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\(đk:a;b\ne\dfrac{5}{3}\)
\(\dfrac{3b-28}{3a-5}-\dfrac{38-3a}{5-3b}=\dfrac{3b-28}{3\left(11+b\right)-5}-\dfrac{38-3\left(11+b\right)}{5-3b}=1-1=0\)
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: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
BT1 : Tính giá trị của biểu thức ;
Thay 7 = a -b vào biểu thức B ,có :
\(\dfrac{3a-b}{2a+\left(a-b\right)}+\dfrac{3b-a}{2b-\left(a-b\right)}\)
\(=\dfrac{3a-b}{3a-b}+\dfrac{3b-a}{3a-a}\)
\(=1+1\)
= 2
Vậy giá trị của biểu thức B là 2 với a- b=7
Lời giải:
a)\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow4a=3b\)
Và \(4a.5=3b.5\Leftrightarrow20a=15b\Leftrightarrow\dfrac{20a}{3}=5b\)
Khi đó:
\(A=\dfrac{2a-5b}{a-3b}=\dfrac{2a-\dfrac{20}{3}a}{a-4a}=\dfrac{-\dfrac{14}{3}a}{-3a}=\dfrac{-14}{\dfrac{3}{-3}}=14\)
b) Ta có:
\(a-b=7\Leftrightarrow b=a-7\)
\(B=\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}=\dfrac{3a-\left(a-7\right)}{2a+7}+\dfrac{3\left(a-7\right)-a}{2\left(a-7\right)-7}\)
\(B=\dfrac{3a-a+7}{2a+7}+\dfrac{3a-21-a}{2a-14-7}\)
\(B=\dfrac{2a+7}{2a+7}+\dfrac{2a-21}{2a-21}=1+1=2\)
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: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
ta có : \(a-b=15\Leftrightarrow a=15+b\)
thay vào \(P\) ta có \(P=\dfrac{3\left(15+b\right)-b}{2\left(15+b\right)+15}+\dfrac{3b-\left(15+b\right)}{2b-15}\)
\(P=\dfrac{45+3b-b}{30+2b+15}+\dfrac{3b-15-b}{2b-15}=\dfrac{2b+45}{2b+45}+\dfrac{2b-15}{2b-15}\)
\(P=1+1=2\) vậy \(P=2\) với \(a-b=15\)
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: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
Đặt \(a=\dfrac{10}{3}b\Rightarrow\dfrac{3.\dfrac{10}{3}b-2b}{\dfrac{10}{3}b-3b}=\dfrac{10b-2b}{\dfrac{1}{3}b}=\dfrac{8}{\dfrac{1}{3}}=24\)
Giải:
\(\dfrac{a}{b}=\dfrac{10}{3}\Rightarrow\dfrac{a}{10}=\dfrac{b}{3}.\)
Đặt \(\dfrac{a}{10}=\dfrac{b}{3}=k\Rightarrow a=10k;b=3k.\)
Ta có:
\(A=\dfrac{3a-2b}{a-3b}=\dfrac{3.10k-2.3k}{10k-3.3k}=\dfrac{30k-6k}{10k-9k}=\dfrac{\left(30-6\right)k}{\left(10-9\right)k}=\dfrac{24}{1}=24.\)
Vậy \(A=24.\)
a, Theo bài ta có :
\(\dfrac{a}{b}=\dfrac{10}{3}\Leftrightarrow\dfrac{a}{10}=\dfrac{b}{3}\)
Đặt :
\(\dfrac{a}{10}=\dfrac{b}{3}=k\left(k\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=10k\\b=3k\end{matrix}\right.\)
Ta có :
\(Q=\dfrac{3a-2b}{a-3b}=\dfrac{3.10k-2.3k}{10k-3.3k}=\dfrac{30k-6k}{10k-9k}=\dfrac{24k}{1k}=24\)
Vậy ...........
a-b=3=>a=b+3 Thay a=b+3 vào B
\(\Rightarrow B=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\)
\(\Rightarrow B=1-\dfrac{4b-b+12}{3b+9+3}=1-1=0\)