đề kiểu j đây bn?

a/b = b/c = c/d = (a+b+c)/(b+c+d)

=> (a+b+c/b+c+d)^6054 = (a/b)^6054

thử bài bất :D 

Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)

Hoàn toàn tương tự: 

\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)

\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)

Cộng (*),(**),(***) vế theo vế ta được:

\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)

Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )

Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1

1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D 

Đặt A/B=C/D=k

=>A=k*B; C=D*k

A/B=k*B/B=k

\(\dfrac{A+C}{B+D}=\dfrac{k\cdot B+k\cdot D}{B+D}=k\)

=>\(\dfrac{A}{B}=\dfrac{A+C}{B+D}\)

Bài 1: Ta có:

\(M=\frac{ad}{abcd+abd+ad+d}+\frac{bad}{bcd.ad+bc.ad+bad+ad}+\frac{c.abd}{cda.abd+cd.abd+cabd+abd}+\frac{d}{dab+da+d+1}\)

\(=\frac{ad}{1+abd+ad+d}+\frac{bad}{d+1+bad+ad}+\frac{1}{ad+d+1+abd}+\frac{d}{dab+da+d+1}\)

$=\frac{ad+abd+1+d}{ad+abd+1+d}=1$

Bài 2:

Vì $a,b,c,d\in [0;1]$ nên

\(N\leq \frac{a}{abcd+1}+\frac{b}{abcd+1}+\frac{c}{abcd+1}+\frac{d}{abcd+1}=\frac{a+b+c+d}{abcd+1}\)

Ta cũng có:
$(a-1)(b-1)\geq 0\Rightarrow a+b\leq ab+1$

Tương tự:

$c+d\leq cd+1$

$(ab-1)(cd-1)\geq 0\Rightarrow ab+cd\leq abcd+1$

Cộng 3 BĐT trên lại và thu gọn thì $a+b+c+d\leq abcd+3$

$\Rightarrow N\leq \frac{abcd+3}{abcd+1}=\frac{3(abcd+1)-2abcd}{abcd+1}$

$=3-\frac{2abcd}{abcd+1}\leq 3$

Vậy $N_{\max}=3$

3.

Hình vẽ:

Lời giải:

a) △AMC và △BNC có: \(\widehat{AMC}=\widehat{BNC}=90^0;\widehat{ACB}\) là góc chung.

\(\Rightarrow\)△AMC∼△BNC (g-g).

\(\Rightarrow\dfrac{AC}{BC}=\dfrac{CM}{CN}\Rightarrow AC.CN=BC.CM\left(1\right)\)

b) △AMB và △CPB có: \(\widehat{AMB}=\widehat{CPB}=90^0;\widehat{ABC}\) là góc chung.

\(\Rightarrow\)△AMB∼△CPB (g-g)

\(\Rightarrow\dfrac{AB}{CB}=\dfrac{BM}{BP}\Rightarrow AB.BP=BC.BM\left(2\right)\)

Từ (1) và (2) suy ra:

\(AC.CN+AB.BP=BC.CM+BC.BM=BC.\left(CM+BM\right)=BC.BC=BC^2\left(đpcm\right)\)b) Gọi \(M_0\) là trung điểm BC, giả sử \(AB< AC\).

\(\widehat{HBM}=90^0-\widehat{BHM}=90^0-\widehat{AHN}=\widehat{CAM}\)

△HBM và △CAM có: \(\widehat{HBM}=\widehat{CAM};\widehat{HMB}=\widehat{CMA}=90^0\)

\(\Rightarrow\)△HBM∼△CAM (g-g) 

\(\Rightarrow\dfrac{MH}{CM}=\dfrac{BM}{MA}\Rightarrow MH.MA=BM.CM\)

Ta có: \(BM.CM=\left(BM_0-MM_0\right)\left(CM_0+MM_0\right)=\left(BM_0-MM_0\right)\left(BM_0+MM_0\right)=BM_0^2-MM_0^2\le BM_0^2=\dfrac{BC^2}{4}\)

\(\Rightarrow MH.MA\le\dfrac{BC^2}{4}\).

Vì \(BC\) không đổi nên: \(max\left(MH.MA\right)=\dfrac{BC^2}{4}\), đạt được khi △ABC cân tại A hay A nằm trên đường trung trực của BC.

c) Sửa đề: \(S_1.S_2.S_3\le\dfrac{1}{64}.S^3\)

△AMC∼△BNC \(\Rightarrow\dfrac{AC}{BC}=\dfrac{MC}{NC}\Rightarrow\dfrac{AC}{MC}=\dfrac{BC}{NC}\)

△ABC và △MNC có: \(\dfrac{AC}{MC}=\dfrac{BC}{NC};\widehat{ACB}\) là góc chung.

\(\Rightarrow\)△ABC∼△MNC (c-g-c)

\(\Rightarrow\dfrac{S_{MNC}}{S_{ABC}}=\dfrac{S_1}{S}=\dfrac{MC}{AC}.\dfrac{NC}{BC}\left(1\right)\)

Tương tự: 

△ABC∼△MBP \(\Rightarrow\dfrac{S_{MBP}}{S_{ABC}}=\dfrac{S_2}{S}=\dfrac{MB}{AB}.\dfrac{BP}{BC}\left(2\right)\)

△ABC∼△ANP \(\Rightarrow\dfrac{S_{ANP}}{S_{ABC}}=\dfrac{S_3}{S}=\dfrac{AN}{AB}.\dfrac{AP}{AC}\left(3\right)\)

Từ (1), (2), (3) suy ra:

\(\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}=\left(\dfrac{MC}{AC}.\dfrac{NC}{BC}\right).\left(\dfrac{MB}{AB}.\dfrac{BP}{BC}\right).\left(\dfrac{AN}{AB}.\dfrac{AP}{AC}\right)\) 

\(\Rightarrow\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}=\left(\dfrac{MC.MB}{AC.AB}\right).\left(\dfrac{BP.AP}{AC.BC}\right).\left(\dfrac{AN.CN}{AB.BC}\right)\) (*)

Áp dụng câu b) ta có:

\(\left\{{}\begin{matrix}BM.CM\le\dfrac{1}{4}BC^2\\AP.BP\le\dfrac{1}{4}AB^2\\AN.CN\le\dfrac{1}{4}AC^2\end{matrix}\right.\)

Từ (*) suy ra:

\(\dfrac{S_1}{S}.\dfrac{S_2}{S}.\dfrac{S_3}{S}\le\left(\dfrac{\dfrac{1}{4}BC^2}{AC.AB}\right).\left(\dfrac{\dfrac{1}{4}AC^2}{AC.BC}\right).\left(\dfrac{\dfrac{1}{4}AB^2}{AB.BC}\right)=\dfrac{1}{64}\)

\(\Rightarrow S_1.S_2.S_3\le\dfrac{1}{64}.S^3\)

Dấu "=" xảy ra khi △ABC đều.

??????????????????????!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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