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

b)\(x^2-8x+12\\ =x^2-4x-4x+12\\ =\left(x^2-4x\right)+\left(-4x+12\right)\\ =x\left(x-4\right)+4\left(x+3\right)\\ =\left(x-4\right)\left(x+4\right)\left(x+3\right)\)

\(#20/10\)

a)\(\left\{{}\begin{matrix}x-2\ne0\Leftrightarrow x\ne2\\x+2\ne0\Leftrightarrow x\ne-2\end{matrix}\right.\)

b)\(A=\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\\ =\dfrac{x^2}{x^2-2^2}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\\ =\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2-\left(x^2+2x\right)-\left(2x-4\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-4x+4}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{4.\left(-x+1\right)}{\left(x-2\right)\left(x+2\right)}\)

c)\(A=\dfrac{4.\left(-x+1\right)}{\left(x-2\right)\left(x+2\right)}=2\\ \Leftrightarrow-4x+4=2x^2-8\\ \Leftrightarrow4x+4-2x^2+8=0\\ \Leftrightarrow-4x+12-2x^2=0\\ \Leftrightarrow-2x+6-x^2=0\\ \Leftrightarrow\left(-2x-x^2\right)+6=0\\ \Leftrightarrow-\left(x^2+2x\right)+6=0\\ \Leftrightarrow-\left(x^2+2x+1-1\right)+6=0\\ \Leftrightarrow-\left(x+1\right)^2+1+6=0\\ \Leftrightarrow-\left(x+1\right)^2+7=0\\ \Leftrightarrow\left(\sqrt{7}\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left[\sqrt{7}-\left(x+1\right)\right]\left[\sqrt{7}+\left(x+1\right)\right]=0\\ \Leftrightarrow\left(\sqrt{7}-x-1\right)\left(\sqrt{7}+x+1\right)=0\)

+)\(\sqrt{7}-x-1=0\\ x=\sqrt{7}-1\)

+)\(\sqrt{7}+x+1=0\\ x=-\sqrt{7}-1\)

Vậy A = 2 khi x =\(\sqrt{7}-1\) hoặc x = \(-\sqrt{7}-1\)

\(#20/10\)

a)Thể tích không khí trong lều là:

\(V=\dfrac{1}{3}. S_{đáy}.h=\dfrac{1}{3}.\left(3.3\right).2,8=8,4\left(m^3\right) \)

b) Diện tích đáy:

3.3 = 9 (m²)

Độ dài cạnh bên của lều:

\(\sqrt{\text{(2,8² + 1,5²) }}\)≈ 3,18 (m)

Diện tích vải lều:

\(\dfrac{1}{2}\) .4 . 3 . 3,18 + 9 = 28,08 (m²)

Số tiền mua vải:

28,08 . 15000 - 28,08 . 15000 . 5% = 400140 (đồng)

a)Xét tứ giác ABCD có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

hay\(102^o+102^o+102^o+\widehat{D}=360^o\\ =>\widehat{D}=360^o-\left(102^o.3\right)\\ =360^o-306^o\\ =54^o\)

\(=>\widehat{D}=54^o\)

b)Áp dụng định lí Pythagore vào \(\Delta AOD\),có:

\(AD^2=OD^2+AO^2\\ =>AO^2=AD^2-OD^2\)

hay\(AO^2=30^2-26,7^2\)

\(=>AO^2=187,11cm\)

\(=>AO=13,7cm\)

Áp dụng định lí Pythagore vào \(\Delta BAO\),có:

\(AB^2=AO^2+BO^2\\ =>BO^2=AB^2-AO^2\)

hay \(BO^2=17,5^2-13,7^2\)

\(=>BO^2=118,56cm\\ =>BO=10,9cm\)

\(->BD=26,7+10,9=37,6cm\)

a)Xét tứ giác ABCD có:

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

hay\(102^o+102^o+102^o+\widehat{D}=360^o\\ =>\widehat{D}=360^o-\left(102^o.3\right)\\ =360^o-306^o\\ =54^o\)

\(=>\widehat{D}=54^o\)

b)Áp dụng định lí Pythagore vào \(\Delta AOD\),có:

\(AD^2=OD^2+AO^2\\ =>AO^2=AD^2-OD^2\)

hay\(AO^2=30^2-26,7^2\)

\(=>AO^2=187,11cm\)

\(=>AO=13,7cm\)

Áp dụng định lí Pythagore vào \(\Delta BAO\),có:

\(AB^2=AO^2+BO^2\\ =>BO^2=AB^2-AO^2\)

hay \(BO^2=17,5^2-13,7^2\)

\(=>BO^2=118,56cm\\ =>BO=10,9cm\)

\(->BD=26,7+10,9=37,6cm\)

\(a.xy+y^2-x-y\\ =\left(xy-x\right)+\left(y^2-y\right)\\ =x\left(y-1\right)+y\left(y-1\right)\\ =\left(y-1\right)\left(x+y\right)\)

\(b.\left(x^2y^2-8\right)^2-1\\ =\left[\left(x^2y^2-8\right)-1\right]\left[\left(x^2y^2-8\right)+1\right]\\ =\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\\ =\left(x^2y^2-9\right)\left(x^2y^2-7\right)\\ =\left(xy-3\right)\left(xy+3\right)\left(xy-\sqrt{7}\right)\left(xy+\sqrt{7}\right)\)

\(a.\left(-12x^{13}y^{15}+6x^{10}y^{14}\right):\left(-3x^{10}y^{14}\right)\\ =4x^3y-2\)

\(b.\left(x-y\right)\left(x^2-2x+y\right)-x^3+x^2y\\ =x^3-2x^2+xy-x^2y+2xy-y^2-x^3+x^2y\\ =x^3-2x^2-x^2y+3xy-y^2-x^3+x^2y\\ =-2x^2+3xy-y^2\)

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