1. a)

\(h\left(0\right)=1+0+0+....+0=1\)

\(h\left(1\right)=1+\left(1+1+....+1\right)\)

( x thừa số 1)

\(=x+1\)

Với x là số chẵn

\(h\left(-1\right)=1+\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+...+\left(-1\right)^{x-1}+\left(-1\right)^x=1-1+1-1+...-1+1-1=-1\)

Với x là số lẻ

\(h\left(-1\right)=1-1+1-1+1-....+1-1\) =0

b) Tương tự

143. a) \(-6x^n.y^n.\left(-\dfrac{1}{18}x^{2-n}+\dfrac{1}{72}y^{5-n}\right)\)

\(=-6.\left(-\dfrac{1}{18}\right)x^n.x^{2-n}.y^n+\left(-6\right).\dfrac{1}{27}x^n.y^n.y^{5-n}\)

\(=\dfrac{1}{3}x^{n+2-n}y^n-\dfrac{2}{9}x^n.y^{n+5-n}\)

\(=\dfrac{1}{3}x^2y^n-\dfrac{2}{9}x^ny^5\)

b) Ta có: \(\left(5x^2-2y^2-2xy\right)\left(-xy-x^2+7y^2\right)\)

\(=5x^2\left(-xy\right)+5x^2.\left(-x^2\right)+5x^2.7y^2-2y^2.\left(-xy\right)-2y^2.\left(-x^2\right)-2y^2.7y^2-2xy.\left(-xy\right)-2xy\left(-x^2\right)-2xy.7y^2\)

\(=-5x^3y-5x^4+35x^2y^2+2xy^3+2x^2y^2-14y^4+2x^2y^2+2x^3y-14xy^3\)

Rút gọn các đa thức đồng dạng, ta có kết quả:

\(-5x^4-3x^3y+39x^2y^2-12xy^3-14y^4\)

Kết quả đã được xếp theo lũy thừa giảm dần của x

<=>5xⁿ.x²-3xⁿ+2xⁿ.x²-4xⁿ+xⁿ.x²-xⁿ=0

<=>(5xⁿ.x²+2xⁿ.x²)-(3xⁿ+4xⁿ)+(xⁿ.x²-xⁿ)=0

<=>7xⁿ.x²-7xⁿ+xⁿ(x²-1)=0

<=>7xⁿ(x²-1)+xⁿ(x²-1)=0

<=>8xⁿ(x²-1)=0<=>x²-1=0<=>x=-1 hoặc x=1

\(A=\left\{x\in R|\left(x-2x^2\right)\left(x^2-3x+2\right)=0\right\}\)

Giải phương trình sau :

 \(\left(x-2x^2\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow x\left(1-2x\right)\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1-2x=0\\x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\\x=2\end{matrix}\right.\)

\(\Rightarrow A=\left\{0;\dfrac{1}{2};1;2\right\}\)

\(B=\left\{n\in N|3< n\left(n+1\right)< 31\right\}\)

Giải bất phương trình sau :

\(3< n\left(n+1\right)< 31\)

\(\Leftrightarrow\left\{{}\begin{matrix}n\left(n+1\right)>3\\n\left(n+1\right)< 31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n^2+n-3>0\\n^2+n-31< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n< \dfrac{-1-\sqrt[]{13}}{2}\cup n>\dfrac{-1+\sqrt[]{13}}{2}\\\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1-5\sqrt[]{5}}{2}< n< \dfrac{-1-\sqrt[]{13}}{2}\\\dfrac{-1+\sqrt[]{13}}{2}< n< \dfrac{-1+5\sqrt[]{5}}{2}\end{matrix}\right.\)

Vậy \(B=\left(\dfrac{-1-5\sqrt[]{5}}{2};\dfrac{-1-\sqrt[]{13}}{2}\right)\cup\left(\dfrac{-1+\sqrt[]{13}}{2};\dfrac{-1+5\sqrt[]{5}}{2}\right)\)

\(\Rightarrow A\cap B=\left\{2\right\}\)

\(5x^{n+2}-3x^n+2x^{n+2}-4x^n+x^{n+2}-x^n=0\)

\(\Rightarrow5x^n.x^2-3x^n+2x^n.x^2-4x^n+x^n.x^2-x^n=0\)

\(\Rightarrow x^2\left(5x^n+2x^n+x^n\right)-\left(3x^n+4x^n+x^n\right)=0\)

\(\Rightarrow x^2.8x^n-8x^n=0\)

\(\Rightarrow\left(x^2-1\right)8x^n=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-1=0\\8x^n=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=1\\x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1;0\right\}\)

Tìm x

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\(M\left(x\right)=4x^3+x^2-7x+3x^2-x^3+9\)

=\(\left(4x^3-x^3\right)+\left(x^2+3x^2\right)-7x+9\)

=\(3x^3+4x^2-7x+9\)

\(N\left(x\right)=6+5x^3+6x^2+3x-2x^2-2x^3\)

=\(\left(5x^3-2x^3\right)+\left(6x^2-2x^2\right)+3x+6\)

=\(3x^3+4x^2+3x+6\)

\(M\left(x\right)=3.x^3+4x^2-7x+9\)

\(N\left(x\right)=3.x^3+4.x^2+3x+6\)

M(x)=4x³+x²-7x-3x²+x^3-9=5x³-2x²-7x-9

N(x)=6+5x³+6x²+3x-2x²-2x³⁼3x³+4x²+3x+6