alo

ko ai trả lời à   

ai tra lời cho 2 like

Bài 1:

\(a.A=23,12+45,56+76,88+54,44\\ =\left(23,12+76,88\right)+\left(45,56+54,44\right)\\ =100+100\\ =200\\ b.201,5\cdot9+201,5\cdot2-201,5\\ =201,5\cdot\left(9+2-1\right)\\ =201,5\cdot10\\ =2015\\ c.C=\dfrac{1}{2}:0,5+\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+2014\\ =\dfrac{1}{2}\cdot2+\dfrac{1}{4}\cdot4+\dfrac{1}{8}\cdot8+2014\\ =1+1+1+2014\\ =2017\\ d.D=2\dfrac{2}{3}+\dfrac{5}{9}+\dfrac{4}{9}:\left(30\%-\dfrac{1}{10}\right)-\dfrac{2}{9}\\ =\dfrac{8}{3}+\left(\dfrac{5}{9}-\dfrac{2}{9}\right)+\dfrac{4}{9}:\left(\dfrac{3}{10}-\dfrac{1}{10}\right)\\ =\dfrac{8}{3}+\dfrac{1}{3}+\dfrac{4}{9}:\dfrac{1}{5}\\ =\dfrac{9}{3}+\dfrac{4}{45}\\ =3+\dfrac{4}{9}\cdot5=3+\dfrac{20}{9}=\dfrac{47}{9}\)

Bài 2: 

a: y+2=2017

=>y=2017-2=2015

b: \(3y-2\dfrac{2}{7}=3\dfrac{5}{7}\)

=>\(3y=3+\dfrac{5}{7}+2+\dfrac{2}{7}=6\)

=>\(y=\dfrac{6}{3}=2\)

c: \(1\dfrac{3}{4}-\dfrac{3}{4}y=75\%\)

=>\(\dfrac{7}{4}-\dfrac{3}{4}y=\dfrac{3}{4}\)

=>7-3y=3

=>3y=7-3=4

=>\(y=\dfrac{4}{3}\)

d: \(\dfrac{2}{3}+\dfrac{1}{3}y+3\dfrac{2}{3}y=\dfrac{8}{3}\)

=>\(4y=\dfrac{8}{3}-\dfrac{2}{3}=\dfrac{6}{3}=2\)

=>\(y=\dfrac{2}{4}=\dfrac{1}{2}\)

e: \(y-14=25\)

=>y=25+14=39

f: \(5y-25=35\)

=>5y=25+35=60

=>y=60/5=12

g: 9,34-y=1,28

=>y=9,34-1,28=8,06

h: y:1,2=2,4

=>\(y=2,4\cdot1,2=2,88\)

i: 2,4:y=0,2

=>y=2,4:0,2=12

k: (y+1)+(y+3)=24

=>y+1+y+3=24

=>2y=20

=>y=10

        Bài 6:

\(\dfrac{9^5.9^7}{3^{22}}\) = \(\dfrac{3^{15}.3^{21}}{3^{22}}\) = \(\dfrac{3^{36}}{3^{22}}\)  = 3¹⁴

  Bài 7:

\(\dfrac{9^{16}.8^{11}}{6^{33}}\) =  \(\dfrac{3^{32}.2^{33}}{3^{33}.2^{33}}\) = \(\dfrac{1}{3}\) 

a; (\(\dfrac{1}{x}\) - 5)(\(\dfrac{1}{x}\) + 5)

= (\(\dfrac{1}{x}\))^2 -  5²

=  \(\dfrac{1}{x^2}\) - 25

b; (\(\dfrac{x}{3}\) - \(\dfrac{y}{4}\))(\(\dfrac{x}{3}\) + \(\dfrac{y}{4}\))

 = \(\left(\dfrac{x}{3}\right)^2\) - \(\left(\dfrac{y}{4}\right)^2\)

  =  \(\dfrac{x^2}{9}\) - \(\dfrac{y^2}{16}\)

d; (\(\dfrac{x}{y}\) - \(\dfrac{2}{3}\) (\(\dfrac{x}{y}\)+\(\dfrac{2}{3}\))

= (\(\dfrac{x}{y}\))² - (\(\dfrac{2}{3}\))²

= \(\dfrac{x^2}{y^2}\) - \(\dfrac{4}{9}\)

e; (2\(x\) - \(\dfrac{2}{3}\))(\(\dfrac{2}{3}\) + 2\(x\))

= (2\(x\))² - (\(\dfrac{2}{3}\))²

= 4\(x^2\) - \(\dfrac{4}{9}\)

Nam đếm được 129 cái gì.

Đổi:\(\dfrac{39}{6}=\dfrac{13}{2}\)

Chiều rộng mảnh đất là:

\(\dfrac{13}{2}\times\dfrac{1}{3}=\dfrac{13}{6}\left(cm\right)\)

Chu vi mảnh đất là:

\(2\times\left(\dfrac{13}{6}+\dfrac{13}{2}\right)=\dfrac{52}{3}\left(m\right)\)

Diện tích mảnh đất là:

\(\dfrac{13}{6}\times\dfrac{13}{2}=\dfrac{169}{12}\left(m^2\right)\)

Đ/S:...

\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{200}}\)

\(\Rightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{200}}\right)\)

\(\Rightarrow2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{199}}\)

\(\Rightarrow2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{199}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{200}}\right)\)

\(\Rightarrow S=1-\dfrac{1}{2^{100}}< 1\)

\(\Rightarrow S< 1\)

Vậy \(S< 1\)

\(1.2\left(x+2\right)^2< 2x\left(x+2\right)+4\\ \Leftrightarrow2\left(x^2+4x+4\right)-2x\left(x+2\right)-4< 0\\ \Leftrightarrow2x^2+8x+4-2x^2-4x-4< 0\\ \Leftrightarrow4x< 0\\ \Leftrightarrow x< 0\\ 2.\left(x-1\right)^2+x^2< \left(x+1\right)^2+\left(x+2\right)^2\\ \Leftrightarrow x^2-2x+1+x^2< x^2+2x+1+x^2+4x+4\\ \Leftrightarrow2x^2-2x+1-2x^2-6x-5< 0\\ \Leftrightarrow-8x-4< 0\\ \Leftrightarrow8x>-4\\ \Leftrightarrow x>-\dfrac{1}{2}\\ 3.\left(x^2+1\right)\left(x-6\right)< \left(x-2\right)^3\\ \Leftrightarrow x^3-6x^2+x-6< x^3-6x^2+12x-8\\ \Leftrightarrow x-6< 12x-8\\ \Leftrightarrow12x-x>-6+8\\ \Leftrightarrow11x>2\\ \Leftrightarrow x>\dfrac{2}{11}\)

a: Xét ΔBAD và ΔABC có

AB chung

BD=AC

AD=BC

Do đó: ΔBAD=ΔABC

=>\(\widehat{ABD}=\widehat{BAC}\)

=>\(\widehat{TAB}=\widehat{TBA}\)

=>ΔTAB cân tại T

=>TA=TB

b: Ta có: TA+TC=AC

TB+TD=BD

mà TA=TB và AC=BD

nên TC=TD

nối t với m sao cho tm vuông góc ab 

xét tam giác AMT và tam giác BMT có

amt=bmt=90 độ

mt chung 

am=mb

suy ra hai tam giác bằng nhau 

suy ra ta=tb

CMTT ta có tam giác TDN và TCN 

suy ra TD=TC

Bài 18:

a: \(\left(-\dfrac{40}{51}\cdot0,32\cdot\dfrac{17}{20}\right):\dfrac{64}{75}\)

\(=-\dfrac{40}{20}\cdot\dfrac{17}{51}\cdot\dfrac{8}{25}\cdot\dfrac{75}{64}\)

\(=-\dfrac{2}{3}\cdot\dfrac{8}{64}\cdot\dfrac{75}{25}=-\dfrac{2}{3}\cdot3\cdot\dfrac{1}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)

b: \(-\dfrac{10}{11}\cdot\dfrac{8}{9}+\dfrac{7}{18}\cdot\dfrac{10}{11}=\dfrac{10}{11}\left(-\dfrac{8}{9}+\dfrac{7}{18}\right)\)

\(=\dfrac{10}{11}\left(-\dfrac{16}{18}+\dfrac{7}{18}\right)=\dfrac{10}{11}\cdot\dfrac{-9}{18}=\dfrac{10}{11}\cdot\dfrac{-1}{2}=-\dfrac{5}{11}\)

c: \(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)

\(=\left(\dfrac{3}{14}-\dfrac{13}{21}+\dfrac{29}{42}\right):\dfrac{1}{28}-8\)

\(=\left(\dfrac{9}{42}-\dfrac{6}{42}+\dfrac{29}{42}\right):\dfrac{1}{28}-8\)

\(=\dfrac{32}{42}\cdot28-8=32\cdot\dfrac{2}{3}-8=\dfrac{64}{3}-\dfrac{24}{3}=\dfrac{40}{3}\)

d: \(-1\dfrac{5}{7}\cdot15+\dfrac{2}{7}\cdot\left(-15\right)+\left(-105\right)\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)

\(=-\dfrac{12}{7}\cdot15+\dfrac{2}{7}\cdot\left(-15\right)+\left(-105\right)\left(\dfrac{70}{105}-\dfrac{84}{105}+\dfrac{15}{105}\right)\)

\(=\dfrac{-180-30}{7}+\left(-105\right)\cdot\dfrac{1}{105}\)

=-30-1=-31

Bài 19:

a: \(A=7x-2x-\dfrac{2}{3}y+\dfrac{7}{9}y=5x+y\left(\dfrac{7}{9}-\dfrac{2}{3}\right)=5x+\dfrac{y}{9}\)

Khi x=-1/10;y=4,8 thì \(A=5\cdot\dfrac{-1}{10}+\dfrac{4.8}{9}\)

\(=-\dfrac{1}{2}+\dfrac{8}{15}=\dfrac{-15+16}{30}=\dfrac{1}{30}\)

b: \(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)

\(=x+\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{-\dfrac{3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}=x+\dfrac{2\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\)

\(=x-\dfrac{2}{3}\)

Khi x=-1/3 thì \(B=-\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{3}{3}=-1\)