Ta có: x²y + xy - 2x² - 3x + 4 = 0

=> x²(y - 2) + x(y - 2) - (x + 1) = -5

=> (x² + x)(y - 2) - (x + 1) = -5

=> x(x + 1)(y - 2) - (x + 1) = -5

=> (x - 1)[x(y - 2) - 1] = -5

=>  x - 1; x(y - 2) - 1 \(\in\)Ư(-5) = {1; -1; 5; -5}

Với : x - 1 = 1           => x = 2

      x(y - 2)  - 1 = -5 => x(y - 2) = -4   => y - 2 = -2      => y = 0

x - 1 = -1                => x = 0

x(y - 2) - 1 = 5  => x(y - 2) = 6       (ktm vì x = 0)

x - 1 = 5                => x = 6

x(y - 2) - 1 = -1      => x(y - 2) = 0                 => y - 2 = 0          => y = 2

x - 1 = -5              => x = -4

x(y - 2) - 1 = 1    => x(y - 2) = 2              => y - 2 = -1/2             => y = 3/2

Vậy ...

Câu 9.

a) Ta có: \(\left(a-1\right)^2\ge0\)(điều hiển nhiên)

\(\Leftrightarrow a^2-2a+1\ge0\)

\(\Leftrightarrow a^2+2a+1\ge4a\)

\(\Leftrightarrow\left(a+1\right)^2\ge4a\left(đpcm\right)\)

b) Áp dụng BĐT Cauchy cho 2 số không âm:

\(a+1\ge2\sqrt{a}\)

\(b+1\ge2\sqrt{b}\)

\(c+1\ge2\sqrt{c}\)

\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)(Vì abc = 1)

Câu 10. 

a) Ta có: \(-\left(a-b\right)^2\le0\)(điều hiển nhiên)

\(\Leftrightarrow-a^2+2ab-b^2\le0\)

\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)

\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

Có: \(2ab\le a^2+b^2;2bc\le b^2+c^2;2ac\le a^2+c^2\)(BĐT Cauchy)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3\left(a^2+b^2+c^2\right)\)

Vậy ​​\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

học tốt 

\(2n^2+2n-3=2n\left(n+1\right)-3⋮n+1\)

\(\Rightarrow3⋮n+1\)

Mà \(n\inℤ\Rightarrow n+1\inℤ\)

\(\Rightarrow n+1\in\left\{-1;1;3;-3\right\}\)

\(\Rightarrow n\in\left\{-2;0;-4;2\right\}\)

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

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

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

con chó sì ta poi vn chơi freefire

A C B H I D K

\(a.Xét\Delta ABDvà\Delta KBDcó:\)

\(BÂD\)\(=\widehat{BKD}\)\(\left(=90^O\right)\)

\(BD:cạnhchung\)

\(\widehat{ABD}=\widehat{KBD}\)

\(\Rightarrow\Delta ABD=\Delta KBD\)( cạnh huyền - góc nhọn )

\(c.Tacó:IH\perp BC;DK\perp BC\Rightarrow IH//DK\)