha.

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Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-5c}{2}=\dfrac{5\left(3a-2b\right)\left(2c-5a\right)}{5.5+3.3+}=\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\dfrac{5b-3c}{2}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3c}{5}\\a=\dfrac{2c}{5}\end{matrix}\right.\)

Mà \(a+b+c=-50\)

\(\Leftrightarrow\dfrac{2c}{5}+\dfrac{3c}{5}+c=-50\)

\(\Leftrightarrow2c=-50\)

\(\Leftrightarrow c=-25\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-10\\b=-15\end{matrix}\right.\)

Vậy ...

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\leftrightarrow\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}=\dfrac{5\left(3a-2b\right)+3\left(2c-5a\right)+2\left(5b-3c\right)}{25+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}3a-2b=0\\2c-5a=0\\5b-3c=0\end{matrix}\right.\)
⇔ 15a= 10b = 6c ⇔ \(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}=\dfrac{a+b+c}{\dfrac{1}{15}+\dfrac{1}{10}+\dfrac{1}{6}}=-\dfrac{50}{\dfrac{1}{3}}=-150\)
\(\Rightarrow\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)

`a^2+4ab-5b^2=0`

`<=>a^2+4ab+4b^2-9b^2=0`

`<=>(a+2b)^2-9b^2=0`

`<=>(a+2b-3b)(a+2b+3b)=0`

`<=>(a-b)(a+5b)=0`

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=-5b\end{matrix}\right.\)

`Q={2a-b}/{a-b}+{3a-2b}/{a+b}`

Với `a=b` `=>` giá trị vô nghĩa

Với `a=-5b` 

`Q={-10b-b}/{-5b-b}+{-15b-2b}/{-5b+b}`

`Q={-11b}/{-6b}+{-17b}/{-4b}`

`Q=11/6+17/4`

`Q=73/12`

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

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}=\dfrac{5.\left(3a-2b\right)+3.\left(2c-5a\right)}{5.5+3.3}=\dfrac{-10b+6c}{34}=\)

\(=\dfrac{-5b+3c}{17}\)

Do đó: \(\dfrac{5b-3c}{14}=\dfrac{-5b+3c}{2}\)

Suy ra: \(5b-3c=0\Rightarrow b=\dfrac{3}{5}c\) và \(a=\dfrac{2}{5}c\)

Lại có: \(a+b+c=-50\Rightarrow\dfrac{2}{5}c+\dfrac{3}{5}c+c=-50\Rightarrow c=-25\)

\(\Rightarrow b=\dfrac{3}{5}.\left(-25\right)=-15\)

và \(a=\dfrac{2}{5}.\left(-25\right)=-10\)

Vậy \(\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)

Chúc bạn học tốt!!!

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}=\dfrac{5\left(3a-2b\right)\left(2c-5a\right)}{5.5+3.3}=\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\dfrac{5b-3c}{2}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3c}{5}\\a=\dfrac{2c}{5}\end{matrix}\right.\)

Mà \(a+b+c=-50\)

\(\Leftrightarrow\dfrac{2c}{5}+\dfrac{3c}{5}+c=-50\)

\(\Leftrightarrow2c=-50\)

\(\Leftrightarrow c=-25\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=-15\\a=-10\end{matrix}\right.\)

Vậy ..

Một cách giải khác:

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)

\(\Leftrightarrow \frac{5(3a-2b)}{25}=\frac{3(2c-5a)}{9}=\frac{2(5b-3c)}{4}=\frac{5(3a-2b)+3(2c-5a)+2(5b-3c)}{25+9+4}=0\)

\(\Rightarrow \left\{\begin{matrix} 3a-2b=0\\ 2c-5a=0\\ 5b-3c=0\end{matrix}\right.\)

\(\Leftrightarrow 15a=10b=6c\Leftrightarrow \frac{a}{\frac{1}{15}}=\frac{b}{\frac{1}{10}}=\frac{c}{\frac{1}{6}}=\frac{a+b+c}{\frac{1}{15}+\frac{1}{10}+\frac{1}{6}}=\frac{-50}{\frac{1}{3}}=-150\)

(Áp dụng tính chất dãy tỉ số bằng nhau)

\(\Rightarrow \left\{\begin{matrix} a=-10\\ b=-15\\ c=-25\end{matrix}\right.\)