a) \(A=\left(x^2+x-2\right)\left(x+7\right)-16\)

\(=x^3+8x^2+5x-14-16\)

\(=x^3+8x^2+5x-30\)

\(=x^3+3x^2+5x^2+15x-10x-30\)

\(=x^2\left(x+3\right)+5x\left(x+3\right)-10\left(x+3\right)\)

\(=\left(x^2+5x-10\right)\left(x+3\right)\)

b) \(A=x^4-2x^3-3x^2+4x+4+x^2-4x+4\)

\(=x^4-2x^3-2x^2+8\)

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

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

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

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

c) \(81x^4+4=81x^4+36x^2+4-36x^2\)

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

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

d) \(\left(x^2-3\right)^2+16=x^4-6x^2+25\)

\(=\left(x^4+10x^2+25\right)-16x^2\)

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

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

sửa câu b) xíu nha!

\(A=\left(x-2\right)\left(x^3-2x-4\right)\)

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

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

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

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

\(=x^4-2x^3-2x^2+8\)

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

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

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

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

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

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

Câu a):

ta có (x2-x-2)2+(x-2)2

=((x-2)2(x+1))2+(x-2)2

=(x-2)2(x2+2x+2)

\(\left(x^2-6x\right)^2-2\left(x-3\right)^2-81=\left[\left(x^2-6x\right)^2-81\right]-2\left(x-3\right)^2=\left[\left(x^2-6x\right)^2-9^2\right]-2\left(x-3\right)^2=\left(x^2-6x+9\right)\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x-9\right)-2\left(x-3\right)^2=\left(x-3\right)^2\left(x^2-6x+11\right)\)

=\(\left(x-3\right)^2\left(x^2-6x-11\right)\)

nha

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

\(=x^4-2x^3-2x^2+8\)

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

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

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

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

b) \(=x^4-x+2019\left(x^2+x+1\right)\)

\(=x\left(x^3-1\right)+2019\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2019\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2019\right)\)\

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

1) x²-2xy+y²-x+y
(=) (x-y)²-(x-y)
(=) [(x-y)-1].(x-y)
(=) (x-y-1).(x-y)
C= (x-y)(x²+xy+y²)-x(x²-y)+y(y²-x)
(=) x³-y³-x³+xy+y³-xy

(=)(x³-x³)+(-y³+y³)+(xy-xy)
(=) 0

đúng nha

\(=\dfrac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\dfrac{1}{16}x^3+1\right)\)

\(=\dfrac{1}{4}\left(x^3-64y^3\right)+4\left(4y^3-\dfrac{1}{16}x^3+1\right)\)

\(=\dfrac{1}{4}x^3-16y^3+16y^3-\dfrac{1}{4}x^3+4=4\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)

\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)

\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

Ta có:

\(\left\{{}\begin{matrix}\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\ge\left|x+2+1-x\right|=3\\3-\left(y+2\right)^2\le3\end{matrix}\right.\)

\(\left|x+2\right|+\left|x-1\right|=3-\left(y+2\right)^2\) khi: \(\left\{{}\begin{matrix}-2\le x\le1\\y=-2\end{matrix}\right.\)