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a)ta có: a/b +1 = c/d +1 =>a/b+ b/b = c/d +d/d
=>a+b/b= c+d/d
a) Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{ma}{mc}=\frac{nb}{nd}\)
áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{ma}{mc}=\frac{nb}{nd}=\frac{ma+nb}{mc+nd}=\frac{ma-nb}{mc-nd}\)
\(\Rightarrow\frac{ma+nc}{ma-nb}=\frac{mc+nd}{mc-nd}\left(đpcm\right)\)
sai đề mb=nb TL:
a)đặt \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\)
=>a=kb ;c=kd
=>\(\frac{ma+nb}{ma-nb}=\frac{m.k.b+n.b}{m.k.b-n.b}=\frac{b\left(m.k+n\right)}{b\left(m.k-n\right)}=\frac{m.k+n}{m.k-n}\)
Mặt khác:
\(\frac{mc+nd}{mc-nd}=\frac{m.k.d+n.d}{m.k.d-n.d}=\frac{d\left(m.k+n\right)}{d\left(m.k-n\right)}=\frac{m.k+n}{m.k-n}\)
=>\(\frac{ma+nb}{ma-nb}=\frac{mc+nd}{mc-nd}\) (đpcm)
hc tốt
đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)
a, ta có
+) \(\frac{ma+nc}{mb+nd}=\frac{mck+nc}{mdk+nd}=\frac{c\left(mk+n\right)}{d\left(mk+n\right)}=\frac{c}{d}\)
+) \(\frac{pa+qc}{pb+qd}=\frac{pck+qc}{pdk+qd}=\frac{c\left(pk+q\right)}{d\left(pk+q\right)}=\frac{c}{d}\)
Vậy...........
b, Ta có
+) \(\frac{ma+nd}{mc+nd}=\frac{mck+ndk}{mc+nd}=\frac{k\left(mc+nd\right)}{mc+nd}=k\)
+) \(\frac{pa+qb}{pc+qd}=\frac{pck+pdk}{pc+qd}=\frac{k\left(pc+qd\right)}{pc+qd}=k\)
Vậy.............
c, ta có
+) \(\frac{ma+nc}{pa+qc}=\frac{mck+nc}{pck+qc}=\frac{c\left(mk+n\right)}{c\left(pk+q\right)}=\frac{mk+n}{pk+q}\)
+) \(\frac{mb+nd}{pb+qd}=\frac{mdk+nd}{pdk+qd}=\frac{d\left(mk+n\right)}{d\left(pk+q\right)}=\frac{mk+n}{pk+q}\)
vậy.........
d, ta có
+) \(\frac{ma+nb}{pa+qb}=\frac{mck+ndk}{pck+qdk}=\frac{k\left(mc+nd\right)}{k\left(pc+qd\right)}=\frac{mc+nd}{pc+qd}\)
Vậy.........
Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{ac}{c^2}\)=\(\dfrac{bd}{d^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{d^2}{c^2}\)=\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ac}{bd}\)=\(\dfrac{2d^2}{2c^2}\)= \(\dfrac{2c^2-ac}{2c^2-bd}\)
=> \(\dfrac{a}{b}\)=\(\dfrac{2c^2-ac}{2c^2-bd}\)=>\(\dfrac{a^2}{b^2}\)=\(\dfrac{2c^2-ac}{2d^2-bd}\)
b) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)= \(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{ma}{mc}\)=\(\dfrac{nb}{nd}\)=\(\dfrac{ma+nb}{mc+nd}\)=\(\dfrac{ma-nb}{mc-nd}\)
=> \(\dfrac{ma+nb}{ma-nb}\)=\(\dfrac{mc+nd}{mc-nd}\)
c) Theo đề bài ta có:
\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a^3}{c^3}\)=\(\dfrac{b^3}{d^3}\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a-b}{c-d}\)=\(\left(\dfrac{a-b}{c-d}\right)^3\)(2)
Từ (1) và (2) suy ra:
\(\left(\dfrac{a-b}{c-d}\right)^3\)=\(\dfrac{a^3+b^3}{c^3+d^3}\)
Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a\cdot b}{c\cdot d}=\frac{a^2-b^2}{c^2-d^2}\left(đpcm\right)\)
đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)=k =>a=bk; c=dk
xét: \(\frac{ab}{cd}\)=\(\frac{bk.b}{dk.d}\)=\(\frac{b^2}{d^2}\)
\(\frac{a^2-b^2}{c^2-d^2}\)=\(\frac{b^2k^2-b^2}{d^2k^2-d^2}\)=\(\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}\)=\(\frac{b^2}{d^2}\)
=> \(\frac{ab}{cd}\)=\(\frac{a^2-b^2}{c^2-d^2}\)đpcm
tương tự
xét: \(\left(\frac{a+b}{c+d}\right)^2\)=\(\left(\frac{bk+b}{dk+d}\right)^2\)=\(\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2\)=\(\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{b^2k^2+b^2}{d^2k^2+d^2}\)=\(\frac{b^2\left(k+1\right)}{d^2\left(k+1\right)}\)=\(\frac{b^2}{d^2}\)
=> \(\left(\frac{a+b}{c+d}\right)^2\)=\(\frac{a^2+b^2}{c^2+d^2}\)đpcm
a) sai đề rồi bn
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\left(\frac{a+b}{c+d}\right)^3\)(tính chất dãy tỉ số bằng nhau) (1)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3-b^3}{c^3-d^3}\)(2)
từ (1) và (2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\left(đpcm\right)\)
a, Có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\) (tính chất dãy tỉ số bằng nhau) (1)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{ma}{mc}=\frac{nb}{nd}=\frac{ma+nb}{mc+nd}\) (tính chất dãy tỉ số bằng nhau) (2)
Từ (1),(2)=> \(\frac{a+b}{c+d}=\frac{ma+nb}{mc+nd}\Rightarrow\frac{a+b}{ma+nb}=\frac{c+d}{mc+nd}\)
b, tương tự a