Xét \(x=-1;0\) đều thỏa mãn 

Xét \(x>0\)ta có \(x^{2018}>0;\left(x+1\right)^{2018}>1\Rightarrow x^{2018}+\left(x+1\right)^{2018}>1\)(loại)

Xét \(x< -1\) ta có \(x^{2018}>1;\left(x+1\right)^{2018}>0\Rightarrow x^{2018}+\left(x+1\right)^{2018}>1\) (loại)

Xét \(-1< x< 0\) ta có : \(x^{2008}< -x;\left(x+1\right)^{2018}< -x+1\Rightarrow x^{2008}+\left(x+1\right)^{2018}< 1\) (loại)

Vậy \(PT\) có 2 nghiệm là \(x=-1\) và \(x=0\)

minh la WHY

vay con truong hop 0<x<1 dau

a)\(C\left(x\right)=9x^2-6x+14\)

\(\Rightarrow C\left(x\right)=\left(3x\right)^2-2.3x+1+13\)

\(\Rightarrow C\left(x\right)=\left(3x-1\right)^2+13\ge13\)

Dấu "=" xảy ra khi

\(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

\(\Rightarrow MinC\left(x\right)=13\Leftrightarrow x=\frac{1}{3}\)

\(b,D\left(x\right)=3x^2+12x+15\)

\(\Rightarrow D\left(x\right)=3x^2+6x+6x+12+3\)

\(\Rightarrow D\left(x\right)=3x\left(x+2\right)+6\left(x+2\right)+3\)

\(\Rightarrow D\left(x\right)=3\left(x+2\right)^2+3\ge3.Với\forall x\in R\)

Dấu "=" xảy ra khi

\(x+2=0\Leftrightarrow x=-2\)

\(\Rightarrow MinD\left(x\right)=3\Leftrightarrow x=-2\)

a) C(x)= 9x^2 -6x + 14

       = (3x -1)^2 +13 >  13

vậy C_{min = 13 <=> x= 1/3}

b)D(x)= 3x^2 +12x +15

          = 3(x^2 +4x +5)

          = 3((x+2)^2 +1)

          = 3(x+2)^2 +3 > 3

vậy D_min=3 <=> x=-2

\(4x^2-x+\frac{1}{2}\)

\(=\left(2x\right)^2-x.2.\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\)

\(=\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}.Với\forall x\in R\)

\(\RightarrowĐPCM\)

   4x^2-x +1/2

= (2x -1/2)^2 +1/4 > 1/4 với mọi x

vậy 4x^2 -x +1/2 luôn có giá trị dương với mọi x

a) Ta có: \(x^2-20x+101=x^2-2.x.10+10^2+1=\left(x-10\right)^2+1\)

Vì \(\left(x-10\right)^2\ge0\left(\forall x\in Z\right)\)

\(\Rightarrow\left(x-10\right)^2+1>1>0\)

Vậy x²-20x+101 >0 với mọi x

b) \(4a^2+4a+2=\left(2a\right)^2+2.2a.1+1+1=\left(2a+1\right)^2+1\)

Vì \(\left(2a+1\right)^2\ge0\left(\forall a\in Z\right)\)

\(\Rightarrow\left(2a+1\right)^2+1>1>0\)

Vậy 4a²+4a+2 > 0 với mọi a

c) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)

\(=\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+16\)

\(=\left(x^2+10x+20\right)^2\) \(\ge0\left(\forall x\right)\)

Giúp mình với !!

Có \(a^2+b^2=ab+ba\)

\(\Rightarrow a^2+b^2=2ab\)

\(\Rightarrow a^2-2ab+b^2=0\)

\(\Rightarrow\left(a-b\right)^2=0\)

\(\Rightarrow a-b=0\Leftrightarrow a=b\)

\(\RightarrowĐPCM\)

\(x^2+4x+3=0\)

\(\Leftrightarrow x^2+x+3x+3=0\)

\(\Leftrightarrow x.\left(x+1\right)+3.\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right).\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)

Vậy  x = -1 hoặc x = -3

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Link đây nhé mn 

x=7 => x+1=8

\(x^{2006}-8x^{2005}+8x^{2004}-...+8x^2-8x-5\)

\(=x^{2006}-\left(x+1\right)x^{2005}+\left(x+1\right)x^{2004}-...+\left(x+\right)x^2-\left(x+1\right)x-5\)

\(=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\)

\(=-x-5=-7-5=-12\)

Vậy...

Cho n số dương a₁;a₂;a₃;...;a_n ta có BĐT:
\(a_1+a_2+a_3+...+a_n\ge n\sqrt[n]{a_1a_2a_3...a_n}\)
Từ BĐT trên ta suy ra:
\(\frac{a_1+a_2+a_3+...+a_n}{n}\ge\sqrt[n]{a_1a_2a_3...a_n}\)
=> Trung bình cộng \(\ge\)Trung bình nhân 

bn chỉ mk tách cái phân thức đầu bài cho để đưa về dạng tổng quát đi

Đặt \(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\)

\(=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+ac+bc\right)}+2+2+2\) (BĐT Cosi vs Svac sơ)

\(\ge\frac{\left(a+b+c\right)^2}{\frac{2.\left(a+b+c\right)^2}{3}}+6\)(BĐT Svac sơ) \(=\frac{3}{2}+6=\frac{15}{2}\)