Vì \(ab+bc+ac=3\)  =>   \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{abc}\)

Đặt \(\frac{1}{a}=x\):  \(\frac{1}{b}=y\):  \(\frac{1}{c}=z\)=> x+y+z=3xyz

Ta có   \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{xyz}\ge13\)

AD BĐT  \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) dấu = khi a=b=c ta có 

  \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{36}{x+y+z}\)=\(\frac{36}{3xyz}=\frac{12}{xyz}\)

=> \(\frac{12}{xyz}+\frac{1}{xyz}\ge13\)

=>  \(\frac{13}{xyz}\ge13\)

mà \(3xyz=x+y+z\ge3\sqrt[3]{xyz}\)dấu = khi x=y=z 

=> xyz\(\le1\)

=> đpcm 

Câu hỏi của Thiên Ân - Toán lớp 8 - Học toán với OnlineMath

tương tự như câu này đều thay số thôi

\(A=\left(n^2+3n+2\right)\left(2n-1\right)-2\left(n^3-2n-1\right)\)

\(A=2n^3+6n^2+4n-n^2-3n-2-2n^3+4n+2\)

\(A=5n^2+5n\)

\(A=5n\left(n+1\right)\)

\(\text{Vì 5⋮5 nên 5n(n+1)⋮5}\)(1)

\(\text{Vì n;n+1 là hai số tự nhiên liên tiếp nên n(n+1)⋮2}\)

\(\Rightarrow5n\left(n+1\right)⋮2\)(2)

\(\text{Từ (1) và (2)}\Rightarrow5n\left(n+1\right)⋮10\text{ vì (2,5)=1}\)

\(\text{Vậy A⋮10}\)

=X⁴-3X³ +6X³-18X²+11X²-33X+6X-18

=(X-3)(X³+6X²+11X+6)

=(X-3)(X+3)(X+1)(X+2)

\(x^4+3x^3-7x^2-27x-18.\)

\(=\left(x^4-9x^2\right)+\left(3x^3-27x\right)+\left(2x^2-18\right)\)

\(=x^2\left(x-3\right)\left(x+3\right)+3x\left(x-3\right)\left(x+3\right)+2\left(x-3\right)\left(x+3\right).\)

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

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

\(=\left(x-3\right)\left(x+3\right)\left(x+1\right)\left(x+2\right).\)

a) x(x+1)=0

=> \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

b)\(x\left(x^2+1\right)\)

=> \(\orbr{\begin{cases}x=0\\x^2=-1\end{cases}=>x=0}\)

k cho ik nha

\(a,x^2+x=0\)

\(\Rightarrow x\left(x+1\right)=0\) 

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\Leftrightarrow x=-1\end{cases}}\)

Vậy...

\(b,x^3+x=0\)

\(\Rightarrow x\left(x^2+1\right)=0\)

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

Vậy..

x^4+4

=(x^2)^2+4x^2+4-4x^2

=(x^2+2)^2-4x^2

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

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