Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b, \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c, \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
CÁC BẠN GIÚP BẠN Heo Mách VỚI NHA!!!!!
1) Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR(với giả thiết các tỉ số đều có nghĩa)
a)\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
b)\(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
c)\(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
d)\(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
e)\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
b: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
c: \(\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{k-1}{k+1}\)
\(\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{k-1}{k+1}\)
Do đó: \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
Bài 7: Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng ta có các tỉ lệ thức sau( giả thiết các tỉ lệ thức phải chứng minh đều có nghĩa):
a)\(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\) b)\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c)\(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\) d)\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
ai hộ mik vs
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Cho tỉ lệ thức : \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh
a) \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
b) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
b) \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\frac{2a}{2c}=\frac{5b}{5d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\)
\(\Rightarrow\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Giúp mik nhé mí bạn.
1) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . CM :
b) \(\dfrac{5a-3b}{3a+2b}=\dfrac{5c-3d}{3c+2d}\)
c) \(\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
d) \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
e) \(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)
f) \(\dfrac{\left(a+c\right)^2}{a^2-c^2}=\dfrac{\left(b+d\right)^2}{b^2-d^2}\)
Làm được câu nào thì trả lời nhé . Thanks trước
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)=>\(\dfrac{a}{c}=\dfrac{b}{d}\)
<=>\(\dfrac{5a}{5c}=\dfrac{3b}{3d}=\dfrac{3a}{3c}=\dfrac{2b}{2d}\)
<=>\(\dfrac{5a-3b}{5c-3d}=\dfrac{3a-2b}{3c-2d}\)(đpcm)
Các câu sau tương tự
c/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\left(\dfrac{a+c}{b+d}\right)^2=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{ac}{bd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
d/ tương tự câu b/
e/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ac}{c^2}=\dfrac{bd}{d^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}
\)(1)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{ac}{bd}=\dfrac{c^2}{d^2}\)(đpcm)
f/ Theo đề bài ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(1\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2-c^2}{b^2-d^2}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(a+c\right)^2}{a^2-c^2}=\dfrac{\left(b+d\right)^2}{b^2-d^2}\)
cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). chứng minh rằng ta có các tỉ lệ thức sau( giả thiết rằng các tỉ lệ thức phải chứng minh đều có nghĩa)
a) \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c) \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\)\(\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Từ \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\) \(\Rightarrow\dfrac{c-d}{c+d}=\dfrac{a-b}{a+b}\)
b) Từ \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2a}{2c}=\dfrac{3a}{3c}=\dfrac{4b}{4d}=\dfrac{5b}{5d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\)
Từ \(\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\) \(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh :
a) \(\dfrac{3a+5b}{2a-7b}=\dfrac{3c+5d}{2c-7d}\)
b) \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\)
1) Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR(với giả thiết các tỉ số đều có nghĩa)
a)\(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b)\(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
c)\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
2) Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR ta có các tỉ lệ thức sau
a)\(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b)\(\dfrac{7a1^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\)
3) CMR nếu \(a^2=bc\) thì \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\). Đảo lại có đúng không?
4) CMR nếu \(\dfrac{a}{b}=\dfrac{b}{d}\) thì \(\dfrac{a^2+b^2}{b^2+d^2}=\dfrac{a}{d}\)
5) Cho tỉ lệ thức \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}.CMR\dfrac{a}{b}=\dfrac{c}{d}\)
các bn giúp bn Heo Mách với nha
Bài 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b: \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7\cdot b^2k^2+5\cdot bk\cdot dk}{7\cdot b^2k^2-5\cdot bk\cdot dk}\)
\(=\dfrac{7b^2k^2+5bdk^2}{7b^2k^2-5bdk^2}=\dfrac{7b^2+5bd}{7b^2-5bd}\)(đpcm)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) CMR:
\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+5b}{2c+5d}\)
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3a-4b}{3c-4d}\)
\(\Rightarrow\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}=\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\left(dpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left[{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) \(\Rightarrow\dfrac{2bk+5b}{3bk-4b}=\dfrac{2dk+5d}{3dk-4d}\)
\(VT=\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(1\right)\)
\(VP=\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\) Đpcm.
Cho a+b+c+d ≠ 0 thỏa mãn:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{b+a+d}=\dfrac{d}{c+b+a}\)
Tính P = \(\dfrac{2a+5b}{3c+4d}+\dfrac{2b+5c}{3d+4a}+\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)