Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) . Chứng minh rằng \((\dfrac{a+b+c}{b+c+d})^3=\dfrac{a}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) Chứng minh rằng \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\b=ck\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{b^3k^3+c^3k^3+d^3k^3}{b^3+c^3+d^3}=k^3\)
\(\dfrac{a}{d}=\dfrac{bk}{d}=\dfrac{ck^2}{d}=\dfrac{dk^3}{d}=k^3\)
Do đó: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
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 minh rằng : \(\dfrac{a^3}{b^3}=\dfrac{a}{d}\)
Chứng minh rằng : Nếu \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) thì
a.\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\) b.\(\dfrac{a}{b}\)=\(\dfrac{a+c}{b+c}\) c.\(\dfrac{a}{c}\)=\(\dfrac{a-b}{c-d}\) d.\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)
a: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
d: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
hay \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
hay \(\dfrac{a}{c}=\dfrac{a-b}{c-d}\)
Cho hai phân số \(\dfrac{a}{b}\) và \(\dfrac{c}{d}\) thỏa mãn b, d > 0 và \(\dfrac{a}{b}\) < \(\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
:)
- Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\) (gt)
=>\(ad< bc\)
=>\(ad+ab< bc+ab\)
=>\(a\left(b+d\right)< b\left(a+c\right)\)
=>\(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) (1)
- Ta có: \(\dfrac{c}{d}>\dfrac{a}{b}\) (gt)
=>\(bc>ad\)
=>\(bc+cd>ad+cd\)
=>\(c\left(b+d\right)>d\left(a+c\right)\)
=>\(\dfrac{c}{d}>\dfrac{a+c}{b+d}\) (2)
- Từ (1) và (2) suy ra: \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
a,Tìm x,y,z biết: \(\dfrac{y+z+1}{x}\)=\(\dfrac{x+z+2}{y}\)=\(\dfrac{x+y-3}{z}\)=\(\dfrac{1}{x+y+z}\)
b,Cho \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\). Chứng minh rằng: (\(\dfrac{a+b+c}{b+c+d}\))3=\(\dfrac{a}{d}\)
c,Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng: \(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)
d,Cho \(\dfrac{3x-2y}{4}\)=\(\dfrac{2z-4x}{3}\)=\(\dfrac{4y-3z}{2}\).Chứng minh rằng: \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{4}\)
b/ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\dfrac{a}{d}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
=> \(\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{c+d+b}\right)^3\) (2)Từ (1) và (2)=>đpcm
Chứng minh rằng \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) nếu:
a, \(\dfrac{a}{c}\) = \(\dfrac{a+b}{c+d}\)
b, \(\dfrac{b}{d}\) = \(\dfrac{a-b}{c-d}\)
a) \(\dfrac{a}{c}=\dfrac{a+b}{c+d}\)
=> a(c + d) = c(a + b)
=> ac + ad = ac + bc
=> ad = bc \(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
b) \(\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
=> b(c - d) = d(a - b)
=> bc - bd = ad - bd
=> bc = ad \(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\). Chứng minh rằng :
\(\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=(\dfrac{a+b-c}{b+c-d})^3\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
Vậy \(\dfrac{a}{b}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\left(dpcm\right)\)
Cho các số hữu tỉ \(\dfrac{a}{b}\) và \(\dfrac{c}{d}\) với mẫu dương, trong đó \(\dfrac{a}{b}< \dfrac{c}{d}\) . Chứng minh rằng :
\(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
`a/b<(a+c)/(b+d)`
`<=>a(b+d)<b(a+c)`
`<=>ab+ad<ad<bc`
`<=>ad<bc`
`<=>a/b<c/d`(theo giả thiết)
`(a+c)/(b+d)<c/d`
`<=>d(a+c)<c(b+d)`
`<=>ad+cd<bc+dc`
`<=>ad<bc`
`<=>a/b<c/d`(theo giả thiết)`
`=>a/b<(a+c)/(b+d)<c/d`
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.