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

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

$\Rightarrow [\begin{array}{c}x+1=0\\ 7x+3=0\end{array}\Rightarrow [\begin{array}{c}x=-1\\ x=-\frac{3}{7}\end{array}$

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 => $[\begin{array}{c}x+8=0\\ 3-x=0\end{array}\Rightarrow [\begin{array}{c}x=-8\\ x=3\end{array}$

cứu mị :<

\(1,\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\left(B\right)\\ 2,Giống.1\\ 3,=5x^2\left(B\right)\\ 4,x^3=x\Leftrightarrow x^3-x=0\\ \Leftrightarrow x\left(x^2-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\\ \Leftrightarrow A\)

\(a,80-\left(10x-5\right)=45\\ \Rightarrow80-10x+5=45\\ \Rightarrow-10x=-40\\ \Rightarrow x=4\)

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

\(c,\left|5+5x\right|=2^2.5\\ \Rightarrow\left|5+5x\right|=20\\ \Rightarrow\left[{}\begin{matrix}5+5x=20\\5+5x=-20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}5x=15\\5x=-25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

\(d,-10-\left(-5\right)+\left(3-x\right)=-8\\ \Rightarrow-10+5+3-x=-8\\ \Rightarrow-x=-6\\ \Rightarrow x=6\)

a) 80-(10x-5)=45

=> 80 - 10 x + 5 =45

=>-10x = -40

=>x=4

b)(x+1)×(x-2)=0

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

c) |5+5x|=2^2×5

=>|5+5x|=20

\(\Rightarrow\left[{}\begin{matrix}5+5x=-20\\5+5x=20\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

d) -10-(-5)+(3-x)=-8

=>-10+5 +3-x=-8

=>2+x=8

=>x=6

\(a,=2x^2y+4xy^2\\ b,=2x^2-x+2x-1=2x^2+x-1\\ c,\dfrac{5}{3}x^2\cdot\dfrac{1}{y^2}\\ d,=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x^2+2x+4\right)=x-2\)

6B, 7A, 8D

x=9

=>x+1=10

\(A=x^{10}-10x^9+10x^8-...+10x^2-10x+1\)

\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+1\)

\(=x^{10}-x^{10}-x^9+x^8+...+x^3+x^2-x^2-x+1\)

=-x+1

=-9+1=-8