Bài 6: Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

chtt: Câu hỏi của Hồ Quế Ngân - Toán lớp 8 | Học trực tuyến

\(a^6+a^4+a^2b^2+b^4-b^6\)

\(=(a^2)^3-(b^2)^3+(a^4+a^2b^2+b^4)\)

\(=(a^2-b^2)(a^4+a^2b^2+b^4)+(a^4+a^2b^2+b^4)\)

\(=(a^2-b^2+1)(a^4+a^2b^2+b^4)\)

\(=(a^4+2a^2b^2+b^4-a^2b^2)(a^2-b^2+1)\)

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

\(a^6+a^2b^2+a^4+b^2-b^6\)

\(=a^4\left(a^2+b^2\right)+a^2\left(a^2+b^2\right)-b^6\)

\(=\left(a^2+b^2\right)+\left(a^4+a^2\right)-b^6\)

\(x\left(x+y\right)-5x-5y\)

\(=x\left(x+y\right)-\left(5x+5y\right)\)

\(=x\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x-5\right)\left(x+y\right)\)

\(x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\)

a)\(3xy+5xy^2-xy^2\)

\(=3xy+4xy^2=xy\left(4y+3\right)\)

b)\(x\left(y-x\right)+3\left(y-x\right)\)

\(=\left(x+3\right)\left(y-x\right)\)

Ta có : $x^3+x=0$

$=>x(x^2+1)=0$

\(=>\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=0\\x^2=-1\left(vo-ly\right)\end{matrix}\right.\)

$=>x=0$

\(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x\in\varnothing\end{matrix}\right.\)

Vậy \(x=0\)

0

a, \(xy+2x+2y+y^2\)

\(=\left(xy+2x\right)+\left(y^2+2y\right)\)

\(=x.\left(y+2\right)+y.\left(y+2\right)\)

\(=\left(y+2\right).\left(x+y\right)\)

b, Sửa đề:

\(-6x^2-9xy+15y^2\)

\(=-\left(6x^2+9xy-15y^2\right)\)

\(=-\left(6x^2-6xy+15xy-15y^2\right)\)

\(=-\left[\left(6x^2-6xy\right)+\left(15xy-15y^2\right)\right]\)

\(=-\left[6x.\left(x-y\right)+15y.\left(x-y\right)\right]\)

\(=-\left[\left(x-y\right).\left(6x+15y\right)\right]\)

c, \(2x\left(x-3\right)+y\left(x-3\right)+\left(3-x\right)\)

\(=2x\left(x-3\right)+y\left(x-3\right)-\left(x-3\right)\)

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

Chúc bạn học tốt!!!

a, \(\left(x+1\right)^2=3\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)^2-3\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right).\left(x+1-3\right)=0\)

\(\Rightarrow\left(x+1\right).\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;2\right\}\)

b, \(\left(2x-7\right)^3=8\left(7-2x\right)^2\)

\(\Rightarrow\left(2x-7\right)^3-8\left(2x-7\right)^2=0\) (do \(\left[A\left(x\right)\right]^2=\left[-A\left(x\right)\right]^2\))

\(\Rightarrow\left(2x-7\right)^2.\left(2x-7-8\right)=0\)

\(\Rightarrow\left(2x-7\right)^3.\left(2x-15\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-7\right)^3=0\\2x-15=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-7=0\\2x=15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x=7\\x=7,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3,5\\x=7,5\end{matrix}\right.\)

Vậy \(x\in\left\{3,5;7,5\right\}\)

Chúc bạn học tốt!!!

Ta có:

\(55^{n+1}-55^n=55^n.\left(55-1\right)=55^n.54\)

Vì \(54⋮54\) nên \(55^n.54⋮54\)

\(\Rightarrow55^{n+1}-55^n\) chia hết cho 54 (đpcm)

Chúc bạn học tốt!!!

\(55^{n-1}-55^n\) \(=55^n.55-55^n\)

\(=55^n\left(55-1\right)\)

\(=55^n.54\)

Vì 54 : 54 nên \(55^n.54:54\)

=> \(55^{n+1}-55^n\) chia hết cho 54 (đccm)

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