a: ĐKXĐ: \(x^4+9x^3-9x^2+9x-10<>0\)

=>\(x^4+10x^3-x^3-10x^2+x^2+10x-x-10<>0\)

=>\(\left(x+10\right)\left(x^3-x^2+x-1\right)<>0\)

=>\(\left(x+10\right)\left(x-1\right)\left(x^2+1\right)<>0\)

=>(x+10)(x-1)<>0

=>x∉{-10;1}

b: \(B=\frac{\left|x+10\right|}{x^4+9x^3-9x^2+9x-10}\)

\(=\frac{\left|x+10\right|}{\left(x+10\right)\left(x-1\right)\left(x^2+1\right)}\)

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

Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)

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

\(=\dfrac{\left(x+2\right)\left(x+1\right)+6x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{-8x^2+2}{3x}\cdot\dfrac{1}{-4x+2}+\dfrac{x^2-3x-1}{3x}\)

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

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

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

c: Khi x=1 thì \(D=\dfrac{1-1}{3}=0\)

Bạn thay x vào biểu thức rồi tính thôi

a)(x-10)²-x(x+80)

(x²-2x10+100)-x²-80x

=x²-20x+100-x_²-80x=-100x+100 
khi x = 0.98 
ta có 
(-100*0.98)+100=-98+100=2
b)x³-9x+27x-27
 hình như là -27x :))

Bài 1:

a: \(2x^2-8x=0\)

=>\(x^2-4x=0\)

=>x(x-4)=0

=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b: \(\left(x+2\right)^2-x\left(x-1\right)=10\)

=>\(x^2+4x+4-x^2+x=10\)

=>5x+4=10

=>5x=6

=>\(x=\dfrac{6}{5}\)

c: \(x^3-6x^2+9x=0\)

=>\(x\left(x^2-6x+9\right)=0\)

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

=>\(\left[{}\begin{matrix}x=0\\\left(x-3\right)^2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Câu 1:

a: \(2x^2\left(x^2+3x+\frac12\right)\)

\(=2x^2\cdot x^2+2x^2\cdot3x+2x^2\cdot\frac12\)

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

b: \(\left(x+1\right)\left(x-2\right)-\left(x+2\right)^2\)

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

\(=x^2-x-2-x^2-4x-4=-5x-6\)

c: \(\left(3x+1\right)^2-9x\left(x+3\right)\)

\(=9x^2+6x+1-9x^2-27x\)

=-21x+1

Câu 2:

a: \(\left(x+2\right)^2-x\left(x+4\right)+10\)

\(=x^2+4x+4-x^2-4x+10\)

=4+10

=14

=>Biểu thức này không phụ thuộc vào biến

b: \(\left(x+3\right)\left(4x-1\right)-\left(2x+1\right)^2-7x+3\)

\(=4x^2-x+12x-3-4x^2-4x-1-7x+3\)

=12x-x-7x-4x-3-1+3

=x(12-1-7-4)-1

=-1

=>Biểu thức này không phụ thuộc vào biến

Câu 3:

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

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

=>5x+4=2

=>5x=2-4=-2

=>\(x=-\frac25\)

b: \(\left(2x+1\right)^2-\left(x+1\right)\left(4x-3\right)=-3\)

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

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

=>\(4x^2+4x+1-4x^2-x+3=-3\)

=>3x+4=-3

=>3x=-7

=>\(x=-\frac73\)

Câu 5

Xét ΔADC và ΔBCD có

AD=BC

DC chung

AC=BD

Do đó: ΔADC=ΔBCD

=>\(\hat{ACD}=\hat{BDC}\)

=>\(\hat{OCD}=\hat{ODC}\)

=>ΔOCD cân tại O

=>OC=OD

Ta có: OC+OA=AC

OD+OB=BD

mà OC=OD và AC=BD

nên OA=OB

x=8 nên x+1=9

\(F=x^{13}-9x^{12}+9x^{11}-9x^{10}+...-9x^2+9x-2\)

\(=x^{13}-x^{12}\left(x+1\right)+x^{11}\left(x+1\right)-x^{10}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)-2\)

\(=x^{13}-x^{13}-x^{12}+x^{12}+...-x^3-x^2+x^2+x-2\)

=x-2

=8-2

=6

các bạn đi học chưa hả?

Chưa bạn

\(\dfrac{x^2-2x-8}{2x^2+9x+10}\)

\(=\dfrac{x^2-4x+2x-8}{2x^2+4x+5x+10}\)

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

\(=\dfrac{x-4}{2x+5}\)