mọi người giúp mình với

\(A=x^2-4^2-\left(x+3\right).\left(-2x+x+3\right)=x^2-4^2-\left(x+3\right).\left(-x+3\right)\)

\(=x^2-16+9-x^2=-7\)

=> đpcm

\(a,\left(\dfrac{1}{x-1}-\dfrac{x}{x-1^2}.\dfrac{x^2+1+x}{x+1}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1}{x-1}-\dfrac{x\left(x^2+1+x\right)}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\\ =\left(\dfrac{1\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^3+x+x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1}{x^2-1}\)

\(\dfrac{x+1-x^3-x-x^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{\left(x+1-x^3-x-x^2\right)\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=1-x^3-x^2\)

b,

thay x=\(\dfrac{1}{2}\) vào bt M ta được:

\(1-\left(\dfrac{1}{2}\right)^3-\left(\dfrac{1}{2}\right)^2=\dfrac{5}{8}\)

Câu b đâu bạn

mk nhịu

a)    \(\frac{28\times7-45\times7+7\times18}{45\times14}\)

\(=\frac{7\left(28-45+7\right)}{45\times14}\)

\(=\frac{7\times\left(-10\right)}{45\times14}=\frac{-1}{9}\)

b)   \(\frac{12.3-2.6}{4.5.6}\)

\(=\frac{2.6.3-2.6}{4.5.6}\)

\(=\frac{2.6\left(3-1\right)}{2.2.5.6}\)

\(=\frac{2.6.2}{2.2.5.6}\)\(=\frac{1}{5}\)

 A = x + | x |

có ; \(\left|x\right|\ge0\forall x\)

=> \(x+\left|x\right|\ge x\forall x\)

dấu ''='' xảy ra <=> x =0

vậy gtnn của A là x tại x=0

b) ta có : \(\left|x-3\right|\ge0\forall x\in Z\)

dấu ''='' xảy ra <=> x-3=0

=>  x=3

vậy gtnn  của bt B là 0 tại x=3

c) | x - 2 | + | x - 4 |

\(C=\left|x-2\right|+\left|x-4\right|\ge\left|x-2\right|+\left|4-x\right|\ge\left|x-2+4-x\right|\ge2\)

dấu ''='' xảy ra <=> \(\left(x-2\right)\left(4-x\right)\ge0\)

\(\orbr{\begin{cases}x-2=0\\x-4=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

vậy gtnn của bt C là 2 tại x ={2;4}

a)\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^8-1\right)\left(3^8+1\right)\)

\(=\dfrac{1}{2}.\left(3^{16}-1\right)\)

\(=\dfrac{1}{2}3^{16}-\dfrac{1}{2}\)

b) \(48\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^4-1\right)\left(5^4+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^8+1\right).....\left(5^{32}+1\right)\)

\(=2.\left(5^{32}-1\right)\)

\(=2.5^{32}-2\)

Tham khảo nhé~

3 chấm ở giữa để kia làm j z

a, Với x khác 1 

\(A=\dfrac{x^2+x+1-3x^2+2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)

b, Ta có \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\dfrac{-1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< 0\)

Vậy với x khác 1 thì bth A luôn nhận gtri âm