\(\left(x+1\right)\left(x-5\right)=\)\(x^2-4x-5\)\(\)

\(=x^2-4x+4-9=\left(x-2\right)^2-9\)

Mà \(\left(x-2\right)^2\ge0\) với mọi x \(\Rightarrow\left(x-2\right)^2-9\ge-9\) với mọi x

Vậy GTNN của  \(\left(x+1\right)\left(x-5\right)=-9\)đạt được khi \(x=2\)

1. Thực hiện phép chia đa thức: ta có kết quả:

\(x^3+5x^2+3x+a=\left(x+3\right)\left(x^2+2x+b\right)+\left(-3-b\right)x+a-3b\)

Để f(x) chia hết cho x²+2x+b thì -3-b=0 và a-3b=0 <=> b=-3; a=-9

      \(3a^2+4b^2=7ab\)

\(\Rightarrow3a^2+4b^2-7ab=0\)

\(\Rightarrow3a^2-3ab-4ab+4b^2=0\)

\(\Rightarrow3a\left(a-b\right)-4b\left(a-b\right)=0\)

\(\Rightarrow\left(a-b\right)\left(3a-4b\right)=0\)

Mà \(a\ne b\Rightarrow a-b\ne0\) 

Từ đó \(3a-4b=0\Rightarrow3a=4b\Rightarrow a=\frac{4}{3}b\)

\(E=\frac{a+2b}{3a-b}=\frac{\frac{4}{3}b+2b}{3.\frac{4}{3}b-b}=\frac{10}{9}\)

\(A=4x^2+10y^2-4xy-32y+4x+27\)       

\(=\left(4x^2-4xy+y^2\right)+4x-2y+1+9y^2-30y+25+1\)

\(=\left(2x-y\right)^2+2\left(2x-y\right)+1+\left(3y\right)^2-2.3y.5+5^2+1\)

\(=\left(2x-y+1\right)^2+\left(3y-5\right)^2+1>0\forall x;y\)

Pham Van Hung

A=4x^2+10y^2-4xy-32y+4x+27A=4x2+10y2−4xy−32y+4x+27       

=\left(4x^2-4xy+y^2\right)+4x-2y+1+9y^2-30y+25+1=(4x2−4xy+y2)+4x−2y+1+9y2−30y+25+1

=\left(2x-y\right)^2+2\left(2x-y\right)+1+\left(3y\right)^2-2.3y.5+5^2+1=(2x−y)2+2(2x−y)+1+(3y)2−2.3y.5+52+1

=\left(2x-y+1\right)^2+\left(3y-5\right)^2+1&gt;0\forall x;y=(2x−y+1)2+(3y−5)2+1>0∀x;y

     \(a^3+b^3=3ab-1\)

\(\Rightarrow a^3+b^3+1-3ab=0\)

\(\Rightarrow\left(a+b\right)^3+1-3ab\left(a+b\right)-3ab=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b\right)=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2-ab+b^2-a-b+1\right)=0\)

Mà \(a,b>0\Rightarrow a+b+1>0\)

\(\Rightarrow a^2-ab+b^2-a-b+1=0\)

\(\Rightarrow2a^2-2ab+2b^2-2a-2b+2=0\)

\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)

\(\Rightarrow a=b=1\Rightarrow a^{2018}+b^{2019}=1+1=2\)

\(\frac{x+9}{x^2-9}-\frac{3}{x^2-3x}=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\left(x-3\right)}\)

                                  \(=\frac{x^2+9x-3\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}\)

                                     \(=\frac{x^2+9x-3x-9}{x\left(x-3\right)\left(x+3\right)}\)

                                      \(=\frac{x^2+6x-9}{x\left(x-3\right)\left(x+3\right)}\)

\(\frac{x+9}{x^2-9}-\frac{3}{x^2-3x}\)

\(=\frac{x+9}{\left(x+3\right)\left(x-3\right)}-\frac{3}{x\left(x-3\right)}\)

\(=\frac{\left(x+9\right)x}{x\left(x+3\right)\left(x-3\right)}-\frac{3\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}\)

\(=\frac{x^2+9x-3x-9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x-9}{x\left(x+3\right)\left(x-3\right)}\)

\(=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)

hok tốt ...

đề là j

Đề: C/M A không phải là số nguyên ak?

Bổ sung thêm điều kiện: \(a,b,c>0\)

Để C/m A không phải là số nguyên ta sẽ C/m 1<A<2 ( phần này ko cần phải ghi)

 \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)

Ta có:

 \(\frac{a}{a+b}>\frac{a}{a+b+c}\)

\(\frac{b}{b+c}>\frac{b}{a+c+b}\)

\(\frac{c}{c+a}>\frac{c}{a+c+b}\)

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

Ta chứng minh bất đẳng thức phụ:  

Nếu \(a,b,c>0\); \(a< b\)thì \(\frac{a}{b}< \frac{a+c}{b+c}\)

Ta có: \(a< b\)

\(\Rightarrow ac< bc\)

\(\Rightarrow ac+ab< bc+ab\)

\(a\left(b+c\right)< b\left(a+c\right)\)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)

             đpcm

Áp dụng:

 \(\frac{a}{a+b}< \frac{a+c}{a+b+c}\)

\(\frac{b}{b+c}< \frac{b+a}{a+c+b}\)

\(\frac{c}{c+a}< \frac{c+b}{a+c+b}\)

\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{b+c+a}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=\frac{2.\left(a+b+c\right)}{a+b+c}=2\)(2)

Từ (1) và (2)

\(\Rightarrow1< A< 2\)

\(\Rightarrow\)A không phải là số nguyên

                       đpcm

Tham khảo nhé~