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                                        GIẢI:

Q=3+3²+3³+...+3²⁰²⁴

Q=3+3²+(3³+3⁴+3⁵)+(3⁶+3⁷+3⁸)+...+(3²⁰²²+3²⁰²³+3²⁰²⁴)

Q=12+3³(1+3+3²)+3⁶(1+3+3²)+...+3²⁰²²(1+3+3²)

Q=12+3³.13+3⁶.13+...+3²⁰²².13

Q=12+13(3³+3⁶+...+3²⁰²²)

mà [13(3³+3⁶+...+3²⁰²²)] chia hết cho 13

do đó Q:13 dư 12

vậy số dư khi cha Q cho 13 là 12

Bài 1 : \(A=1+3+3^2+...+3^{31}\)

a. \(A=\left(1+3+3^2\right)+...+3^9.\left(1.3.3^2\right)\)

\(\Rightarrow A=13+3^9.13\)

\(\Rightarrow A=13.\left(1+...+3^9\right)\)

\(\Rightarrow A⋮13\)

b. \(A=\left(1+3+3^2+3^3\right)+...+3^8.\left(1+3+3^2+3^3\right)\)

\(\Rightarrow A=40+...+3^8.40\)

\(\Rightarrow A=40.\left(1+...+3^8\right)\)

\(\Rightarrow A⋮40\)

Bài 2:

Ta có: \(C=3+3^2+3^4+...+3^{100}\)

\(\Rightarrow C=(3+3^2+3^3+3^4)+...+(3^{97}+3^{98}+3^{99}+3^{100})\)

\(\Rightarrow3.(1+3+3^2+3^3)+...+3^{97}.(1+3+3^2+3^3)\)

\(\Rightarrow3.40+...+3^{97}.40\)

Vì tất cả các số hạng của biểu thức C đều chia hết cho 40

\(\Rightarrow C⋮40\)

Vậy \(C⋮40\)

1) \(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)

\(=\left(x+3\right).x^2-5\left(x+3\right)+\left(x+4\right)\left(x-1x^2\right)\)

\(=x^3+3x^2-5x-15+\left(x+4\right)\left(x-x^2\right)\)

\(=x^3+3x^2-5x-15-x^3+x^2-4x^2+4x\)

\(=3x^2-5x-15-3x^2+4x\)

\(=-x-15\)

2) Đặt đa thức là \(N\left(x\right)\)ta được: \(3x^3+2x^2-x+k=N\left(x\right)\left(x-1\right)\)

Để \(3x^3+2x^2-x+K⋮x-1\Leftrightarrow x=1\)

Thay vào ta được

\(\Rightarrow3.1^3+2.1^2-1+K=0\)

\(\Rightarrow3+2-1+K=0\)

\(\Rightarrow K=-4\)

3) \(Q=x^2-x+3\)

\(=x^2-x+\frac{1}{4}+\frac{11}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\)

Với \(\forall x\) ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\Rightarrow N=\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\)

chỗ 32015 là 3²⁰¹⁵ nha

Bài này làm từng câu thôi :

 \(A=1+3^1+3^2+.......+3^{2014}+3^{2015}\)

\(\Rightarrow3A=3+3^2+3^3+......+3^{2015}+3^{2016}\)

\(\Rightarrow3A-A=\left(3+3^2+......+3^{2016}\right)-\left(1+3^1+.....+3^{2015}\right)\)

\(\Rightarrow2A=3^{2016}-1\)

\(\Rightarrow A=\frac{3^{2016}-1}{2}\)

a) \(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(P=\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1^3}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)

\(P=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)\(P=\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}\)

\(P=\dfrac{2}{x+\sqrt{x}+1}\)

b) Mà với \(x\ge0\) và \(x\ne1\) thì 

\(x+\sqrt{x}+1\ge0\) và \(2>0\) nên \(P>0\)

a: \(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2}=\dfrac{2}{x+\sqrt{x}+1}\)

b: x+căn x+1+1>=1>0

2>0

=>P>0 với mọi x thỏa mãn x>=0 và x<>1