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Từ giả thiết ta có \(\frac{15a-10b}{25}=\frac{6c-15}{9}=\frac{10b-6c}{4}\)
\(=\frac{0}{38}=0\)
(Theo t/c day ti so bang nhau)
Suy ra \(\hept{\begin{cases}15a-10b=0\\6c-15a=0\end{cases}\Rightarrow\hept{\begin{cases}3a-2b=0\\2c-5a=0\end{cases}}}\Rightarrow\hept{\begin{cases}b=\frac{3}{2}a\\c=\frac{5}{2}a\end{cases}}\)
Mà a^2+275=bc Suy ra \(^{a^2+275=\frac{15}{4}a^2\Rightarrow a^2=100\Rightarrow a=\pm10}\)
ĐS: a=10; b=15; c=25 và a=-10; b=-15; c=-25
Sửa chút \(\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)
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
Lời giải:
\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)
\(\Leftrightarrow \frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)
Áp dụng TC dãy tỉ số bằng nhau:
\(\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=0\)
\(\Rightarrow 15a=10b=6c\)
\(\Leftrightarrow \frac{a}{6}=\frac{b}{9}=\frac{c}{15}\)
Tiếp tục áp dụng TCDTSBN:
$\frac{a}{6}=\frac{b}{9}=\frac{c}{15}=\frac{a+b+c}{6+9+15}=\frac{50}{30}=\frac{5}{3}$
$\Rightarrow a=10; b=15; c=25$
bạn tham khảo tại: câu hỏi của nguyễn thị thanh mai- onlinemath
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:
\(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}=\dfrac{a+b+c}{\dfrac{1}{15}+\dfrac{1}{10}+\dfrac{1}{6}}=-\dfrac{50}{\dfrac{1}{3}}=-150\)
\(\Rightarrow\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)
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 ..
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: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
\(3a=\dfrac{b}{\dfrac{2}{5}}=\dfrac{8c}{3}\Rightarrow\dfrac{6a}{2}=\dfrac{5b}{\dfrac{2}{5}\cdot5}=\dfrac{16c}{6}\\ \Rightarrow\dfrac{6a}{2}=\dfrac{5b}{2}=\dfrac{16c}{6}\)
Áp dụng t/c dtsbn:
\(3a=\dfrac{b}{\dfrac{2}{5}}=\dfrac{8c}{3}=\dfrac{6a}{2}=\dfrac{5b}{2}=\dfrac{16c}{6}=\dfrac{6a+5b+16c}{2+2+6}=\dfrac{10}{10}=1\\ \Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{2}{5}\\c=\dfrac{3}{8}\end{matrix}\right.\)