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Đặt a/2=b/-3=c/-4,5=k
=>a=2k; b=-3k; c=-4,5k
\(P=\dfrac{3a-2b}{8a-b+3c}=\dfrac{6k+6k}{16k+3k-13.5k}=\dfrac{12k}{5.5k}=\dfrac{24}{11}\)
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
\(\Rightarrow B=\dfrac{b+3-8}{b-5}-\dfrac{4\left(b+3\right)-b}{3\left(b+3\right)+3}\)
\(\Rightarrow B=1-\dfrac{4b-b+12}{3b+9+3}=1-1=0\)
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a/ \(VT=\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1=\left(1\right)\)
\(VP=\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b/ \(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
c/ \(VT=\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5d}=\dfrac{b\left(2k-5\right)}{d\left(2k-5\right)}=\dfrac{b}{d}\left(1\right)\)
\(VP=\dfrac{3a+4b}{3c+4d}=\dfrac{3bk+4b}{3dk+4d}=\dfrac{b\left(3k+4\right)}{d\left(3k+4\right)}=\dfrac{b}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2a-5b}{2c-5đ}=\dfrac{3a+4b}{3c+4d}\)
d/ \(VT=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-k^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\left(1\right)\)
\(VP=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
Theo đề bài, ta có: \(\dfrac{a}{b}=\dfrac{10}{3}\Rightarrow\dfrac{a}{10}=\dfrac{b}{3}\)
Đặt \(\dfrac{a}{10}=\dfrac{b}{3}=k\) \(\left(k\ne0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a=10k\\b=3k\end{matrix}\right.\)
Ta có: \(Q=\dfrac{3a-2b}{a-3b}=\dfrac{3\times10k-2\times3k}{10k-3\times3k}=\dfrac{30k-6k}{10k-9k}=\dfrac{24k}{1k}=24\)
Vậy \(Q=24\).
\(Q=\dfrac{3a-2b}{a-3b}=\dfrac{3.\dfrac{a}{b}-2}{\dfrac{a}{b}-3}=\dfrac{3.\dfrac{10}{3}-2}{\dfrac{10}{3}-3}=\dfrac{8.3}{10-9}=24\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
hỏi mỗi từng câu 1 thôi nhé ! Vậy mình giải cho . Mình k có ý kiếm GP + SP đâu . Nhưng nhìn 8 câu này hoa hết cả mắt :v
Đúng thật. Tớ nhìn cũng thấy ngán mà. Nhiều quá nên hơi nản 
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(\dfrac{2a+7b}{3a-4b}=\dfrac{2bk+7b}{3bk-4b}=\dfrac{b\left(2k+7\right)}{b\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\left(1\right)\)
\(\dfrac{2c+7d}{3c-4d}=\dfrac{2dk+7d}{3dk-4d}=\dfrac{d\left(2k+7\right)}{d\left(3k-4\right)}=\dfrac{2k+7}{3k-4}\)\(\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ tương tự
Không có điều kiện gì à ( Kiểu \(\dfrac{a}{b}=\dfrac{c}{d}\) ấy )
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{5}{2}\\ \Rightarrow\dfrac{a}{5}=\dfrac{b}{2}\\ \Rightarrow\dfrac{3a}{15}=\dfrac{4a}{20}=\dfrac{b}{2}=\dfrac{3a-b}{15-2}=\dfrac{4a-b}{20-2}\\ \Rightarrow\dfrac{3a-b}{4a-b}=\dfrac{13}{18}\)
Chúc bạn học tốt nha
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