biết 2a-b=5; a\(\ne\)\(\dfrac{-10}{3}\);b\(\ne2\)
tính giá trị của biểu thức A=\(\dfrac{7a-2b}{3a+10}-\dfrac{7b-4a}{15b-30}\)
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Lời giải:
Vì \(2a-b=5\Rightarrow b=2a-5\Rightarrow 2b=4a-10\)
\(\Rightarrow 7a-2b=7a-(4a-10)=3a+10\)
\(\Rightarrow \frac{7a-2b}{3a+10}=\frac{3a+10}{3a+10}=1\)
Lại có:
\(2a-b=5\Rightarrow 2a=b+5\Rightarrow 4a=2b+10\)
\(\Rightarrow 7b-4a=7b-(2b+10)=5b-10\)
\(\Rightarrow \frac{7b-4a}{15b-30}=\frac{5b-10}{15b-30}=\frac{5b-10}{3(5b-10)}=\frac{1}{3}\)
Vậy: \(A=1-\frac{1}{3}=\frac{2}{3}\)
\(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{2a-5b}{-14}=\dfrac{a-3b}{-9}=\dfrac{4a+b}{16}=\dfrac{8a-2b}{16}\\ \Leftrightarrow A=\dfrac{-14}{-9}-\dfrac{16}{16}=\dfrac{14}{9}-1=\dfrac{5}{9}\)
\(\dfrac{a}{6}=\dfrac{b}{9}\)
\(\Leftrightarrow9a=6b\)
\(\Rightarrow3a=2b\)(chia cả 2 vế cho 3)
\(\Rightarrow3a-2b=0\Rightarrow\dfrac{3a-2b}{3a+2b}=0\)
Chúc bn học tốt
Ta có: `a/6 = b/9` `-> 9a = 6b`
`-> 3a = 2b`
Vì `3a = 2b` nên `3a - 2b = 0`.
`-> A = (3a - 2b)/(3a + 2b) = 0/(3a + 2b) = 0`
Vậy giá trị biểu thức `A` là `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: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Lời giải:
$a-b=3\Rightarrow b=a-3$. Khi đó:
$A=\frac{a-8}{a-3-5}-\frac{4a-(a-3)}{3a+3}=\frac{a-8}{a-8}-\frac{3a+3}{3a+3}=1-1=0$
2a-b=5 nên b=2a-5
\(A=\dfrac{7a-2b}{3a+10}-\dfrac{7b-4a}{15b-30}\)
\(=\dfrac{7a-2\left(2a-5\right)}{3a+10}-\dfrac{7\left(2a-5\right)-4a}{15\left(2a-5\right)-30}\)
\(=\dfrac{7a-4a+10}{3a+10}-\dfrac{14a-35-4a}{30a-75-30}\)
\(=1-\dfrac{5\left(2a-7\right)}{15\left(2a-7\right)}=1-\dfrac{1}{3}=\dfrac{2}{3}\)