a) (x+9)(x-9)-x²=x²-81-x²=-81

b) (10x-1)(10x+1)-(10x-1)²=100x²-1-100x²+20x-1=20x-2

d) (x-1)(x-2)-(x-2)(x+2)=x²-3x+2-x²+4=-3x+6

\(\left(3-x\right)\left(3+x\right)+\left(x-5\right)^2\\ =9-x^2+x^2-10x+25\\ =34-10x\)

e) \(E=x^5-15x^4+16x^3-29x^2+13x\) tại x = 14

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

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

\(E=-x\)

\(E=-14\)

d)  \(D=x^3-30x^2-31+1\) tại x = 31

\(D=31^3-30.31^2-31+1\)

\(D=31^2\left(31-30-1\right)+1\)

\(D=0+1\)

\(D=1\)

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`

2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`

\(G=x^4+10x^3+10x^2+10x+10\)
\(=x^4+10\left(x^3+x^2+x+1\right)\)
\(=\left(-9^4\right)+10\left[\left(-9\right)^3+\left(-9\right)^2+-9+1\right]\)
\(=6561+10\cdot-656\)
\(=6561-6560\)
\(=1\)

Thay `x=-9` vào biểu thức G:

`G=(-9)^4+10.(-9)^3+10.(-9)^2+10.(-9)+10`

`=6561-7290+810-90+10`

`=1`

Do \(x=-9\Rightarrow x+9=0\)

Ta có:

\(G=\left(x^4+9x^3\right)+\left(x^3+9x^2\right)+\left(x^2+9x\right)+\left(x+9\right)+1\)

\(=x^3\left(x+9\right)+x^2\left(x+9\right)+x\left(x+9\right)+\left(x+9\right)+1\)

\(=x^3.0+x^2.0+x.0+0+1=1\)

1 ) Nếu \(x=9\Rightarrow10=x+1\)

Thay \(10=x+1\) vào B , ta được :

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

\(\Leftrightarrow B=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)

\(\Leftrightarrow B=1\)

2 ) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)

\(=x^3+ax^2+bx^2+abx+x^2c+axc+bxc+abc\)

\(=x^3+\left(ax^2+bx^2+cx^2\right)+\left(abx+axc+bcx\right)+abc\)

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

\(\left(đpcm\right)\)

:D

a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)

mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)

nên \(x^2-10x-2000=0\)

\(\Leftrightarrow x^2+40x-50x-2000=0\)

\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)

\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)

Vậy: S={-40;50}

a: \(=\dfrac{5}{2x^2y}+\dfrac{2}{3xy}-\dfrac{y}{x^3}\)

\(=\dfrac{5\cdot3\cdot x}{6x^3y}+\dfrac{2\cdot2\cdot x^2}{6x^3y}-\dfrac{6y^2}{6x^3y}\)

\(=\dfrac{15x+4x^2-6y^2}{6x^3y}\)

b: \(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)

c: \(=\dfrac{x^4-1-x^4+3x^2}{x^2-1}=\dfrac{3x^2-1}{x^2-1}\)