cho \(\frac{a}{b}\)= \(\frac{c}{d}\)chứng minh rằng:
a) \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
b)\(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
c) \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
cho \(\frac{a}{b}=\frac{c}{d}\)chung minh rang:
\(\frac{a}{a-b}=\frac{c}{c-d}\) \(\frac{a}{b}=\frac{a+c}{b+d}\) \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) \(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)\(\frac{a.c}{b.d}=\frac{a^2-c^2}{b^2-d^2}\)
+ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
+ \(\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\) \(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
+ \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\Rightarrow\frac{a\cdot b}{c\cdot d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{b}=\frac{a^2+c^2}{b^2+d^2}\Rightarrow\frac{a\cdot c}{b\cdot d}=\frac{a^2+c^2}{b^2+d^2}\)
câu cuối lm tương tự
1. Chứng minh tỉ lệ thức:
Cho \(\frac{a}{b}=\frac{c}{d}\) , chứng minh rằng :
a, \(\frac{a}{3a+b}\) \(=\frac{c}{3c+d}\)
b, \(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
c, \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
Có \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}\)
Mà \(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
Nên \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
1. a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)
\(\frac{c}{3c+d}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)
Từ (1) và (2) => \(\frac{a}{3a+b}=\frac{c}{3c+d}\)
c,
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{ab}{cd}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\) (3)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(4\right)\)
Từ (3) và (4) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
@@ Học tốt
Chiyuki Fujito
1) a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{a}{3a+b}=\frac{bk}{3bk+b}=\frac{bk}{b\left(3k+1\right)}=\frac{k}{3k+1}\left(1\right)\)
\(\frac{c}{3a+b}=\frac{dk}{3dk+d}=\frac{dk}{d\left(3k+1\right)}=\frac{k}{3k+1}\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\left(đpcm\right)\)
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)chứng minh rằng
a)\(\frac{a}{a-b}\)=\(\frac{c}{c-d}\)
b)\(\frac{a}{b}=\frac{a+c}{b+d}\)
c)\(\frac{a}{3a+b}=\frac{c}{3c+d}\)
d)\(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
f)\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a) \(\frac{a}{b}=\frac{c}{d}\)
\(\frac{a}{b}=\frac{c}{d}\)<=>\(\frac{a}{c}=\frac{b}{d}\)
áp dụng t/c dãy tỉ số = nhau :
\(\frac{a}{c}=\frac{b}{d}\)\(=\frac{a-b}{c-d}\) <=> \(\frac{a}{c}\)\(=\frac{a-b}{c-d}\)<=> \(\frac{a}{a-b}=\frac{c}{c-d}\)
mấy bài kia cũng tương tự em ạ !
gợi ý: đặt chung cho cả 4 phần a/b = c/d = k( k khác 0)
=> a=bk; c=dk
rồi thay vào các biểu thức
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Chứng minh rằng \(\frac{a}{b}=\frac{c}{d}\) biết:
a) \(\frac{a}{a-b}=\frac{c}{c-d}\) b)\(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
c) \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}.\)Chứng minh.
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b)\(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a/ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{5b}{5d}=\frac{3a+5b}{3c+5d}=\frac{3a-5b}{3c-5d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b/ \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\left(\frac{a+b}{c+d}\right)^2\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{b^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
cho \(\frac{a}{b}=\frac{c}{d}\). chứng minh \(\frac{a.c}{b.d}=\frac{a^2-c^2}{b^2-d^2}=\frac{a^2+c^2}{b^2+d^2}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a.c}{b.d}\left(1\right)\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2-c^2}{b^2-d^2}=\frac{a^2+c^2}{b^2+d^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a.c}{b.d}=\frac{a^2-c^2}{b^2-d^2}=\frac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
1/ cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a) \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b)\(\frac{a,d}{c.b}=\frac{\left(a+b\right).\left(a-b\right)}{\left(c+d\right).\left(c-d\right)}\)
2/ cho \(a.b=c^2\)chứng minh : \(\frac{a}{b}=\frac{\left(2a+3c\right)^2}{\left(2c+3b\right)^2}\)
a, Ta có: \(\frac{a}{b}=\frac{c}{d}=k\left(k\ne0\right)\Rightarrow a=kb;c=kd\)
Thay:
\(\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)
=> đpcm
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh: \(\frac{2.a^2-3.a.b+3.b^2}{2.b^2+3.a.b}=\frac{2.c^2-3.c.d+5.d^2}{2.d^2+3.c.d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\), suy ra \(a=bk;c=dk\)
\(VT=\frac{2b^2k^2-3b^2k+3b^2}{2b^2+3b^2k}=\frac{b^2\left(2k^2-3k+3\right)}{b^2\left(2+3k\right)}=\frac{2k^2-3k+3}{3k+2}\left(1\right)\)
\(VP=\frac{2d^2k^2-3d^2k+3d^2}{2d^2+3d^2k}=\frac{d^2\left(2k^2-3k+3\right)}{d^2\left(2+3k\right)}=\frac{2k^2-3k+3}{3k+2}\left(2\right)\)
Từ (1) và (2) suy ra ĐPcm
Cho \(\frac{a}{b}=\frac{c}{d}\)CMR:
\(a,\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\) \(b,\frac{a.c}{b.d}=\frac{a^2-c^2}{b^2-d^2}\)\(c,\frac{a.c}{b.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
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