tính B=4a-b\3a+20-4b-a\3b-5
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`Answer:`
a. Ta có: \(\frac{a}{b}=\frac{1}{3}\Rightarrow\frac{a}{1}=\frac{b}{3}\)
Đặt \(k=\frac{a}{1}=\frac{b}{3}\Rightarrow\hept{\begin{cases}a=k\\b=3k\end{cases}}\)
\(E=\frac{3a+2b}{4a-3b}\)
\(=\frac{3k+2.3k}{4k-3.3k}\)
\(=\frac{3k+6k}{4k-9k}\)
\(=\frac{9k}{-5k}\)
\(=-\frac{9}{5}\)
b. Thay `a-b=5` vào biểu thức `F`, ta được:
\(F=\frac{3a-\left(a-b\right)}{2a+b}-\frac{4b+\left(a-b\right)}{a+3b}\)
\(=\frac{3a-a+b}{2a+b}-\frac{4b+a-b}{a+3b}\)
\(=\frac{2a+b}{2a+b}-\frac{3b+a}{a+3b}\)
\(=1+1\)
\(=0\)
a) \(G=\frac{\frac{3a}{b}-\frac{2b}{b}}{\frac{a}{b}-\frac{3b}{b}}=\frac{3.\frac{10}{3}-2}{\frac{10}{3}-3}=\frac{10-2}{\frac{1}{3}}=24\)
b) \(H_1=\frac{\frac{2a-3b}{b}}{\frac{4a+3b}{b}}=\frac{\frac{2a}{b}-\frac{3b}{b}}{\frac{4a}{b}+\frac{3b}{b}}=\frac{2.\frac{10}{3}-3}{4.\frac{10}{3}+3}=\frac{\frac{11}{3}}{\frac{49}{3}}=\frac{11}{49}\)
\(H_2=\frac{\frac{5a-4b}{b}}{\frac{3a+b}{b}}=\frac{5.\frac{a}{b}-4}{3.\frac{a}{b}+1}=\frac{5.\frac{10}{3}-4}{3.\frac{10}{3}+1}=\frac{\frac{38}{3}}{\frac{33}{3}}=\frac{38}{33}\)
=> \(H=\frac{11}{49}-\frac{38}{33}=\frac{-1499}{1617}\)
\(\dfrac{a}{b}=\dfrac{1}{3}\)
nên b=3a
\(E=\dfrac{3a+2b}{4a-3b}=\dfrac{3a+6a}{4a-9a}=\dfrac{9}{-5}=-\dfrac{9}{5}\)
a-b=5 nên a=b+5
\(F=\dfrac{3\left(b+5\right)-5}{2\left(b+5\right)+b}-\dfrac{4b+5}{b+5+3b}\)
\(=\dfrac{3b+10}{3b+10}-1=1-1=0\)
a)
`a<b`
`<=>3a<3b`
`<=>3a-5<3b-5`
b)
`a<b`
`<=>-8a> -8b`
`<=>-8a-3> -8b-3`
c)
`a<b`
`<=>4a<4b`
`<=>4a+9<4b+9`
mà `4a-7<4a+9`
`<=>4a-7<4b+9`
4a-b=6 nên b=4a-6
\(\dfrac{6a-b}{3a+5}-\dfrac{4a-4b}{3b-5}\)
\(=\dfrac{6a-\left(4a-6\right)}{3a+5}-\dfrac{4a-4\left(4a-6\right)}{3\left(4a-6\right)-5}\)
\(=\dfrac{6a-4a+6}{3a+5}-\dfrac{4a-16a+24}{12a-18-5}\)
\(=\dfrac{2a+6}{3a+5}-\dfrac{-12a+24}{12a-23}\)
\(=\dfrac{2a+6}{3a+5}+\dfrac{12a-24}{12a-23}\)
\(=\dfrac{\left(2a+6\right)\left(12a-23\right)+\left(12a-24\right)\left(3a+5\right)}{\left(3a+5\right)\left(12a-23\right)}\)
\(=\dfrac{24a^2-46a+72a-138+36a^2+60a-72a-120}{\left(3a+5\right)\left(12a-23\right)}\)
\(=\dfrac{60a^2+14a-258}{\left(3a+5\right)\left(12a-23\right)}\)
Câu 5:
\(D\left(2\right)=21a+9b-6a-4b\)
\(D\left(2\right)=\left(21a-6a\right)+\left(9b-4b\right)\)
\(D\left(2\right)=15a+5b\)
Mà: \(3a+b=18\Rightarrow b=18-3b\)
\(\Rightarrow D\left(2\right)=15a+5\left(18-3b\right)\)
\(D\left(2\right)=15a+90-15a\)
\(D\left(2\right)=90\)
Vậy: ...