a/ \(\left(5x+1\right)^2=\left(3x-2\right)^2\)

<=> \(\left(5x+1\right)^2-\left(3x-2\right)^2=0\)

<=> \(\left(5x+1-3x+2\right)\left(5x+1+3x-2\right)=0\)

<=> \(\left(2x+3\right)\left(8x-3\right)=0\)

<=> \(\orbr{\begin{cases}2x+3=0\\8x-3=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{3}{2}\\x=\frac{3}{8}\end{cases}}\)

a ) 

\(\left(5x+1\right)^2=\left(3x-2\right)^2\)

\(\Rightarrow\left(5x\right)^2+2.5x.1+1=\left(3x\right)^2-2.3x.2+2^2\)

\(\Rightarrow25x^2+10x+1=9x^2-12x+4\)

\(\Rightarrow25x^2+10x+1-9x^2+12x-4=0\)

\(\Rightarrow16x^2+22x-3=0\)

\(\Rightarrow\left(4x\right)^2+2.4x.2,75+\left(2,75\right)^2-10,5625=0\)

\(\Rightarrow\left(4x+2,75\right)^2=10,5625\)

\(\Rightarrow4x+2,75=3,25\)

\(\Rightarrow4x=0,5\)

\(\Rightarrow x=0,125\)

Vậy \(x=0,125\)

\(\left(5n-1\right)\left(n+3\right)-9n+3=5n^2+15n-n-3-9n+3=5n^2+5n=5n\left(n+1\right)⋮5\)

Mà n(n+1) là tích 2 số nguyên liên tiếp => \(n\left(n+1\right)⋮2\)

\(\Rightarrow5n\left(n+1\right)⋮5.2=10\) (đpcm)

\(\left(5n-1\right)\left(n+3\right)-9n+3\)

\(=5n^2+15n-n-3-9n+3\)

\(=5n^2+5n=5n\left(n+1\right)⋮5\)

Lại có \(n\left(n+1\right)⋮2\)

\(\Rightarrow5n^2+5n⋮\left(2.5\right)=10\)

\(\RightarrowĐPCM\)

\(A=\frac{2x-1}{x+2}\)

Để A \(\in\)\(ℤ\)thì \(2x-1\) \(⋮\)\(x+2\) ; \(x+2\) \(\ne\)0; \(2x-1,x+2\inℤ\)

Ta có: \(2x-1=2\left(x+2\right)-5\)

Vì \(2\left(x+2\right)⋮x+2\)

nên để \(2x-1⋮x+2\)

thì \(5⋮x-2\)

=> \(x-2\in\left\{\pm1;\pm5\right\}\)

Ta có bảng sau:

Vì \(x\inℤ\)=>\(x\in\left\{1;\pm3;7\right\}\)

Còn 2 ý còn lại làm tương tự như ý này

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 !!