Chứng minh rằng \(A=\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)\)chia hết cho 3 với mọi n
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^3+2x^2-2x-12=\left(x^3-2x^2\right)+\left(4x^2-8x\right)+\left(6x-12\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+6\left(x-2\right)=\left(x-2\right)\left(x^2+4x+6\right)\)
Vậy B = x2+4x+6
tôi chưa học về cái này nên chua trả lời được
xin lỗi nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Chả biết đúng hay sai. Làm bừa! =(((
Dự đoán dấu "=" xảy ra khi a = 3
Ta có: \(VT=\left(\frac{1}{a}+\frac{a}{9}\right)+\frac{8a}{9}\ge2\sqrt{\frac{1a}{9a}}+\frac{8a}{9}\) (BĐT AM-GM)
\(=2\sqrt{\frac{1}{9}}+\frac{8a}{9}=\frac{2}{3}+\frac{8a}{9}\ge\frac{2}{3}+\frac{8}{3}=\frac{10}{3}^{\left(đpcm\right)}\) (do \(a\ge3\))
Dấu "=" xảy ra \(\Leftrightarrow a=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2x^2-2y^2=2\left(x^2-y^2\right)=2\left(x-y\right)\left(x+y\right)\)
b) \(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\)
c) \(x^2-4x=x\left(x-4\right)\)
d) \(x^2+10x+25=x^2+2.5x+5^2=\left(x+5\right)^2\)
e) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y+3\right)\left(x-y-3\right)\)
\(2x^2-2y^2=2.\left(x^2-y^2\right)=2.\left(x-y\right)\left(x+y\right)\)
\(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(x^2-4x=x.\left(x-4\right)\)
\(x^2+10x+25=x^2+2.x.5+5^2=\left(x+5\right)^2\)
\(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
Tham khảo nhé~
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+1\right)^2-\left(3-x\right)^2\)
\(=4x^2+4x+1-9+6x-x^2\)
\(=3x^2+10x+\left(1-9\right)=3x^2+10x-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=4b^2c^2-\left(b^2+c^2-a^2\right)^2=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)
\(=\left[a^2-\left(b^2-2bc+c^2\right)\right]\left[\left(b^2+2bc+c^2\right)-a^2\right]\)
\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)
Vì a,b,c là 3 cạnh của 1 t/g nên
a+c>b => a-b+c >0
a+b>c => a+b-c > 0
b+c>a => b+c-a > 0
b+c+a > 0
=> M > 0
\(A=\left(n-1\right)\left(3-2n\right)-n\left(n+5\right)\)
\(=3n-2n^2-3+2n-\left(n^2+5n\right)\)
\(=3n-2n^2-3+2n-n^2-5n\)
\(=\left(3n-5n+2n\right)-\left(2n^2-n^2\right)-3\)
\(=-3\)
\(\Rightarrowđpcm\)
em ms hok lớp 1