Bài cúi thả like :) Cho a/b=c/d CMR a) a+b/c+d=a-b/c-d b) 2017a+2018b/2017c+2018d =2017a-2018b/2017c-2018d
a) cho: \(\dfrac{a}{b}=\dfrac{c}{d}\)
và a, b, c, d \(\ne\) 0 ; \(2017a-2018b\ne0;2017c-2018d\ne0\)
CMR: \(\dfrac{2017a+2017b}{2017a-2017b}=\dfrac{2017c+2018d}{2017c-2018d}\)
b) cho \(a^2=bc\)
CMR: \(\dfrac{c}{b}=\dfrac{a^2+c^2}{b^2+a^2}\)
b) Ta có: [tex]\frac{a^{2} + c^{2}}{b^{2} + a^{2}}[/tex]= [tex]\frac{bc + c^{2}}{b^{2} + bc}= \frac{c(b +c)}{b(b + c)}= \frac{c}{b}[/tex] (đpcm)
Cho \(\frac{a}{b}\) =\(\frac{c}{d}\)CMR \(\frac{2017a+2018b}{2017c+2018d}\)tất cả mũ 3=\(\frac{a^3-b^3}{c^3-d^3}\)
Giúp mik nhé sắp thi HK1 rùi!
\(Cho\dfrac{a}{b}=\dfrac{c}{d}\)
a) \(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
b) \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(VT=\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{a\left(7a+5c\right)}{a\left(7a-5c\right)}=\dfrac{7ck+5c}{7ck-5c}=\dfrac{c\left(7k+5\right)}{c\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(1\right)\)
\(VP=\dfrac{7b^2+5bd}{7b^2-5bd}=\dfrac{b\left(7b+5d\right)}{b\left(7b-5d\right)}=\dfrac{7dk+5d}{7dk-5d}=\dfrac{d\left(7k+5\right)}{d\left(7k-5\right)}=\dfrac{7k+5}{7k-5}\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)
\(\Rightarrow\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\left(đpcm\right)\)
Vậy \(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7b^2+5bd}{7b^2-5bd}\)
Bài 1 cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh
d) \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
e) \(\frac{2016a-2017b}{2017c+2018d}=\frac{2016c-2017d}{2017a+2018b}\)
Bài 1: Cho x2 = y.z; z2 = x.y và x+y+z khác 0
Tính x2017
y2000. z17
Bài 2: Cho 2x = 3y = 4z. Tính giá trị của biểu thức
M = 7x - 2y +8z
4x + 3y - 2z
Bài 3:
Cho a = c
b d .
Chứng minh
a) 3a - 7b = 3c - 7d
b b
b) 2017a - 2018b = 2017a + 2018b
2017c - 2018d 2017c + 2018d
c) (4a + 5b) . (7c - 11d) = (7a - 11b) . (4c + 5d)
Cho \(\frac{a}{b}=\frac{c}{d}\).CMR \(\frac{2017-2018b}{2018a+2019b}=\frac{2017c-2018d}{2018c+2019d}\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2017a-2018b}{2017c-2018d}=\frac{2018a+2019b}{2018c+2019d}\)
<=>\(\left(2017a-2018b\right)\left(2018c+2019d\right)=\left(2018a+2019b\right)\left(2017c-2018d\right)\)
<=>\(\frac{2017a-2018b}{2018a+2019b}=\frac{2017c-2017d}{2018x+2019d}\)(đpcm)
Cho:\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)(a,b,c,d > 0)tính giá trị của biểu thức P=\(\dfrac{2018a-2017b}{c+d}+\dfrac{2018b-2017}{a+d}+\dfrac{2018c-2017a}{b+c}+\dfrac{2018d-2017a}{b+c}\)
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:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)
\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)
\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)
\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)
\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)
\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)
Hình như sai sai
Cho a, b, c, d là các số dương thỏa mãn \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\)
Tính giá trị biểu thức: \(M=\frac{2020a-2018b}{c+d}-\frac{2019b+2017c}{a+d}+\frac{2017c-2019d}{a+b}-\frac{2018d+2020a}{b+c}\)