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22 tháng 10 2021

a) \(24x^2-4xy\)

\(=4x\left(6x-y\right)\)

b) \(5x^3-10x^2+5x-20xy^2\)

\(=5x\left(x^2-10x+5-20y^2\right)\)

22 tháng 10 2021

nhầm câu b nha 

\(5x^3-10x^2+5x-20xy^2\)

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

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

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

4 tháng 11 2017

\(\text{a) }\left(\dfrac{1}{2}a^2x^4+\dfrac{4}{3}\:ax^3-\dfrac{2}{3}ax^2\right):\left(-\dfrac{2}{3}\:ax^2\right)\\ =-3ax^2-2x+1\)

\(\text{b) }4\left(\dfrac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\\ =3x-4-4x+1-2x-1\\ =-3x-4\)

4 tháng 11 2017

kết quả cuối cùng là: a. -\(\dfrac{3}{4}ax^2-2x+1\)

b. \(\)-\(3x-4\)

a: \(9x^2-6x+3\)

\(=\left(9x^2-6x+1\right)+2\)

\(=\left(3x-1\right)^2+2\ge2\)

b: \(6x-x^2+1\)

\(=-\left(x^2-6x-1\right)\)

\(=-\left(x^2-6x+9-10\right)\)

\(=-\left(x-3\right)^2+10\le10\)

24 tháng 7 2017

\(x^2+4xy+2y^2-22y+173\)

\(=\left(x^2+4xy+4y^2\right)-2\left(y^2+11y+\dfrac{121}{4}\right)+\dfrac{467}{2}\)\(=\left(x+2y\right)^2-2\left(y+\dfrac{11}{2}\right)^2+\dfrac{467}{2}\ge\dfrac{467}{2}\forall x;y\)Vậy GTNN của biểu thức là; \(\dfrac{467}{2}\) khi \(\left\{{}\begin{matrix}x+2y=0\\y+\dfrac{11}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-11=0\\y=-\dfrac{11}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=-\dfrac{11}{2}\end{matrix}\right.\)Học tốt nha<3

16 tháng 9 2017

a,(5x-2y)(x2-xy+1)=5x3-5x2+5x-2yx2+2xy2-2y

=5x3-7x2y+2xy2+5x-2y

b,(x-2)(x+2)(\(\dfrac{1}{2}\) x-5)=x2-4.\(\left(\dfrac{1}{2}x-5\right)\)

=\(\dfrac{1}{2}x^3-5x^2-2x+20\)

16 tháng 9 2017

c,\(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)\)

=\(\dfrac{1}{2}x^3-5x^2-1x^2+10x+\dfrac{3}{2}x-15\)

=\(\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

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

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

=\(-5x+4x-15\)

=\(-x-15\)

Chúc bạn học tốt(mỏi tay quá)

29 tháng 10 2017

a.) \\(\\left(a+b+c\\right)^3-a^3-b^3-c^3\\)

\\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc-a^3-b^3-c^3\\)\\(=3\\left(3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc\\right)\\)

\\(=3\\left(abc+a^2b+a^2c+ac^2+b^2c+ab^2+abc+bc^2\\right)\\)

\\(=3\\left[ab\\left(a+c\\right)+ac\\left(a+c\\right)+b^2\\left(a+c\\right)+bc\\left(a+c\\right)\\right]\\)

\\(=3\\left(a+c\\right)\\left(ab+ac+bc+b^2\\right)\\)

\\(=3\\left(a+c\\right)\\left[a\\left(b+c\\right)+b\\left(b+c\\right)\\right]\\)

\\(=3\\left(a+c\\right)\\left(a+b\\right)\\left(b+c\\right)\\)

b) 4a2b2-(a2  +b2-c2)2

=(2ab+a2+b2-c2)(2ab-a2-b2+c2

=[(a+b)2-c2][c2-(a-b)2]

=(a+b+c)(a+b-c)(c+a-b)(c-a+b)

 

30 tháng 10 2017

a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc-a^3-b^3-c^3\)

\(=3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)

\(=3\left(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\right)\)

\(=3\left(ab\left(a+b\right)+b^2c+abc+bc^2+c^2a+ca^2+abc\right)\)

\(=3\left(ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\right)\)

\(=3\left(a+b\right)\left(ab+bc+c^2+ac\right)\)

\(=3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

18 tháng 7 2017

a, Theo bài ra ta có:

\(=x^3-x-2x+2\)

\(=x\left(x^2-1\right)-2\left(x-1\right)\)

\(=x\left(x+1\right)\left(x-1\right)-2\left(x-1\right)\)

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

b, theo bài ra ta có:

\(=x^3-3x^2-\left(2x^2-6x\right)-\left(3x-9\right)\)

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

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

c,Theo bài ra ta có:

\(=x^3+5x^2+3x^2+15x+2x+10\)

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

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

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

\(=\left(x+5\right)\left(x+1\right)\left(x+2\right)\)

CHÚC BẠN HỌC TỐT...........

18 tháng 7 2017

a) \(x^3-3x+2\)

= \(x^3-x^2+x^2-x-2x+2\)

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

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

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

= \(\left(x-1\right)\left[x\left(x+2\right)-\left(x+2\right)\right]\)

= \(\left(x-1\right)\left(x+2\right)\left(x-1\right)\)

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

b) \(x^3-5x^2+3x+9\)

= \(x^3+x^2-6x^2-6x+9x+9\)

= \(x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2-6x+9\right)\)

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

c) \(x^3+8x^2+17x+10\)

= \(x^3+x^2+7x^2+7x+10x+10\)

= \(x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2+7x+10\right)\)

= \(\left(x+1\right)\left(x^2+2x+5x+10\right)\)

= \(\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]\)

= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

d) \(x^3-3x^2+6x+4\)

Câu này đúng là sai đề rồi, mình sửa + làm bên dưới:

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

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

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

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

Học tốt nhé :))

31 tháng 7 2016

Ta có x=9 => 10=x+1

Thay vào ta có:

\(Q\left(x\right)=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}...-\left(x+1\right)x+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...-x^2-x+x+1=1\)

31 tháng 7 2016

Toán lớp 8Hi!!!

Bài 2:

a: \(A=-3\left(x^2-\dfrac{4}{3}x+\dfrac{1}{3}\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{9}\right)\)

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

Dấu '=' xảy ra khi x=2/3

b: \(B=-x^2+5x+3\)

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

\(=-\left(x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{37}{4}\right)\)

\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)

Dấu '=' xảy ra khi x=5/2

26 tháng 9 2017

a) \(x^3-\dfrac{1}{9}x=0\)

\(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)

\(\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\\x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(x\left(x-3\right)+x-3=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)

c) \(2x-2y-x^2+2xy-y^2=0\) (thêm đề)

\(\Rightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)

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

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

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

d) \(x^2\left(x-3\right)+27-9x=0\)

\(\Rightarrow x^2\left(x-3\right)+\left(x-3\right).9=0\)

\(\Rightarrow\left(x-3\right)\left(x^2+9\right)=0\)

\(\Rightarrow x-3=0\Rightarrow x=3.\)

4 tháng 10 2017

\(\dfrac{2}{5}\)

29 tháng 8 2017

Tự làm đê em ơi cô Viết cho xong lên mạng chứ j

30 tháng 8 2017

thg kia m nói ai là em hả