\(\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)(t/c dãy tỉ số bằng nhau)

\(\Rightarrow\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)\(\Rightarrow y+z=\dfrac{1}{2}-x\)

Ta có: \(\dfrac{\dfrac{1}{2}-x+1}{x}=2\Rightarrow x=\dfrac{1}{2}\)

\(\Rightarrow y+z=0\Rightarrow y=-z\)

Ta có:

\(\dfrac{z+x+2}{y}=2\)\(\Rightarrow\dfrac{-y+\dfrac{1}{2}+2}{y}=2\Rightarrow y=\dfrac{5}{6}\Rightarrow z=-\dfrac{5}{6}\)

\(\sqrt{x+1}+\sqrt{x-2}=3\left(Đk:x\ge2\right)\)

\(x+1+x-2+2\sqrt{\left(x+1\right)\left(x-2\right)}=9\)

\(2\sqrt{x^2-2x+x-2}=10-2x\)

\(\sqrt{x^2-x-2}=5-x\)

\(x^2-x-2=x^2-2x+25\)

\(x=27\)

\(P\left(x\right)=5x^3-3x+7\)

a) Thay x=-1

=> \(P\left(-1\right)=5.\left(-1\right)^3-3.\left(-1\right)+7\)

\(\Rightarrow P\left(-1\right)=-5+3+7=5\)

b) Thay x=-2

=> \(P\left(-2\right)=5.\left(-2\right)^3-3.\left(-2\right)+7\)

\(\Rightarrow P\left(-2\right)=-40+6+7=-27\)

\(\left(x+y\right)^2-2\left(x+y\right)z+z^2\)

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

\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)

\(A=2^{21}-2\)

B tương tự câu A

\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)

\(C=\dfrac{5^{51}-5}{4}\)

\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)

\(D=\dfrac{3^{101}-1}{2}\)

a) \(\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{IB}-\overrightarrow{IB}=\overrightarrow{0}\)

b) \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\)

\(=2\overrightarrow{GE}+\overrightarrow{GC}=2\overrightarrow{GE}-2\overrightarrow{GE}=\overrightarrow{0}\)

Ta có:

\(2x=3y=5z\)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{3y}{30}=\dfrac{2z}{12}=\dfrac{x+3y-2z}{15+30-12}=\dfrac{66}{33}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=30\\y=20\\z=12\end{matrix}\right.\)

a) Xét tứ giác ADHE có:

\(\left\{{}\begin{matrix}\widehat{A}=90^o\\\widehat{HDA}=90^o\\\widehat{HEA}=90^o\end{matrix}\right.\)

=> ADHE là h.c.n

b) Ta có:

\(\left\{{}\begin{matrix}\widehat{BID}=2\widehat{IHD}\\\widehat{IKE}=2\widehat{KCE}\end{matrix}\right.\)

mà \(\widehat{IHD}=\widehat{KCE}\)

=> \(\widehat{BID}=\widehat{IKE}\) mà 2 góc có vị trí đồng vị

=> DI//EK

=> DEKI là hình thang

\(1,8=\dfrac{18}{10}=\dfrac{9}{5}\)