Làm đi đăng gì giỏi toán lắm màk :"))))))))))

-.-

Đa thức \(x^3+ax+b\) có bậc 3 chia cho đa thức \(x^2-x-2\) có bậc 2 nên sẽ đợc thương có bậc 1

Thương của phép chia có dạng : cx + d

\(\Rightarrow\left(x^2-x-2\right)\left(cx+d\right)=x^3+ax+b\)

\(\Leftrightarrow cx^3+dx^2-cx^2-dx+2cx+2d=x^3+ax+b\)

\(\Leftrightarrow cx^3+\left(d-c\right)x^2-\left(d-2c\right)x+2d=x^3+ax+b\)

\(\Rightarrow\left\{{}\begin{matrix}cx^3=x^3\\\left(d-c\right)x^2=0\\-\left(d-2c\right)x=ax\\2d=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c=1\\d-c=0\\-d+2c=a\\2d=b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=1\\d=1\\a=-1+2.1=1\\b=2.1=2\end{matrix}\right.\)

Vậy a = 1 , b = 2

Lời giải:

Ta có:

\(\left\{\begin{matrix} x=by+cz\\ y=ax+cz\\ z=ax+by\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x-y=by-ax\\ z=ax+by\end{matrix}\right.\)

\(\Rightarrow x-y+z=2by\Rightarrow b=\frac{x+z-y}{2y}\)

Hoàn toàn tương tự ta nhận được:

\(a=\frac{y+z-x}{2x};c=\frac{x+y-z}{2z}\)

Suy ra:

\(\left\{\begin{matrix} a+1=\frac{x+y+z}{2x}\\ b+1=\frac{x+y+z}{2y}\\ c+1=\frac{x+y+z}{2z}\end{matrix}\right.\)

\(\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2\) (ĐPCM)

Áp dụng BĐT AM-GM ta có:

\(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{b^2\left(a+1\right)}{b^2+1}\ge\left(a+1\right)-\frac{b^2\left(a+1\right)}{2b}\)

\(=\left(a+1\right)-\frac{ab+b}{2}\). Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\ge3+\left(a+b+c\right)-\frac{ab+bc+ca+a+b+c}{2}\)

\(\ge3+\left(a+b+c\right)-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)

Dấu "=" <=> \(a=b=c=1\)

\(Áp dụng BĐT AM-GM ta có: \(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{b^2\left(a+1\right)}{b^2+1}\ge\left(a+1\right)-\frac{b^2\left(a+1\right)}{2b}\) \(=\left(a+1\right)-\frac{ab+b}{2}\). Tương tự cho 2 BĐT còn lại rồi cộng theo vế: \(VT\ge3+\left(a+b+c\right)-\frac{ab+bc+ca+a+b+c}{2}\) \(\ge3+\left(a+b+c\right)-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\) Dấu "=" <=> \(a=b=c=1\)\)

\(\Leftrightarrow-\left(4x^2-y^2-7\right)^2=0\)

SURPRISE MOTHERFUKA !!

Giải câu 1 thôi câu 2 không hứng lắm:

\(P=\dfrac{1}{2a+3b+c+6}+\dfrac{1}{2b+3c+a+6}+\dfrac{1}{2c+3a+b+6}\)

Ta có:

\(\dfrac{1}{2a+3b+c+6}\le\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{b+2}\right)=\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{a+2}+\dfrac{2}{b+2}\right)\left(1\right)\)

Tương tự ta có:

\(\left\{{}\begin{matrix}\dfrac{1}{2b+3c+a+6}\le\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{b+2}+\dfrac{2}{c+2}\right)\left(2\right)\\\dfrac{1}{2c+3a+b+6}\le\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{c+2}+\dfrac{2}{a+2}\right)\left(3\right)\end{matrix}\right.\)

Cộng (1), (2), (3) vế theo vế ta được:

\(P\le\dfrac{3}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\)

\(\le\dfrac{3}{16.3\sqrt[3]{abc}}+\dfrac{3}{16}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\)

\(=\dfrac{1}{16}+\dfrac{3}{16}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\left(4\right)\)

Giờ ta tính Max của \(Q=\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\)

Vì \(abc=1\) nên không mất tính tổng quát ta giả sử \(\left\{{}\begin{matrix}ab\le1\\c\ge1\end{matrix}\right.\)

Ta có: \(Q=\dfrac{1}{2}.\left(\dfrac{1}{\dfrac{a}{2}+2}+\dfrac{1}{\dfrac{b}{2}+2}\right)+\dfrac{1}{c+2}\)

Ta có bổ đề: Với \(x,y>0;xy\le1\) thì

\(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}\le\dfrac{2}{xy+1}\)

Áp dụng vào bài toán ta được:

\(Q\le\dfrac{2}{1+\dfrac{\sqrt{ab}}{2}}+\dfrac{1}{c+2}=\dfrac{2\sqrt{c}}{2\sqrt{c}+1}+\dfrac{1}{c+2}\)

Xét hàm số \(f\left(\sqrt{c}\right)=\dfrac{2\sqrt{c}}{2\sqrt{c}+1}+\dfrac{1}{c+2}\) với \(\sqrt{c}\ge1\) thì hàm số \(f\left(\sqrt{c}\right)\) nghịch biến. Vậy Q đạt GTLN khi c bé nhất.

\(\Rightarrow Q\le f\left(1\right)=1\left(2\right)\)

Từ (4) và (5) ta suy ra

\(P\le\dfrac{1}{16}+\dfrac{3}{16}.1=\dfrac{1}{4}\)

Vậy GTLN là \(P=\dfrac{1}{4}\) đạt được khi \(a=b=c=1\)

2) A = n³ - n² + n - 1

A = n²(n - 1) + (n - 1)

A = (n - 1)(n² + 1)

Để A nguyên tố thì n > 1

=> n² + 1 > 1

Mà A = (n - 1)(n² + 1) là số nguyên tố, chỉ gồm 2 ước là 1 và chính nó

Nên A = n² + 1; n - 1 = 1

=> n = 2 (TM)

b) n⁵ - n + 2

= n(n⁴ - 1) + 2

= n(n² - 1)(n² + 1) + 2

= n(n - 1)(n + 1)(n² + 1) + 2

n(n - 1)(n + 1) là tích 3 số nguyên liên tiếp do n \(\in N\) nên n(n - 1)(n + 1) chia hết cho 3

=> n(n - 1)(n + 1)(n² + 1) + 2 chia 3 dư 2, không là số chính phương

Vậy ...

Đặt độ dài mỗi cạnh của hình vuông ABCD là a (a\(\in\)R⁺)

Theo ĐL Thales, ta có có tỉ lệ sau: \(\frac{BC}{BM}=\frac{AN}{AM}\); \(\frac{ND}{DC}=\frac{AN}{AM}\)

\(\Rightarrow\frac{BC}{BM}=\frac{ND}{DC}\Rightarrow BM.ND=BC.DC=a^2\)(1)

Sau đó chứng minh \(\Delta\)AOD ~ \(\Delta\)DAB (g.g) => \(\frac{AO}{AD}=\frac{AD}{BD}\)\(\Rightarrow AO.BD=AD^2=a^2\)

hay \(BO.BD=a^2\)(2)

Từ (1) và (2) => \(BM.ND=BO.BD\)\(\Rightarrow\frac{BM}{BD}=\frac{BO}{ND}\)

Ta có: \(\widehat{MBO}=\widehat{ABO}+\widehat{MBA}=135^0\), \(\widehat{BDN}=\widehat{ADO}+\widehat{NDA}=135^0\)

\(\Rightarrow\widehat{MBO}=\widehat{BDN}\)

Xét \(\Delta\)MBO và \(\Delta\)BDN: \(\widehat{MBO}=\widehat{BDN};\) \(\frac{BM}{BD}=\frac{BO}{ND}\)(cmt)

=> \(\Delta\)MBO ~ \(\Delta\)BDN (c.g.c) => \(\widehat{M_1}=\widehat{B_1}\)

Ta thấy \(\widehat{BKO}\)là góc ngoài của tam giác MBK

=> \(\widehat{BKO}=\widehat{M_1}+\widehat{MBK}=\widehat{B_1}+\widehat{MBK}=\widehat{MBO}=135^0\)\(\Rightarrow\)\(\widehat{MKB}=45^0\)(Kề bù)

\(\Rightarrow\widehat{MKB}=\widehat{MCO}=45^0\)

\(\Rightarrow\Delta\)MKB ~ \(\Delta\)MCO (g.g) => \(\frac{BK}{OC}=\frac{MK}{MC}\)hay \(\frac{BK}{OB}=\frac{MK}{MC}\)

Xét \(\Delta\)KBO và \(\Delta\)KMC: \(\widehat{B_1}=\widehat{M_1}\);\(\frac{BK}{OB}=\frac{MK}{MC}\)\(\Rightarrow\widehat{BKO}=\widehat{MKC}\).

Mà \(\widehat{BKO}=135^0\)(cmt)\(\Rightarrow\widehat{MKC}=135^0\)

Lại có: \(\widehat{MKC}=\widehat{MKB}+\widehat{BKC}\)

\(\Rightarrow\widehat{BKC}=\widehat{MKC}-\widehat{MKB}=135^0-45^0=90^0\)(Do ^MKB=45⁰(cmt))

=> \(CK\perp BN\)(đpcm).

VP = \(\dfrac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\dfrac{\left(b-c\right)^2}{\left(b+a\right)\left(c+a\right)}+\dfrac{\left(c-a\right)^2}{\left(c+b\right)\left(a+b\right)}\)

\(=\left(a-b\right).\dfrac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\left(b-c\right).\dfrac{\left(b+a\right)-\left(c+a\right)}{\left(b+a\right)\left(c+a\right)}+\left(c-b\right).\dfrac{\left(c+b\right)-\left(a+b\right)}{\left(c+b\right)\left(a+b\right)}\)

\(=\left(a-b\right).\left(\dfrac{1}{b+c}-\dfrac{1}{a+c}\right)+\left(b-c\right)\left(\dfrac{1}{c+a}-\dfrac{1}{b+a}\right)+\left(c-a\right).\left(\dfrac{1}{a+b}-\dfrac{1}{c+b}\right)\)

\(=\left(a-b\right).\dfrac{1}{b+c}-\left(a-b\right).\dfrac{1}{a+c}+\left(b-c\right).\dfrac{1}{c+a}-\left(b-c\right).\dfrac{1}{b+a}+\left(c-a\right).\dfrac{1}{a+b}-\left(c-a\right).\dfrac{1}{c+b}\)

\(=\left(2a-b-c\right).\dfrac{1}{b+c}+\left(2b-c-a\right).\dfrac{1}{c+a}+\left(2c-a-b\right).\dfrac{1}{a+b}\)

\(=\dfrac{2a}{b+c}-\left(b+c\right).\dfrac{1}{b+c}+\dfrac{2b}{c+a}-\left(c+a\right).\dfrac{1}{c+a}+\dfrac{2c}{a+b}-\left(a+b\right).\dfrac{1}{a+b}\)

\(=2\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)-3\left(đpcm\right)\)

\(VT=\dfrac{2a^3-a^2b-a^2c-ab^2-ac^2+2b^3-b^2c-bc^2+2c^3}{(a+b)(b+c)(c+a)} \)

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

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

Nhân hai vào 2 vế thử đi