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Cho tỉ lệ thức : \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh
\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)
Suy ra: \(\frac{2a+3b}{2a-3b}=\frac{2.bk+3b}{2.bk-3b}=\frac{b.\left(2k+3\right)}{b.\left(2k-3\right)}=\)\(\frac{2k+3}{2k-3}\)
\(\frac{2c+3d}{2c-3d}=\frac{2.dk+3d}{2.dk-3d}=\frac{d.\left(2k+3\right)}{d.\left(2k-3\right)}=\)\(\frac{2k+3}{2k-3}\)
Vậy \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Ta có:\(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{a}{c}=\frac{b}{d}\)=>\(\frac{2a}{2c}=\frac{3b}{3d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)
=>\(\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)=>\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Vậy\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Ta có: \(\frac{a+3b}{a-3b}=\frac{c+3d}{c-3d}\)
\(\rightarrow\left(a+3b\right)\left(c-3d\right)=\left(a-3b\right)\left(c+3d\right)\)
\(\rightarrow ac+3bc-3ad-9bd=ac-3bc+3ad-9bd\)
\(\rightarrow3bc-3ad=3ad-3bc\)
\(\rightarrow6bc=6ad\)
\(\rightarrow bc=ad\rightarrow\frac{a}{c}=\frac{b}{d}\left(đpcm\right)\)
Chúc bn học tốt
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Xem ở lick này nhé (mình gửi cho)
Học tốt!!!!!!!!!!!!!
a/ Với
\(\frac{3x-y}{x+y}=\frac{3}{4}=\frac{3\frac{x}{y}-1}{\frac{x}{y}+1}\Rightarrow3\left(\frac{x}{y}+1\right)=4\left(3\frac{x}{y}-1\right)\)
\(\Rightarrow3\frac{x}{y}+3=12\frac{x}{y}-4\Rightarrow9\frac{x}{y}=7\Rightarrow\frac{x}{y}=\frac{7}{9}\)
b/
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)
\(\Rightarrow\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\Rightarrow\frac{2a+3a}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(2)
Từ (1) và (2) \(\Rightarrow\)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c)\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(ck+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
k cho mình nhé
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}.\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\Rightarrow\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)(T/c dãy tỷ số bằng nhau)
\(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
Ta có:
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a) \(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b) \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c) \(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2.\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2\right)+1}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
c) có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a^2}{^{c^2}}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
Lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(2\right)\)
Từ (1) và (2) có \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
các câu còn lại bạn tự làm đi! HI.......
-,-'' theo trí nhớ của miu thì nok là thế nì....
a) Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\left(a;b;c;d\ne0\right)\)
\(CM:\frac{a+b}{b}=\frac{c+d}{d}\)
\(\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Leftrightarrow\)\(\frac{a}{b}=\frac{c}{d}\)hay theo đề bài: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
^^ làm đại khái :V ko cần suy nghĩ... chỉ là mò về kiến thức cũ (nếu có sai mong thánh thông cảm!! :P)
caj câu b bao h nghĩ xong cách làm thì mk đăng (h fai gô-tu-bét r`)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{23a-3b}{23a+3b}=\frac{23bk-3b}{23bk+3b}=\frac{b\left(23k-3\right)}{b\left(23k+3\right)}=\frac{23k-3}{23k+3}\)(1)
\(\frac{23c-3d}{23c+3d}=\frac{23dk-3d}{23dk+3d}=\frac{d\left(23k-3\right)}{d\left(23k+3\right)}=\frac{23k-3}{23k+3}\)(2)
Từ (1)(2) => \(\frac{23a-3b}{23a+3b}=\frac{23c-3d}{23c+3d}\)
Vì a/b=c/d nên a/c=b/d
Áp dụng tính chất dãy tỉ số bằng nhau ta có
a/c=b/d=23a/23c=3b/3d=23a+3b/23c+3d=23a-3b/23c-3d
Từ 23a+3b/23c+3d=23a-3b/23c-3d =>23a-3b/23a+3b=23c-3d/23c+3d
Vậy ....