Cho \(\frac{a}{b}=\frac{c}{d}\), chứng minh rằng:
\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\). Ta có:
\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{\left(bk-b\right)^3}{\left(dk-d\right)^3}=\frac{b^3\left(k-1\right)^3}{d^3\left(k-1\right)^3}=\frac{b^3}{d^3}\)
\(\frac{3a^2+2b^2}{3c^2+2d^2}=\frac{3\left(bk\right)^2+2b^2}{3\left(dk\right)^2+2d^2}=\frac{3b^2k^2+2b^2}{3d^2k^2+2d^2}=\frac{b^2\left(3k^2+2\right)}{d^2\left(3k^2+2\right)}=\frac{b^2}{d^2}\)
Đến đây nhìn có vẻ đề sai
\(\frac{a}{b}=\frac{c}{d}=k\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)ta có:
\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{\left(bk-b\right)^3}{\left(dk-d\right)^3}=\frac{\left[b\left(k-1\right)\right]^3}{\left[d\left(k-1\right)\right]^3}=\frac{b^3}{d^3}\)
\(\frac{2b^2+3a^2}{2d^2+3c^2}=\frac{4.b^2+9.k^2.b^2}{4.d^2+9.d^2.k^2}=\frac{b^2\left(4+k^2.9\right)}{d^2\left(4+9.k^2\right)}=\frac{b^2}{d^2}\)
\(Taco:\frac{b^3}{d^3}=\frac{b^2}{d^2}\Leftrightarrow b=d\)
1. Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)
Thay vào 2 vế là sẽ CM được
1. Đặt \(\frac{a}{b}=\frac{c}{d}=k>a=bk.c=dk\)
Thay vào 2 vế để chứng minh
Đặt A=\(\left(\frac{-a}{2}+\frac{b}{3}+\frac{c}{6}\right)^3+\left(\frac{a}{3}+\frac{b}{6}-\frac{c}{2}\right)^3+\left(\frac{a}{6}-\frac{b}{2}+\frac{c}{3}\right)^3\)
\(=\left(\frac{-3a+2b+c}{6}\right)^3+\left(\frac{2a+b-3c}{6}\right)^3+\left(\frac{a-3b+2c}{6}\right)^3\)
\(=\left(\frac{-3a+2b+c+2a+b-3c+a-3b+2c}{6}\right)^3-\frac{\left(-a+3b-2c\right)\left(3a-2b-c\right)\left(-2a-b+3c\right)}{72}\)
(Hằng đẳng thức)
\(=0-\frac{\left(-a+3b-2c\right)\left(3a-2b-c\right)\left(-2a-b+3c\right)}{72}\)
\(\Rightarrow\frac{\left(a-3b+2c\right)\left(-3a+2b+c\right)\left(2a+b-3c\right)}{72}=\frac{1}{8}\)
\(\Leftrightarrow\left(a-3b+2c\right)\left(2a+b-3c\right)\left(-3a+2b+c\right)=9\)(đpcm).
mk làm câu a thôi, b dài nhưng tương tự
Gọi a/b=c/d=k =>a=bk ; c=dk
=>\(\frac{\left(2a+3b\right)^2}{\left(3a-4b\right)^2}=\frac{\left(2bk+3b\right)^2}{\left(3bk-4b\right)^2}=\frac{\left[b\left(2k+3\right)\right]^2}{\left[b\left(3k-4\right)\right]^2}=\frac{b^2\left(2k+3\right)^2}{b^2\left(3k-4\right)^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(1)
=>\(\frac{\left(2c+3d\right)^2}{\left(3c-4d\right)^2}=\frac{\left(2dk+3d\right)^2}{\left(3dk-4d\right)^2}=\frac{\left[d\left(2k+3\right)\right]^2}{\left[d\left(3k-4\right)\right]^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(2)
Từ (1);(2)=> đpcm
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk,c=dk\)
Ta có:
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\) (1)
\(\frac{3a^2+2b^2}{3c^2+2d^2}=\frac{3.\left(bk\right)^2+2b^2}{3\left(dk\right)^2+2d^2}=\frac{3.b^2.k^2+2b^2}{3.d^2.k^2+2d^2}=\frac{b^2\left(3k^2+2\right)}{d^2\left(3.k^2+2\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) suy ra \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
Mk có sửa đề chút nhé!