Đặt \(\sqrt[3]{2x+1}=a \);\(\sqrt[3]{x}=b;=>a^3-2b^3=1;\left(1\right)\)
=> \(a+b=1;=>a=b-1;\left(2\right)\)
thay (2) vào (1) ta đc pt
\(\left(b-1\right)^3-2b^3=1\)
\(<=>3b^3-3b^2+3b=0\)
\(<=>3b\left(b^2-b+1\right)=0\)
\(<=>b=0;\left(b^2-b+1>0\right)\)
\(<=>x=0\)
 

a) Ke AD  sao cho goc DAB =goc ACD => goc DAB =goc BAD ( cung phu voi DAC)

=> tam giac ABD can tai D => AD=BD

=>Tam giac ADC  can tai D => AD=DC

=>DB=DC=DA => D trung voi M

=> AM =BC/2

b) Nguoc lai :

Neu AM =BC/2 => AM =MB =MC

=> ABM can tai M ; ACM can tai M

=> BAM + CAM = (180- AMB)/2 +(180-AMC)/2 = (360 -(AMB+AMC))/2 =(360-180)/2=180/2=90

=>BAC=90

=> A=90

Ny đấy , cx xih