\(a^2+b^2+c^2\ge ab+ac+bc\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

a)Áp dụng BĐT AM-GM: \(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+a^2\ge2ca\)

Cộng theo vế suy ra \(2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca^{\left(đpcm\right)}\)

Dấu "=" xảy ra tại a = b = c

Đặt \(y=x^2-2x+3=\left(x-1\right)^2+2\ge2\), ta có:

\(x^2-2x+3=\frac{6}{x^2-2x+4}\Leftrightarrow y=\frac{6}{y+1}\Leftrightarrow y\left(y+1\right)=6\Leftrightarrow y^2+y-6=0\)

\(\Leftrightarrow\left(y+3\right)\left(y-2\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=-3\end{cases}\Rightarrow y=2\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1}\)

Vậy \(S=\left\{1\right\}\)

có điều kiện j k thế

đề vậy thôi, nhưng cám ơn nha. mk biết lm oii

\(a.\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)

\(\Leftrightarrow\left(x^2+7x-4x+16-14\right)\left(x^2+7x+4x+16+14\right)-60=0\)

\(\Leftrightarrow\left(x^2+7x+16-4x-14\right)\left(x^2+7x+16+4x+14\right)=0\)

\(\Leftrightarrow\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60=0\)

Vì \(\left(x^2+7x+16\right)^2>0;\left(4x+14\right)^2>0\)

Nên \(\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60\ge-60\)

V...\(S=\varnothing\)

\(b.4^x-12.2^x+32=0\)

\(\Leftrightarrow\left(2^x\right)^2-2.2^x.6+36-4=0\)

\(\Leftrightarrow\left(2^x-6\right)^2-4=0\)

\(\Leftrightarrow\left(2^x-4\right)\left(2^x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2^x-4=0\\2^x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2^x=4\\2^x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}2^x=2^2\\2^x=2^3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

V...\(S=\left\{2;3\right\}\)

^^ đúng ko ta

a) (x+1)(x+2)(x+5)(x+6)-60=0

[(x+1)(x+6)][(x+2)(x+5)]-60=0

(x^2 + 7x + 6)(x^2  + 7x + 10) - 60 = 0

đặt t = x^2 + 7x + 8

pt trở thành

(t-2)(t+2)-60=0

t^2 - 64=0 .....

t=8 hoặc t=-8.

tìm x ....

x³ + 2x² + 2x +1 = 0

(=) x³ + x² +x² + x + x + 1 = 0

(=) x².(x+1) + x.(x+1) + (x+1) = 0

(=) (x² + x +1 ).(x+1) = 0

(=) \(\orbr{\begin{cases}x+1=0\\x^2+x+1=0\left(lo\text{ại}\right)\end{cases}}\)(=) x=-1

Vậy phương trình có nghiệm là x=-1

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sao khó vậy