Cho : ad = bc . Cm :
\(\dfrac{3a^6+c^6}{3b^6+d^6}\) = \(\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\), chứng minh rằng:
a) \(\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
b) \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{5b-3a}{5c-3d}\)
ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}\)
Áp dụng tính chất dãy tỉ sốbằng nhau ta có:
\(\dfrac{a^6}{b^6}=\dfrac{c^6}{d^6}=\dfrac{3a^6}{3b^6}=\dfrac{a^6+c^6}{b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
=\(\dfrac{c^6+3a^6}{d^6+3b^6}\)
\(\Rightarrow\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (ĐPCM)
1) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng \(\dfrac{2a^2-3ab+5b^2}{2a^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2c^2+3cd}\)
2) Cho \(\dfrac{a}{c}=\dfrac{c}{b}\). Chứng minh rằng \(\dfrac{b^2-c^2}{a^2+c^2}=\dfrac{b-a}{a}\)
3) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\).Chứng minh rằng\(\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Bài 3:
Đặt $\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$
Khi đó:
$\frac{3a^6+c^6}{3b^6+d^6}=\frac{3(bt)^6+(dt)^6}{3b^6+d^6}=\frac{t^6(3b^6+d^6)}{3b^6+d^6}=t^6(*)$
Và:
$\frac{(a+c)^6}{(b+d)^6}=(\frac{bt+dt}{b+d})^6=t^6(**)$
Từ $(*); (**)\Rightarrow $ đpcm.
\(cho\frac{a}{b}=\frac{c}{d}\\ chứngminh\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
Đặt \(\frac{a}{b}=\frac{c}{d}\)=k \(\Rightarrow a=bk;c=dk\)
Ta có: \(\frac{3a^6+c^6}{3b^6+d^6}=\frac{3\left(bk\right)^6+\left(dk\right)^6}{3b^6+d^6}=\frac{3b^6.k^6+d^6.k^6}{3b^6+d^6}=\frac{k^6\left(3b^6+d^6\right)}{3b^6+d^6}=k^6\)(1)
\(\frac{\left(a+c\right)^6}{\left(b+d\right)^6}=\frac{\left(bk+dk\right)^6}{\left(b+d\right)^6}=\frac{\left[k\left(b+d\right)\right]^6}{\left(b+d\right)^6}=\frac{k^6.\left(b+d\right)^6}{\left(b+d\right)^6}=k^6\)(2)
Từ (1) và (2), ta có: \(\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
1. Cho \(a,b,c>0\) và \(ab+bc+ca=abc\). Chứng minh rằng:
\(\dfrac{1}{a+3b+2c}+\dfrac{1}{b+3c+2a}+\dfrac{1}{c+3a+2b}\le\dfrac{1}{6}\)
2. Cho \(a,b\ge0\) và \(a+b=2\) Tìm Max
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+20ab\)
Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
2,
\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)
\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(ab=x\Rightarrow0\le x\le1\)
\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)
\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)
Cho \(\frac{a}{c}=\frac{b}{d}\) Chứng minh \(\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
Cho \(\frac{a}{3}=\frac{3}{d}\) Chứng minh \(\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
cho a/b=c/d và \(b+d\ne0\)Chứng minh \(\frac{3a^6+c^6}{3b^6+d^6}\)=\(\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
Ta có :\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow\left(\frac{a}{b}\right)^6=\left(\frac{c}{d}\right)^6=\left(\frac{a+c}{b+d}\right)^6\)
\(\Rightarrow\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (1)
Ta lại có : \(\frac{a^6}{b^6}=\frac{c^6}{d^6}=\frac{3a^6}{3b^6}=\frac{3a^6+c^6}{3b^6+d^6}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) (đpcm)
\(\frac{3a^6+c^6}{3b^6+d^6}=\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
có cả phép tính và lời giải cho mình nhé, ai nhanh mình tick!
đề bài là chứng minh hay tính đây bạn ???
cho \(\frac{a}{b}\)=\(\frac{c}{d}\) chung minh \(\frac{3a^6+c^6}{3b^6+d^6}\)=\(\frac{\left(a+c\right)^6}{\left(b+d\right)^6}\) va b+d khac 0