cho \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{c}{d}\). chứng mk: \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
cho a+b+c+d khác 0 vàti\(\dfrac{b+c+d-a}{a}=\dfrac{c+d+a-b}{b}=\dfrac{d+a+b-c}{c}=\dfrac{a+b+c-d}{d}P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{c}{d}\right)\left(1+\dfrac{a}{d}\right)\)tính P
giúp mk với ạ , xin cảm ơn
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) Chứng minh rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Theo tính chất dãy tỉ số bằng nhau, ta có:
a/b = b/c = c/d = (a + b + c)/(b + c + d)
--> ((a + b + c)/(b + c + d))^3 = a^3/b^3
Cần chứng minh:
a^3/b^3 = a/d
<=> a^3/b^3 = a^3/(a^2.d)
--> b^3 = a^2.d
Mà ad = bc (do a/b = c/d)
--> b^3 = abc
<=> b^2 = ac (luôn đúng do a/b = b/c)
--> đpcm
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) chứng mình rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\) (đpcm)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR : \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\left(\dfrac{a}{b}=\dfrac{c}{d}\ne1\right)\)
Giúp mk vs mai mk phải nộp rồi
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)
\(\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{b^3}{d^3}\)
Do đó: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)
cho tỷ lệ thức \(\dfrac{a}{b}\)= \(\dfrac{b}{c}\)=\(\dfrac{c}{d}\). Chứng minh \(\dfrac{\left(a+b+c\right)}{\left(b+c+e\right)}^3\)=\(\dfrac{a}{d}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) chứng minh \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\left(\dfrac{a}{b}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
mà cần chứng minh: \(\left(\dfrac{a+b+c}{b+c+d}\right)=\dfrac{a}{d}\left(2\right)\)
từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\) \(\dfrac{a^3}{b^3}=\dfrac{a}{d}\Rightarrow a^3.d=b^3.a\)
\(\Rightarrow a^2.d=b^3\)
vì \(\dfrac{a}{b}=\dfrac{b}{c}\Rightarrow a.c=b^2\)
\(\Rightarrow a.b.c=b.c\left(3\right)\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a.d=b.c\left(4\right)\)
từ \(\left(3\right)\) và \(\left(4\right)\) \(\Rightarrow a.a.d=b^3\)
\(\Rightarrow a^2.d=b^3\left(đpcm\right)\)
vậy \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
1. Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\). Chứng minh rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
2. Cho \(\dfrac{a}{2003}=\dfrac{b}{2004}=\dfrac{c}{2005}\). Chứng minh rằng \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
chứng minh \(\left(\dfrac{a+b+c}{b+c+d}\right)_3=\dfrac{a}{d}\)
giúp mình với
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Leftrightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
\(\Leftrightarrow\dfrac{a\cdot a\cdot a}{b\cdot b\cdot b}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
\(\Leftrightarrow\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{a}{b}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
\(\Leftrightarrow\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
\(\Leftrightarrow\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)(đpcm)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)(b, c, d ≠ 0 , b + d ≠ 0). Chứng minh rằng: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Theo đề bài ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\) ( 1 )
Theo tính chất dãy tỉ số bằng nhau ta có :
\(k=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(k^2=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( 2 )
Mà từ ( 1 ) = > \(k^2=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\) ( 3 )
Từ ( 2 ) , ( 3 )
= > \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( đpcm )
a) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) (\(a,b,c,d\ne0\)). Chứng minh rằng:
1) \(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
2) \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
3) \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\) \(\left(\dfrac{a}{b}=\dfrac{c}{d}\ne1\right)\)
b)Cho \(\dfrac{2a+13b}{3a-7b}=\dfrac{2c+13d}{3c-7d}\). Chứng minh rằng:\(\dfrac{a}{b}=\dfrac{c}{d}\)
c)Cho \(\dfrac{cy-bz}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}\). Chứng minh rằng: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
Bài 3:
a) Sửa điều kiện: \(\frac{a}{b}=\frac{c}{d}\neq -1\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Theo đkđb thì \(k\neq -1\) nên \(k^3+1\neq 0\); \(k+1\neq 0\)
Ta có: \(\frac{a^3+b^3}{c^3+d^3}=\frac{(bk)^3+b^3}{(dk)^3+d^3}=\frac{b^3(k^3+1)}{d^3(k^3+1)}=\frac{b^3}{d^3}\)
\(\frac{(a+b)^3}{(c+d)^3}=\frac{(bk+b)^3}{(dk+d)^3}=\frac{b^3(k+1)^3}{d^3(k+1)^3}=\frac{b^3}{d^3}\)
\(\Rightarrow \frac{a^3+b^3}{c^3+d^3}=\frac{(a+b)^3}{(c+d)^3}\) (đpcm)
b)
Đặt \(\frac{a}{b}=k; \frac{c}{d}=t\Rightarrow a=bk; c=dt\)
Ta cần cm \(k=t\)
Khi đó:
\(\frac{2a+13b}{3a-7b}=\frac{2bk+13b}{3bk-7b}=\frac{b(2k+13)}{b(3k-7)}=\frac{2k+13}{3k-7}\)
\(\frac{2c+13d}{3c-7d}=\frac{2dt+13d}{3dt-7d}=\frac{d(2t+13)}{d(3t-7)}=\frac{2t+13}{3t-7}\)
Vì \(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\Rightarrow \frac{2k+13}{3k-7}=\frac{2t+13}{3t-7}\)
\(\Rightarrow (2k+13)(3t-7)=(2t+13)(3k-7)\)
\(-14k+39t=-14t+39k\Rightarrow k=t\)
Ta có đpcm.