Cho \(\frac{a}{b}\)= \(\frac{c}{d}\). CMR:
a) \(\frac{2a+7b}{3a-4b}\)= \(\frac{2c+7d}{3c-4d}\)
b) \(\frac{2015a-2016b}{2016c+2017d}\)= \(\frac{2015c-2016d}{2016a+2017b}\)
Cho tỉ lệ thức: a. \(\frac{2015a-2016b}{2016c+2017d}=\frac{2015c-2016d}{2016a+2017b}\)
b. \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
c. \(\frac{ab}{cd}=\left(\frac{2a+3b}{2c+3d}\right)^2\)
Đề bài phải thêm là \(\frac{a}{b}=\frac{c}{d}\) nhé.
a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{2015a}{2015c}=\frac{2016b}{2016d}.\)
\(\Rightarrow\frac{2016a}{2016c}=\frac{2017b}{2017d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{a}{c}=\frac{2015a}{2015c}=\frac{2016b}{2016d}=\frac{2015a-2016b}{2015c-2016d}\) (1)
\(\frac{a}{c}=\frac{2016a}{2016c}=\frac{2017b}{2017d}=\frac{2016a+2017b}{2016c+2017d}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}.\)
\(\Rightarrow\frac{2015a-2016b}{2016c+2017b}=\frac{2015c-2016d}{2016c+2017d}\left(đpcm\right).\)
Câu a) mình nghĩ phải chứng minh như thế.
Chúc bạn học tốt!
cho ti le thuc \(\frac{a}{b}\)=\(\frac{c}{d}\).Chung minh: \(\frac{2015a-2016b}{2016a+2017b}\)=\(\frac{2015c-2016d}{2016c+2017d}\)
tỉ lệ thức cần chứng minh <=> chứng minh: \(\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}\)
Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\) = \(\frac{2015a}{2015c}=\frac{2016b}{2016d}=\frac{2016a}{2016c}=\frac{2017b}{2017d}\)
Áp dụng t/c của dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}\) => đpcm
Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a},\left(a,b,c>0\right)\)
Tính giá trị biểu thức C=\(\frac{2017a-2016b}{c+d}+\frac{2017b-2016c}{a+d}+\frac{2017c-2016d}{a+b}+\frac{2017d-2016a}{b+c}\)
tham khảo bài tương tự này :
Câu hỏi của so yeoung cheing - Toán lớp 7 - Học toán với OnlineMath
Bài 1 cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh
d) \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
e) \(\frac{2016a-2017b}{2017c+2018d}=\frac{2016c-2017d}{2017a+2018b}\)
CMR:
2015a- 2016b / 2016c+ 2017d = 2015c-2016d / 2016a+2017b
a/b=c/d.chung minh: 2015a-2016b/2016c+2017d=2015c-2016c/2016a+2017b
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{2015a-2016b}{2016c+2017d}=\dfrac{2015bk-2016b}{2016dk+2017d}=\dfrac{2015k-2016}{2016k+2017}\)
\(\dfrac{2015c-2016d}{2016a+2017b}=\dfrac{2015dk-2016d}{2016bk+2017b}=\dfrac{2015k-2016}{2016k+2017}\)
Do đó: \(\dfrac{2015a-2016b}{2016c+2017d}=\dfrac{2015c-2016d}{2016a+2017b}\)
1) Cho \(\frac{a}{b}\)\(=\)\(\frac{c}{d}\)
CMR:
a) \(\left(\frac{a+b}{c+d}\right)^2\)\(=\)\(\frac{a^2+b^2}{c^2+d^2}\)
b) \(\frac{7a^2+5ac}{7a^2+5ac}=\frac{7b^2+5bd}{7b^2+5bd}\)
Sử Dụng Tính Chất Của Dãy Tỉ Số Bằng Nhau
2) Cho \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)
CMR:
\(4\left(a-b\right)\left(b-c\right)=\left(c-d\right)^2\)
3) Cho \(\frac{a}{b}=\frac{c}{d}\)
CMR: \(\frac{2015a-2016b}{2016c+2017d}=\frac{2015c-2016d}{2016a+2017b}\)
Bài 5: Cho \(\frac{a}{b}=\frac{c}{d}\)
a) \(\frac{2016a+2017b}{2016a-2017b}=\frac{2016c+2017d}{2016c-2017d}\)
giải nhanh giúp với mai nộp rồi mình gấp lắm cảm ơn nhiều mình tick cho
TỈ lệ cần chứng minh
<br class="Apple-interchange-newline"><div id="inner-editor"></div>2015a−2016b2015c−2016d =2016a+2017b2016c+2017d
Vì ab =cd ⇒ac =bd = 2015a2015c =2016b2016d =2016a2016c =2017b2017d
Áp dụng t/c của dãy tỉ số bằng nhau ta có: \(\frac{a}{c}\)=\(\frac{2015a-2016b}{2015c-2016d}\)=\(\frac{2016a+2017b}{2016c+2017d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) chứng minh
d) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
e) \(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
g) \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5ab}{7b^2-5ab}\)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)
\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)
\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)
\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)
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