Violympic toán 9

\(P=\sum\sqrt[3]{3a+1}=\dfrac{1}{\sqrt[3]{4}}\sum\sqrt[3]{2\cdot2\cdot\left(3a+1\right)}\le\dfrac{1}{3\sqrt[3]{4}}\sum\left(3a+5\right)=3\sqrt[3]{2}\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}.\)

Nãy em sai nha chứ không phải đề sai:vv Buồn ngủ đọc không kỹ đề:vv

Bài 1.1.8 Khá hay và dễ.

Ta chứng minh: \(\left(1+a^3\right)\left(1+b^3\right)^2\ge\left(1+ab^2\right)^3\)

Áp dụng bất đẳng thức Holder:

\(VT=\left(1+a^3\right)\left(1+b^3\right)\left(1+b^3\right)\ge\left[1+\left(a\cdot b\cdot b\right)\right]^3=\left(1+ab^2\right)^3\)

Thiết lập hai bất đẳng thức còn lại và nhân theo vế ta thu được đpcm.

Dấu đẳng thức xin dành cho bạn đọc.

Ps:  BTV thì BTV, thấy bài là em giải nha:v

Bài 183:

Ta cần chứng minh : \(\dfrac{a^3+b^3}{c}+\dfrac{b^3+c^3}{a}+\dfrac{c^3+a^3}{b}\ge2\left(ab+bc+ca\right)\)

\(\Rightarrow ab\left(a^3+b^3\right)+bc\left(b^3+c^3\right)+ca\left(c^3+a^3\right)\ge2abc\left(ab+bc+ca\right)\Leftrightarrow a^4b+ab^4+b^4c+bc^4+c^4a+ca^4\ge2a^2b^2c+2ab^2c^2+2a^2bc^2\) Áp dụng bđt Cô-si vào các số dương a,b,c có:

\(a^4b+bc^4\ge2a^2bc^2,ab^4+ac^4\ge2ab^2c^2,b^4c+ca^4\ge2a^2b^2c\)

\(\Rightarrow a^4b+ab^4+b^4c+bc^4+ca^4+c^4a\ge2a^2b^2c+2ab^2c^2+2a^2bc^2\)

\(\Rightarrow\dfrac{a^3+b^3}{c}+\dfrac{b^3+c^3}{a}+\dfrac{c^3+a^3}{b}\ge2\left(ab+bc+ca\right)\) 

Dấu bằng xảy ra \(\Leftrightarrow a=b=c\)

`\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=6-18x`

`<=>\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=6-18x`

`<=>(9x-3)/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6(3x-1)=0`

`<=>(3x-1)(3/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6)=0`

Ta thấy `3/(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4})+6>0`

`=>3x-1=0`

`=>3x=1`

`=>x=1/3`

Vậy `S={1/3}`

`1/(x^2+9x+20)=1/15-1/(x^2+5x+4)(x ne -1,-4,-5)`

`=>1/((x+4)(x+5))=1/15-1/((x+1)(x+4))`

`=>1/(x+4)-1/(x+5)=1/15-1/((x+1)(x+4))`

`=>3/(x+4)-3/(x+5)=3/15-3/((x+1)(x+4))`

`=>3/(x+4)-3/(x+5)=3/15-1/(x+1)+1/(x+4)`

`=>2/(x+4)-3/(x+5)+1/(x+1)=3/15`

`=>30(x+1)(x+5)-45(x+1)(x+4)+15(x+4)(x+5)=3(x+1)(x+4)(x+5)`

`=>30(x^2+6x+5)-45(x^2+5x+4)+15(x^2+9x+20)=3(x^2+5x+4)(x+5)`

`<=>90x+270=3(x^3+8x^2+29x+20)`

`<=>x^3+24x^2-3x-210=0`

`=>x=-23\or\x=2,85\or\x=-3`

`A=(10^50+2)/(10^50-1)`

`=1+3/(10^50-1)`

Tương tự:

`B=1+3/(10^50-3)`

`10^50-1>10^50-3>0`

`=>3/(10^50-1)<3/(10^50-3)`

`=>A<B`

`20.2^x+1=10.4^2+1`

`=>20.2^x=10.4^2`

`=>2^x=4^2/2=2^3`

`=>x=3`

Vậy x=3

C180 : 

    20 . 2^x + 1 = 10 . 4² + 1

\(\Leftrightarrow\) 2 . 2^x = 4²

\(\Leftrightarrow\) 2^{x + 1} = 2⁴

\(\Leftrightarrow\) x + 1 = 4

\(\Leftrightarrow\) x = 3

Vậy x = 3

179:

CM bđt phụ \(a>b;a,b,m\in N\) thì \(\dfrac{a}{b}< \dfrac{a-m}{b-m}\Rightarrow a\left(b-m\right)< b\left(a-m\right)\Leftrightarrow ab-am< ab-bm\Leftrightarrow-am< -bm\Leftrightarrow a>b\)\(\Rightarrow\dfrac{a}{b}< \dfrac{a-m}{b-m}\)

Áp dụng bđt phụ  \(\dfrac{a}{b}< \dfrac{a-m}{b-m}\) với a>b,a,b,m\(\in N\) có:

\(\Rightarrow A=\dfrac{10^{50}+2}{10^{50}-1}< \dfrac{10^{50}+2-2}{10^{50}-1-2}=\dfrac{10^{50}}{10^{50}-3}=B\)

 \(\Rightarrow A< B\)

139:

Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\left(a,b,c>0\right)\)

GT \(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{3}{abc}\Rightarrow a+b+c=3\)

\(\Rightarrow\dfrac{y^2}{xy^2+2x^2}=\dfrac{1}{b^2}:\left(\dfrac{1}{ab^2}+\dfrac{2}{a^2}\right)=\dfrac{1}{b^2}:\left(\dfrac{a+2b^2}{a^2b^2}\right)=\dfrac{a^2}{a+2b^2}=a-\dfrac{2ab^2}{a+2b^2}\ge a-\dfrac{2ab^2}{3b\sqrt[3]{ab}}=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}\ge a-\dfrac{2}{9}\left(a+b+ab\right)\) Tương tự ta được: 

\(\dfrac{x^2}{zx^2+2z^2}=\dfrac{c^2}{c+2a^2}=c-\dfrac{2ca^2}{c+2a^2}\ge c-\dfrac{2}{9}\left(c+a+ac\right)\)

\(\dfrac{z^2}{yz^2+2y^2}=\dfrac{b^2}{b+2c^2}=b-\dfrac{2bc^2}{b+2c^2}\ge b-\dfrac{2}{9}\left(b+c+bc\right)\)

\(\Rightarrow\dfrac{y^2}{xy^2+2x^2}+\dfrac{x^2}{zx^2+2z^2}+\dfrac{z^2}{yz^2+2z^2}\ge\left(a+b+c\right)-\dfrac{2}{9}\left(2a+2b+2c+ab+bc+ca\right)\) \(\ge3-\dfrac{2}{9}\left[6+\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{2}{9}\left(6+\dfrac{9}{3}\right)=3-\dfrac{2}{9}\cdot9=1\)

Dấu bằng xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=3\)

câu trả lời :

Đặt ⇒1ab+1bc+1ca=3abc⇒a+b+c=3⇒1ab+1bc+1ca=3abc⇒a+b+c=3

141. x > -5

Đặt \(\left\{{}\begin{matrix}2x^2+5x+12=a\\2x^2+3x+2=b\end{matrix}\right.\left(a,b>0\right)\)

pt \(\Leftrightarrow a+b=\dfrac{a^2-b^2}{2}\)

\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{a-b}{2}\right)=0\)

\(\Leftrightarrow a-b=2\) (vì \(\sqrt{2x^2+5x+12}+\sqrt{2x^2+3x+2}>0\))

\(\Rightarrow\sqrt{2x^2+5x+12}=\sqrt{2x^2+3x+2}+2\)

\(\Rightarrow x+3=2\sqrt{2x^2+3x+2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-1\end{matrix}\right.\) (thỏa mãn)

Vậy .....

c131-136 nhỏ ko đọc đc

Bài 129:

ĐKXĐ: \(x^2-y+1\ge0\)\(\left\{{}\begin{matrix}4x^2-2x+y^2+y-4xy=0\left(1\right)\\x^2-x+y=\left(y-x+3\right)\sqrt{x^2-y+1}\left(2\right)\end{matrix}\right.\)

Từ (1) \(\Rightarrow\left(2x-y\right)^2-\left(2x-y\right)=0\Leftrightarrow\left(2x-y\right)\left(2x-y-1\right)=0\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=2x-1\end{matrix}\right.\)

Nếu y=2x Thay vào (2) ta được: 

\(\Rightarrow x^2-x+2x=\left(2x-x+3\right)\sqrt{x^2-2x+1}\Leftrightarrow x^2+x=\left(x+3\right)\left|x-1\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x=\left(x+3\right)\left(1-x\right)\left(x< 1\right)\left(3\right)\\x^2+x=\left(x+3\right)\left(x-1\right)\left(x\ge1\right)\left(4\right)\end{matrix}\right.\) 

Từ (3) \(\Rightarrow x^2+x=x-x^2+3-3x\Leftrightarrow2x^2+3x-3=0\) \(\Leftrightarrow x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{3}{2}=0\Leftrightarrow\left(x-\dfrac{3}{4}\right)^2=\dfrac{33}{16}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{4}\left(L\right)\\x=\dfrac{3-\sqrt{33}}{4}\left(TM\right)\end{matrix}\right.\)\(\Rightarrow y=\) \(2\cdot\left(\dfrac{3-\sqrt{33}}{4}\right)=\dfrac{3-\sqrt{33}}{2}\)

Từ (4) \(\Rightarrow x^2+x=x^2-x+3x-3\Leftrightarrow-x=-3\Leftrightarrow x=3\left(TM\right)\)\(\Rightarrow y=6\)

Nếu y=2x+1 Thay vào (2) ta được: 

\(\Rightarrow x^2-x+2x+1=\left(2x+1-x+3\right)\sqrt{x^2-2x-1+1}\Leftrightarrow x^2+x+1=\left(x+4\right)\sqrt{x^2-2x}\left(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.;x\ge-4\right)\)

\(\Rightarrow x^2+x+1-\left(x+4\right)\sqrt{x^2-2x}=0\Leftrightarrow2x^2+2x+2-2x\sqrt{x^2-2x}-4\sqrt{x^2-2x}=0\Leftrightarrow x^2-2x+x^2+4-2x\sqrt{x^2-2x}+4x-4\sqrt{x^2-2x}=2\Leftrightarrow\left(-\sqrt{x^2-2x}+x+2\right)^2=2\) \(\Leftrightarrow\left[{}\begin{matrix}-\sqrt{x^2-2x}+x+2=\sqrt{2}\left(5\right)\\-\sqrt{x^2-2x}+x+2=-\sqrt{2}\left(6\right)\end{matrix}\right.\)

Từ (5) \(\Rightarrow\sqrt{x^2-2x}=x+2-\sqrt{2}\Rightarrow x^2-2x=x^2+\left(2-\sqrt{2}\right)^2-2x\left(2-\sqrt{2}\right)\Leftrightarrow2x\left(2-\sqrt{2}-2\right)=4+2-4\sqrt{2}\Leftrightarrow-2\sqrt{2}x=6-4\sqrt{2}\Leftrightarrow x=-\dfrac{3\sqrt{2}}{2}+2\left(TM\right)\) \(\Rightarrow y=2\left(\dfrac{-3\sqrt{2}}{2}+2\right)+1=-3\sqrt{2}+5\)

Từ (6) \(\Rightarrow\sqrt{x^2-2x}=x+2+\sqrt{2}\Rightarrow x^2-2x=x^2+\left(2+\sqrt{2}\right)^2+2x\left(2+\sqrt{2}\right)\Leftrightarrow2x\left(2+\sqrt{2}-2\right)=6+4\sqrt{2}\Leftrightarrow2\sqrt{2}x=6+4\sqrt{2}\Leftrightarrow x=\dfrac{3\sqrt{2}}{2}+2\left(TM\right)\)

 \(\Rightarrow y=2\left(\dfrac{3\sqrt{2}}{2}+2\right)+1=3\sqrt{2}+5\)

Vậy...

Bài 286: Bất đẳng thức neibizt khá nổi tiếng :D 

Bđt <=> \(\dfrac{a}{b+c}+\dfrac{1}{2}+\dfrac{b}{c+a}+\dfrac{1}{2}+\dfrac{c}{a+b}+\dfrac{1}{2}\ge\dfrac{9}{2}\)

\(\Leftrightarrow\left(2a+2b+2c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{c+a}+\dfrac{1}{b+c}\right)\ge9\) ( Có thể đơn giản hóa bất đẳng thức bằng việc đặt biến phụ )

Đặt: \(\left\{{}\begin{matrix}x=b+c\\y=c+a\\z=a+b\end{matrix}\right.\) khi đó ta có: \(\left\{{}\begin{matrix}a=\dfrac{y+z-x}{2}\\b=\dfrac{z+x-y}{2}\\c=\dfrac{x+y-z}{2}\end{matrix}\right.\) Bất đẳng thức trở thành: \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\) ( luôn đúng theo AM-GM )

Vậy bất đẳng thức đã được chứng minh. Dấu "=" xảy ra tại a=b=c

C286.(Cách khác)

Áp dụng BĐT BSC và BĐT \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\):

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(=\dfrac{a^2}{ab+ca}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\)

\(\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(a+b+c\right)^2}{\dfrac{2}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)

Đẳng thức xảy ra khi \(a=b=c\)

Bài 284 

Ta cần CM \(\left(a+b\right)\left(a^4+b^4\right)\ge\left(a^2+b^2\right)\left(a^3+b^3\right)\)

\(\Leftrightarrow a^5+b^5+a^4b+ab^4\ge a^5+b^5+a^2b^3+a^3b^2\)

\(\Leftrightarrow a^4b+ab^4\ge a^3b^2+a^2b^3\) \(\Leftrightarrow a^4b-a^3b^2-a^2b^3+ab^4\ge0\)

\(\Leftrightarrow a^3b\left(a-b\right)-ab^3\left(a-b\right)\ge0\Leftrightarrow\left(a-b\right)\left(a^3b-ab^3\right)\ge0\Leftrightarrow\left(a-b\right)ab\left(a^2-b^2\right)\ge0\) \(\Leftrightarrow\left(a-b\right)^2ab\left(a+b\right)\ge0\) luôn đúng với mọi a,b>0 Vậy...

1: ĐKXĐ: a,b>0, a\(\ne b\)

\(\Rightarrow Q=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^3+2a\sqrt{a}+b\sqrt{b}}{3\sqrt{a}\left(a\sqrt{a}+b\sqrt{b}\right)}+\dfrac{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}{\sqrt{a}\left(a-b\right)}=\dfrac{a\sqrt{a}-3a\sqrt{b}+3b\sqrt{a}-b\sqrt{b}+2a\sqrt{a}+b\sqrt{b}}{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\) \(=\dfrac{3\sqrt{a}\left(a-\sqrt{ab}+b\right)}{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{1}{\sqrt{a}+\sqrt{b}}-\dfrac{1}{\sqrt{a}+\sqrt{b}}=0\) 

\(\Rightarrow Q\) ko phụ thuộc vào a,b Vậy...

2: Ta có \(1\ge x+y\ge2\sqrt{xy}\Rightarrow xy\le\dfrac{1}{4}\) 

\(\Rightarrow P=\dfrac{x+y}{xy}\cdot\sqrt{x^2y^2+\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}}\ge\dfrac{2\sqrt{xy}}{xy}\cdot\sqrt{17}\cdot\sqrt[34]{\dfrac{x^2y^2}{16^{16}}}=\sqrt{17}\cdot\dfrac{2}{\sqrt{xy}}\cdot\sqrt[17]{\dfrac{xy}{16^8}}\) \(=\sqrt{17}\cdot\sqrt[17]{\dfrac{2^{17}}{\sqrt{x^{17}y^{17}}}\cdot\dfrac{\sqrt{x^2y^2}}{2^{32}}=\sqrt{17}\cdot\sqrt[17]{\dfrac{1}{\sqrt{x^{15}y^{15}}\cdot2^{15}}}\ge\sqrt{17}\cdot\sqrt[17]{\dfrac{1}{\sqrt{\dfrac{1}{4^{15}}}\cdot2^{15}}}=\sqrt{ }17}\)

Dấu  = xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\) Vậy...

toán mấy đấy aj??

Bài 5

Giả sử diện tích tam giác ABC là số nguyên. 

\(\Rightarrow S_{ABC}=\dfrac{1}{2}sinA\cdot AB\cdot AC\) là số nguyên 

\(\Rightarrow sinA\cdot AB\cdot AC⋮2\) \(\Rightarrow AB\cdot AC⋮2\)( vì \(sinA< 1\) ) vô lí vì AB,AC,BC đều là số nguyên tố \(\Rightarrow\) giả sử sai Vậy ...

Test chức năng.

Test chức năng 2.