Bài 1
CMR: \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}cmr:\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\)
Cho \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}\)
CMR:\(\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}=\dfrac{a^3}{c^3}=\dfrac{c^3}{b^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}\left(1\right)\)
Từ \(\dfrac{a}{c}=\dfrac{c}{b}=\dfrac{b}{d}\)
Ta xét tích: \(\left(\dfrac{a}{c}\right)^3=\dfrac{a}{c}.\dfrac{a}{c}.\dfrac{a}{c}=\dfrac{a}{c}.\dfrac{c}{b}.\dfrac{b}{d}=\dfrac{a}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{a^3+c^3-b^3}{c^3+b^3-d^3}=\dfrac{a}{d}\left(dpcm\right)\)
a,\(Cho\dfrac{a}{b}=\dfrac{c}{d}CMR,\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3+b^3}{c^3+d^3}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}\)(1)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\)(2)
Từ (1) và (2) \(\Rightarrow\) đpcm
Theo đề đã cho, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\left(\dfrac{a+b}{c+d}\right)^3=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)(1)
\(\Rightarrow\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\)(2)
Từ (1) và (2)\(\Rightarrow\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3+b^3}{c^3+d^3}\)(đpcm)
Đặt:
\(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{\left[b\left(k+1\right)\right]^3}{\left[d\left(k+1\right)\right]^3}=\dfrac{b^3}{d^3}\\\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{bk^3+b^3}{dk^3+d^3}=\dfrac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\dfrac{b^3}{d^3}\end{matrix}\right.\)
Vậy
Cho a,b,c,d > 0.
Cmr: \(\dfrac{a^3}{a^3+3bcd}+\dfrac{b^3}{b^3+3cda}+\dfrac{c^3}{c^3+3dab}+\dfrac{d^3}{d^3+3abc}\ge1\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}\)
CMR \(\dfrac{a^3+c^3+m^3}{b^3-d^3-n^3}\) = \(\left(\dfrac{a+c-m}{b+d-m}\right)^3\)
mọi người ơi giup mik với ai làm đc mik tick cho
Sửa: CMR \(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{m}{n}=k\Rightarrow a=kb;c=kd;m=kn\)
\(\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\dfrac{k^3b^3+k^3d^3+k^3n^3}{b^3+d^3+n^3}=\dfrac{k^3\left(b^3+d^3+n^3\right)}{b^3+d^3+n^3}=k^3\)
\(\left(\dfrac{a+c-m}{b+d-m}\right)^3=\left(\dfrac{kb+kd-kn}{b+d-n}\right)^3=\left(\dfrac{k\left(b+d-n\right)}{b+d-n}\right)^3=k^3\)
\(\Rightarrow\dfrac{a^3+c^3+m^3}{b^3+d^3+n^3}=\left(\dfrac{a+c-m}{b+d-n}\right)^3\left(=k^3\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}.CMR\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Từ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
Có \(\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(dpcm\right)\)
Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) \(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3=\dfrac{abc}{bcd}=\dfrac{a}{d}\)
\(\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a}{d}\) (1)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\) (2)
Từ (1) và (2) suy ra: \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Leftrightarrow\dfrac{a^3}{b^3}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
\(\Leftrightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
\(\Leftrightarrow\dfrac{a}{d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(đpcm\right)\)
Vậy .................
Chúc bạn học tốt!
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . CMR :
\(a,\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
\(b,\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{a^3-b^3}{c^3-d^3}\)
bài này bạn cứ đặt a=bk, c=dk là được dễ tính lắm sao đó thì thay vào rồi rút gọn là được khi đó bạn sẽ chứng minh được dễ dàng hihi
a) Đặt a/b=c/d=k
suy ra a=bk, c=dk
Vậy 4a-3b/4c-3d=4bk-3b/4dk-3d=b(4k-3)/d(4k-3)=b/d
4a+3b/4c+3d=4bk+3b/4dk+3d=b(4k+3)/d(4k+3)=b/d
Nên 4a-3b/4c-3d=4a+3b/4c+3d=b/d
Đặt a/b=c/d=k
suy ra a=bk, c=dk. Ta có:
a^3+b^3/c^3+d^3=(bk)^3+b^3/(dk)^3+d^3=b^3(k^3+1)/d^3(k^3+1)=b^3/d^3
a^3-b^3/c^3-d^3=(bk)^3-b^3/(dk)^3-d^3=b^3(k^3-1)/d^3(k^3-1)=b^3/d^3
Vậy a^3+b^3/c^3+d^3=a^3-b^3/c^3-d^3=b^3/d^3
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: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)
Bài 1 Cho \(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\left(b\ne0\right)\) CMR \(c=0\)
Bài 2 Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}CMR\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Bài 1: Nhân chéo
Bài 2:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(\Rightarrowđpcm\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\)
\(=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}\)
\(=\dfrac{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}{\left(a-a\right)+\left(b+b\right)+\left(c-c\right)}\)
\(=\dfrac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow c=-c\)
\(\Rightarrow c+c=0\)
\(\Rightarrow2c=0\Rightarrow c=0\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3\)
\(=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:
\(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
Bài 1: CMR : Nếu a2 = b.c thì \(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\) Đảo lại có đúng không?
Bài 2: Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) CMR: \(\left(\dfrac{a+b+c}{b+c+d}\right)\)3 =\(\dfrac{a}{d}\)
Bài 3: Cho dãy tỉ số: \(\dfrac{b.z-c.y}{a}=\dfrac{c.x-a.z}{b}=\dfrac{a.y-b.x}{c}\) , CMR: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
2.
Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(1\right)\)
Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(dpcm\right)\)