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a, Vì \(\dfrac{a}{c}=\dfrac{c}{b}\Rightarrow ab=c^2\)
Ta có :
\(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b+a\right)\left(b-a\right)}{a^2+ab}=\dfrac{\left(b+a\right)\left(b-a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\)
Vậy \(\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{b-a}{a}\)
a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)
Áp dụng dãy tỉ số bằng nhau ta có :
\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\frac{4a-3b}{4a+3b}=\frac{4c-3d}{4c+3d}\Rightarrow\frac{4a-3d}{4c-3d}=\frac{4a+3b}{4c+3d}\)
b) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)
Áp dụng dãy tỉ số bằng nhau ta có :
\(\frac{2a}{3b}=\frac{2c}{2d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)
\(\dfrac{1+2a}{15}=\dfrac{7-3a}{20}=\dfrac{3b}{23+7a}\)
\(\Leftrightarrow\dfrac{1+2a}{15}=\dfrac{7-3a}{20}\)
\(\Leftrightarrow20\left(1+2a\right)=15\left(7-3a\right)\)
\(\Leftrightarrow20+40a=105-45a\)
\(\Leftrightarrow40a+20-105+45a=0\)
\(\Leftrightarrow85a-85=0\)
\(\Leftrightarrow a=1\)
Thay a = 1 vào, ta được:
\(\dfrac{7-3}{20}=\dfrac{3b}{23+7}\)
\(\Leftrightarrow\dfrac{1}{5}=\dfrac{3b}{30}\)
\(\Leftrightarrow b=2\)
Vậy a = 1 ; b = 2.
a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)
Ta có:
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b}{d}\) (2)
Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)
áp dụng tính chất dãy tỉ số = nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)
= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)
Từ \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> \(\dfrac{a}{c}=\dfrac{b}{d}\)
=> \(\dfrac{2a}{2c}=\dfrac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{2a}{2c}=\dfrac{3b}{3d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{3c-3d}\)
Vậy \(\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\) (ĐPCM)
\(a,Tacó:\\ \dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{a^3}{2^3}=\dfrac{a\cdot a\cdot a}{2\cdot2\cdot2}=\dfrac{a\cdot b\cdot c}{2\cdot3\cdot5}=\dfrac{810}{30}=27\\ \Rightarrow\left\{{}\begin{matrix}a=27\cdot2=54\\b=27\cdot3=81\\c=27\cdot5=135\end{matrix}\right.\\ Vậy...\)
Các câu khác cx cùng dạng tương tự bn tự làm nha!
a, \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}\) và a . b . c = 810
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=k\)
=> \(\left\{{}\begin{matrix}a=2k\\b=3k\\c=5k\end{matrix}\right.\)
Mà a . b . c = 810
=> 2k . 3k . 5k = 810
=> 30\(k^3\) = 810
=> \(k^3=810:30\)
=> \(k^3=27\)
=> \(k^3=3^3\)
=> k = 3
=> \(a=2.3=6\)
\(b=3.3=9\)
\(c=5.3=15\)
Vậy .....
b, \(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{c}{9}\)và a - 3b + 4c = 62
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{c}{9}=\dfrac{a-3b+4c}{4-3.3+4.9}=\dfrac{62}{31}=2\)
=> \(\dfrac{a}{4}=2\Rightarrow a=8\)
\(\dfrac{b}{3}=2\Rightarrow b=6\)
\(\dfrac{c}{9}=2\Rightarrow c=18\)
Vậy .......
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a-3b}{2a+3b}=\dfrac{2bk-3b}{2bk+3b}=\dfrac{2k-3}{2k+3}\)
\(\dfrac{2c-3d}{2c+3d}=\dfrac{2dk-3d}{2dk+3d}=\dfrac{2k-3}{2k+3}\)
=>\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
c: \(\left(\dfrac{a-b}{c-d}\right)^4=\left(\dfrac{bk-b}{dk-d}\right)^4=\left(\dfrac{b}{d}\right)^4\)
\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{b^4k^4+b^4}{d^4k^4+d^4}=\dfrac{b^4}{d^4}\)
Do đó: \(\left(\dfrac{a-b}{c-d}\right)^4=\dfrac{a^4+b^4}{c^4+d^4}\)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
a) Ta có:
\(\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b(5k+3)}{b(5k-3)}=\frac{5k+3}{5k-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d(5k+3)}{d(5k-3)}=\frac{5k+3}{5k-3}\)
\(\Rightarrow \frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
b)
\(\frac{2a-b}{2a+b}=\frac{2bk-b}{2bk+b}=\frac{b(2k-1)}{bb(2k+1)}=\frac{2k-1}{2k+1}\)
\(\frac{2c-d}{2c+d}=\frac{2dk-d}{2dk+d}=\frac{d(2k-1)}{d(2k+1)}=\frac{2k-1}{2k+1}\)
\(\Rightarrow \frac{2a-b}{2a+b}=\frac{2c-d}{2c+d}\) (đpcm)
thank