=2,5*1,99+2,5*0,01

=2,5*2

=5

gggjjjjjjjjjjjjjjjjjjjjjj

y:2,5=8,5

 hay=> y=21,25
                                             k ckoa mìn nkoaa
                                                  Tkènn kiuuu:3

y : 2,5 + 23,4 = 31,9

y : 2,5 = 31,9 - 23,4

y : 2,5 = 8,5

y         = 8.5 x 2,5

y         = 21,25

\(y\times9,6+y:2,5=45\\ \Rightarrow y\times\dfrac{48}{5}+y:\dfrac{5}{2}=45\\ \Rightarrow y\times\dfrac{48}{5}+y\times\dfrac{2}{5}=45\\ \Rightarrow y\times\left(\dfrac{48}{5}+\dfrac{2}{5}\right)=45\\ \Rightarrow y\times10=45\\ \Rightarrow y=45:10\\ \Rightarrow y=4,5\)

giúp em với ạ

\(a,m_{NaOH}=\dfrac{30.500}{100}=150\left(g\right)\\ b,n_{HCl}=2,5.0,5=1,25\left(mol\right)\\ m_{HCl}=1,25.36,5=45,625\left(g\right)\)

13.

\(\lim\limits_{x\rightarrow1}\left(3x^2-2x-1\right)=3.1^2-2.1-1=0\)

14.

\(\lim\limits_{x\rightarrow9}\sqrt{x+16}=\sqrt{9+16}=5\)

15.

\(\lim\limits_{x\rightarrow+\infty}x^{2021}=+\infty\)

16.

\(\lim\limits_{x\rightarrow-\infty}\left(x-x^3+1\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)\)

Do \(\lim\limits_{x\rightarrow-\infty}x^3=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\left(-1+\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=-1< 0\)

\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=+\infty\)

17.

\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+5x-3}{x^2+6x+3}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(2+\dfrac{5}{x}-\dfrac{3}{x^2}\right)}{x^2\left(1+\dfrac{6}{x}+\dfrac{3}{x^2}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{2+\dfrac{5}{x}-\dfrac{3}{x^2}}{1+\dfrac{6}{x}+\dfrac{3}{x^2}}=\dfrac{2+0-0}{1+0+0}=2\)

18.

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2-x-1}}{x+1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\left|x\right|\sqrt{4-\dfrac{1}{x}-\dfrac{1}{x^2}}}{x+1}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-x\sqrt{4-\dfrac{1}{x}-\dfrac{1}{x^2}}}{x\left(1+\dfrac{1}{x}\right)}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4-\dfrac{1}{x}-\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}\)

\(=\dfrac{-\sqrt{4}}{1}=-2\)

19.

\(\lim\limits_{x\rightarrow-\infty}\left(2x^3-x^2\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(2-\dfrac{1}{x}\right)\)

Do \(\lim\limits_{x\rightarrow-\infty}x^3=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\left(2-\dfrac{1}{x}\right)=2>0\)

\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x^3\left(2-\dfrac{1}{x}\right)=-\infty\)

20.

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\sqrt{x^2+1}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{x^2+1}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(\dfrac{1}{x}\right)}{x\left(\sqrt{1+\dfrac{1}{x^2}}+1\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{0}{1+1}=0\)

a) (-35,8) + (-17,2) + 16,4 + 4,6

= [ (-35,8) + (-17,2) ] + (16,4 + 4,6)

= -53 + 21

= -32

b) (5,3 - 2,8) - (4 + 5,3)

= 2,5 - 9,3

= -6,8

c) (34,72 + 32,28):5 - (57,25 - 36,05):2

= 67:5 - 21,2:2

= 13,4 - 10,6

= 2,8

d) 2,5.(-4,68) + 2,5.(-5,32)

= 2,5.[ (-4,68) + (-5,32) ]

= 2,5. (-10)

= -25

e) 5,36. 12,34 + (-5,36). 2,34

= 5,36. 12,34 - 5,36. 2,34

= 5,36. (12,34 - 2,34)

= 5,36. 10

= 53,6

a,-35,8+(-17,2)+16,4+4,6=(-35,8-17,2)+(16,4+4,6=-53+21=-32

b,(5,3-2,8)-(4+5,3)=5,3-2,8-4-5,3=(5,3-5,3)+(-2,8-4)=0-6,8=-6,8.

c,(34,72+32,28):5-(57,25-36,05):2=67:5-21,2:2=13.4-10.6=2,8

d,2,5(-4,68)+2,5(-5,32)=2,5(-4,68-5,32)=2,5x(-10)=-25

e,5,36x12,34+(-5,36)x2,34=5,36(12,34-2,34)=5,36x10=53,6