\(A=2^0+2^1+2^2+2^3+2^4+2^5+2^6+...+2^{99}\)

\(=\left(2^0+2^1+2^2+2^3+2^4\right)+2^5\left(2^0+2^1+2^2+2^3+2^4\right)+...+2^{95}\left(2^0+2^1+2^2+2^3+2^4\right)=31+31.2^5+...+31.2^{95}=31\left(1+2^5+...+2^{95}\right)⋮31\)

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

b) \(A=4x^2-9y^2-4x-6y=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)=\left(2x+3y\right)\left(2x-3y-2\right)\)

c) \(A=3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\) 

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

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

d) \(A=x^2-y^2-2x-2y=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)

\(\dfrac{x+6}{15}=\dfrac{5-x}{7}\Leftrightarrow7x+42=75-15x\Leftrightarrow22x=33\Leftrightarrow x=\dfrac{33}{22}=\dfrac{3}{2}\)

a) \(\left(2x+1\right)^2-\left(2x-3\right)^2=5\Leftrightarrow\left(2x+1-2x+3\right)\left(2x+1+2x-3\right)=5\Leftrightarrow4\left(4x-2\right)=5\Leftrightarrow16x-8=5\Leftrightarrow16x=13\Leftrightarrow x=\dfrac{13}{16}\)b) \(\left(2x-1\right)\left(2x+1\right)-\left(3x-2\right)\left(3x+2\right)=-2\Leftrightarrow4x^2-1-9x^2+4=-2\Leftrightarrow5x^2=5\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\)

a) Áp dụng tính chất dãy tỉ số bằng nhau:

 \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y+z}{3+4+5}=\dfrac{360}{60}=6\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=6\\\dfrac{y}{4}=6\\\dfrac{z}{5}=6\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=18\\y=24\\z=30\end{matrix}\right.\)

b) Áp dụng tính chất dãy tỉ số bằng nhau:\(\dfrac{x}{-2}=-\dfrac{y}{4}=\dfrac{z}{5}=-\dfrac{2y}{8}=\dfrac{3z}{15}=\dfrac{x-2y+3z}{-2+8+15}=\dfrac{1200}{21}=\dfrac{400}{7}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-2}=\dfrac{400}{7}\\-\dfrac{y}{4}=\dfrac{400}{7}\\\dfrac{z}{5}=\dfrac{400}{7}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{800}{7}\\y=-\dfrac{1600}{7}\\z=\dfrac{2000}{7}\end{matrix}\right.\)

c) Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{5}=\dfrac{y}{1}=\dfrac{z}{-2}=\dfrac{-2z}{4}=\dfrac{x+y-2z}{5+1+4}=\dfrac{160}{10}=16\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=16\\\dfrac{y}{1}=16\\\dfrac{z}{-2}=16\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=80\\y=16\\z=-32\end{matrix}\right.\)

Đáp số: 24

Xét tam giác ADC vuông tại A và tam giác EBC vuông tại B có:

\(\left\{{}\begin{matrix}\widehat{ADC}=90^0-\widehat{ACD}\\\widehat{CEB}=90^0-\widehat{ECB}\end{matrix}\right.\)

\(\Rightarrow\widehat{ADC}+\widehat{CEB}=90^0-\widehat{ACD}+90^0-\widehat{ECB}=180^0-\left(\widehat{ACD}+\widehat{ECB}\right)=180^0-\left(180^0-\widehat{DCE}\right)=\widehat{DCE}\left(đpcm\right)\)

a)\(A=3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)b) \(A=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)

c) \(A=x^2+y^2+2xy+yz+zx=\left(x+y\right)^2+z\left(x+y\right)=\left(x+y\right)\left(x+y+z\right)\)

\(\dfrac{1+8x}{8x+4}=\dfrac{2x}{6x-3}-\dfrac{8x^2}{3-12x^2}\Leftrightarrow\dfrac{1+8x}{4\left(2x+1\right)}-\dfrac{2x}{3\left(2x-1\right)}+\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=0\Leftrightarrow\dfrac{3\left(1+8x\right)\left(1-2x\right)+8x\left(1+2x\right)+32x^2}{12\left(1-2x\right)\left(1+2x\right)}=0\Leftrightarrow3\left(1+8x\right)\left(1-2x\right)+8x\left(1+2x\right)+32x^2=0\Leftrightarrow-48x^2+18x+3+16x^2+8x+32x^2=0\Leftrightarrow26x+3=0\Leftrightarrow x=-\dfrac{3}{26}\)