a³+b³+c³=3abc <=> (a+b)³-3ab(a+b)+c³=3abc

<=> (a+b+c)³-3(a+b)c(a+b+c)-3ab(a+b)-3abc=0

<=> (a+b+c)³-3c(a+b)(a+b+c)-3ab(a+b+c)=0

<=>(a+b+c)(a²+b²+c²+2ab+2bc+2ca-3ac-3bc-3ab)=0

<=> (a+b+c)(a²+b²+c²-ab-bc-ca)=0

<=> (a+b+c)\(\frac{\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]}{2}\)=0

<=> \(\left[\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

mà a,b,c dương nên a+b+c khác 0 => a=b=c

lời giải vắn tắt:

a) \(\widehat{AMB}\)là góc nội tiếp chắn nửa đường tròn => \(\widehat{AMB}=90^o\)

=>\(\widehat{OMA}+\widehat{OMT}=\widehat{AMB}=90^o\)

MF là tiếp tuyến của (O)=> \(\widehat{OMF}=90^o\rightarrow\widehat{OMT}+\widehat{TMF}=\widehat{OMF}=90^o\)

=> \(\widehat{OMA}=\widehat{TMF}\)(1)

\(\Delta MAB\)~\(\Delta OTB\)(g.g)(tự cm)=>\(\widehat{OAM}=\widehat{OTB}\)

mà \(\widehat{OCB}=\widehat{MTF}\)(đối đỉnh) =>\(\widehat{OAM}=\widehat{MTF}\)(2)

từ (1) và (2)=> \(\Delta OMA\)~\(\Delta FMT\)(g.g)\(\rightarrow\frac{MA}{MT}=\frac{OA}{FT}\rightarrow MA.FT=OA.MT\)

b)\(\Delta OMA\)~\(\Delta FMT\)(cmt ) mà \(\Delta OMA\)cân ở O=> \(\Delta FMT\)cân ở F

=> FM=FT

mà \(\Delta TME\) vuông ở M => ..... FM=FE

c) ta cm được TA=TB

lại có:\(\Delta MTE\)~\(\Delta OTB\)(g.g) \(\rightarrow\frac{ME}{OB}=\frac{TE}{TB}\)\(\rightarrow ME.TB=OB.TE\)

\(\rightarrow ME.TA=R.2R=2R^2\)(TE=2FM=2R)

cách 1: đặt a = x+2 ,=> A= (a-3)⁴+(a+3)⁴-120

tách ra là ổn

cách 2 : áp dụng BĐT bunyakovsky:

(1+1)(a²+b²)\(\ge\)(a+b)²=> a²+b²\(\ge\)\(\frac{\left(a+b\right)^2}{2}\)(dấu = xảy ra khi a=b)

A= (x-1)⁴+(x+5)⁴-120=(1-x)⁴+(x+5)⁴-120\(\ge\)\(\frac{1}{2}\left[\left(x-1\right)^2+\left(x+5\right)^2\right]^2-120\)

\(A\ge\frac{1}{2}\left(2x^2+8x+26\right)^2-120=\frac{1}{2}\left[2\left(x+2\right)^2+18\right]^2-120\ge\frac{18^2}{2}-120=42\)

dấu = xảy ra khi 1-x=x+5 và x+2=0

=> x=-2

ta có: \(2P=2x^2-2x\sqrt{y}+2x+2y-2\sqrt{y}+2\)

\(2P=\left(x^2-2x\sqrt{y}+y\right)+\left(x^2+2x+1\right)+\left(y-2\sqrt{y}+1\right)\)

\(2P=\left(x-\sqrt{y}\right)^2+\left(x+1\right)^2+\left(\sqrt{y}-1\right)^2\ge0\forall x,y\)

\(\Rightarrow P\ge0\forall x,y\)

dấu = xảy ra khi \(\left\{\begin{matrix}x=\sqrt{y}\\x=-1\\\sqrt{y}=1\end{matrix}\right.\)(có gì đó sai sai)

chờ Em hai mươi năm :v

\(\left\{\begin{matrix}x^3-3x-2=2-y\\y^3-3y-2=2-z\\z^3-3z-2=2-x\end{matrix}\right.\)

ta có: x³-3x-2=x³-2x²+2x²-4x+x-2

=x²(x-2)+2x(x-2)+(x-2)=(x-2)(x+1)²

tương tự : y³-3y-2=(y-2)(y+1)²; z³-3z-2=(z-2)(z+1)²

ta có hệ pt:\(\left\{\begin{matrix}\left(x-2\right)\left(x+1\right)^2=2-y\\\left(y-2\right)\left(y+1\right)^2=2-z\\\left(z-2\right)\left(z+1\right)^2=2-x\end{matrix}\right.\)

nhân vế vs vế 3 pt trên :\(\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=\left(2-x\right)\left(2-y\right)\left(2-z\right)\)

\(\Leftrightarrow\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=-1\)

vt >= 0, vF <0 vậy hệ pt vô nghiệm

n(n+5)-(n-3)(n+2)= n²+5n-n²-2n+3n+6=6n+6 =6(n+1)\(⋮6\forall n\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)

\(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{y^3z^3+x^3z^3+x^3y^3}{x^2y^2z^2}=\frac{\left(xy+yz+xz\right)\left(...\right)}{x^2y^2z^2}=0\)

\(x^4+\sqrt{x^2+2002}=2002\) (DKXĐ: xác định vs mọi x)

\(\Leftrightarrow x^4+x^2+\frac{1}{4}+\sqrt{x^2+2002}=x^2+2002+\frac{1}{4}\)

\(\Leftrightarrow x^4+x^2+\frac{1}{4}=x^2+2002-\sqrt{x^2+2002}+\frac{1}{4}\)

\(\Leftrightarrow\left(x^2+\frac{1}{2}\right)^2=\left(\sqrt{x^2+2002}-\frac{1}{2}\right)^2\)

xét \(x^2+\frac{1}{2}=\sqrt{x^2+2002}-\frac{1}{2}\Leftrightarrow x^2+1=\sqrt{x^2+2002}\)

\(\Leftrightarrow x^4+2x^2+1=x^2+2002\Leftrightarrow x^4+x^2-2001=0\)

đặt x²=a(a>0) => a²+a-2001=0

\(\Delta=1+4.2001=8005\rightarrow\left[\begin{matrix}a=\frac{\sqrt{8005}-1}{2}\\a=\frac{-\sqrt{8005}-1}{2}\end{matrix}\right.\)

mà a>0 \(\rightarrow a=\frac{\sqrt{8005}-1}{2}\Leftrightarrow x=\pm\sqrt{\frac{\sqrt{8005}-1}{2}}\)

xét\(x^2+\frac{1}{2}=\frac{1}{2}-\sqrt{x^2+2002}\Leftrightarrow x^2=-\sqrt{x^2+2002}\)(vô nghiệm)

vậy pt có 2 nghiệm là...