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TL :
Ta có : \(\frac{1+2a}{15}=\frac{7-3a}{20}=\frac{3b}{23+7a}\)
Vì \(\frac{1+2a}{15}=\frac{7-3a}{20}\)
\(\Rightarrow20\left(1+2a\right)=15\left(7-3a\right)\)
\(\Leftrightarrow20+40a=105-45a\Leftrightarrow40a+45a=105-20\)
\(\Leftrightarrow95a=95\Rightarrow a=1\)
Thay a = 1 vào phương trình \(\frac{7-3a}{20}=\frac{3b}{23+7a}\); ta có : \(\frac{7-3.1}{20}=\frac{3b}{23+7.1}\)
\(\Leftrightarrow\frac{4}{20}=\frac{3b}{30}\Leftrightarrow\frac{1}{5}=\frac{b}{10}\Leftrightarrow5b=10\Rightarrow b=2\)
Vậy a = 1 ; b = 2
Có:
\(\frac{1+2a}{15}=\frac{7-3a}{20}\Leftrightarrow20\left(1+2a\right)=15\left(7-3a\right)\Rightarrow a=1\)
Thay a=1 vào\(\frac{1+2a}{15}=\frac{3b}{23+7a}=\frac{1}{5}=\frac{b}{10}\Rightarrow b=2\)
Ta có: \(\dfrac{1+2a}{15}=\dfrac{7-3a}{20}\Rightarrow20\left(1+2a\right)=15\left(7-3a\right)\Leftrightarrow20+40a=105-45a\Leftrightarrow40a+45a=105-20\Leftrightarrow85a=85\Leftrightarrow a=1\)
Ta có: \(\dfrac{7-3a}{20}=\dfrac{3b}{23+7a}\Rightarrow\dfrac{7-3.1}{20}=\dfrac{3b}{23+7.1}\Rightarrow\dfrac{4}{20}=\dfrac{3b}{30}\Rightarrow\dfrac{1}{5}=\dfrac{b}{10}\Rightarrow b=2\) Vậy a=1;b=2
Ta có: \(\frac{1+2a}{15}=\frac{7-3a}{20}\Rightarrow20\left(1+2a\right)=15\left(7-3a\right)\Rightarrow20+40a=105-45a\)
\(\Rightarrow85a=85\Rightarrow a=1\)
Thay a = 1 vào \(\frac{7-3a}{20}=\frac{3b}{23+7a}\), ta được:
\(\frac{3b}{23+7}=\frac{7-3}{20}\Rightarrow\frac{3b}{30}=\frac{1}{5}\Rightarrow b=\frac{30}{3.5}=2\)
Vậy a = 1 , b = 2
\(\frac{1+2a}{15}=\frac{7-3a}{20}=\frac{3b}{23+7a}=\frac{3\left(1+2a\right)}{45}=\frac{2\left(7-3a\right)}{40}=\frac{17}{85}=\frac{1}{5}.\)
Vậy 1 + 2a = 3 => a = 1
Thay vào: \(\frac{3b}{23+7\cdot1}=\frac{1}{5}\Rightarrow\frac{3b}{30}=\frac{1}{5}\Rightarrow b=2.\)
Vậy, a = 1 và b = 2.
Từ \(\frac{1+2a}{15}=\frac{7-3a}{20}\)\(\Rightarrow20+40a=105-45a\)
\(\Rightarrow a=1\)
Lại có \(\frac{1+2a}{15}=\frac{3b}{23+7a}\Rightarrow\frac{1}{5}=\frac{3b}{30}\Rightarrow b=2\)
Đáp số a=1,b=2
\(20\left(1+2a\right)=15\left(7-3a\right)\)
4(1+2a)=3(7-3a) => 8a +9a= 21-4 => 17a=17 => a=1
với a=1 ta có \(\frac{1+2a}{15}=\frac{3b}{23+7a}\)
\(\frac{1+2}{15}=\frac{3b}{23+7}\)1/5=b/10 => b=2
Vậy a= 1; b= 2
Ta có: a-b=6 => a=6+b thế vào BT trên ta có:
D=\(\frac{3\left(6+b\right)-6}{2\left(6+b\right)+b}-\frac{4b+6}{6+b+3b}\)
= \(\frac{18+3b-6}{12+2b+b}-\frac{4b+6}{6+4b}\)
= \(\frac{3b+12}{3b+12}-\frac{4b+6}{4b+6}\)
= 1-1 =0
a/
Đặt $\frac{a-1}{2}=\frac{b-2}{3}=\frac{c-3}{4}=k$
$\Rightarrow a=2k+1; b=3k+2; c=4k+3$
Khi đó:
$3a+3b-c=50$
$\Rightarrow 3(2k+1)+3(3k+2)-(4k+3)=50$
$\Rightarrow 11k+6=50$
$\Rightarrow 11k=44\Rightarrow k=4$
Ta có:
$a=2k+1=2.4+1=9$
$b=3k+2=3.4+2=14$
$c=4k+3=4.4+3=19$
b/
$2a=3b; 5b=7c\Rightarrow \frac{a}{3}=\frac{b}{2}; \frac{b}{7}=\frac{c}{5}$
$\Rightarrow \frac{a}{21}=\frac{b}{14}=\frac{c}{10}$
Áp dụng TCDTSBN:
$\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{45}{15}=3$
$\Rightarrow a=21.3=63; b=14.3=42; c=10.3=30$
\(\dfrac{1+2a}{15}=\dfrac{7-3a}{20}=\dfrac{3b}{23+7a}\)
\(\Leftrightarrow\dfrac{1+2a}{15}=\dfrac{7-3a}{20}\)
\(\Leftrightarrow20\left(1+2a\right)=15\left(7-3a\right)\)
\(\Leftrightarrow20+40a=105-45a\)
\(\Leftrightarrow40a+20-105+45a=0\)
\(\Leftrightarrow85a-85=0\)
\(\Leftrightarrow a=1\)
Thay a = 1 vào, ta được:
\(\dfrac{7-3}{20}=\dfrac{3b}{23+7}\)
\(\Leftrightarrow\dfrac{1}{5}=\dfrac{3b}{30}\)
\(\Leftrightarrow b=2\)
Vậy a = 1 ; b = 2.
Ta có : \(\frac{1+2a}{15}=\frac{7-3a}{20}=\frac{3b}{23+7a}\)
- Vì \(\frac{1+2a}{15}=\frac{7-3a}{20}\)
=> \(20\left(1+2a\right)=15\left(7-3a\right)\)
\(\Leftrightarrow20+40a=105-45a\Leftrightarrow40a+45a=105-20\)
\(\Leftrightarrow95a=95\Leftrightarrow a=1\)
- Thay a = 1 vào phương trình \(\frac{7-3a}{20}=\frac{3b}{23+7a}\) , ta có : \(\frac{7-3.1}{20}=\frac{3b}{23+7.1}\)
\(\Leftrightarrow\frac{4}{20}=\frac{3b}{30}\Leftrightarrow\frac{1}{5}=\frac{b}{10}\Leftrightarrow5b=10\Leftrightarrow b=2\)
Vậy a =1 , b = 2