= .

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Đặt \(q\left(x\right)=0\)

`=>3-2x=0`

`=>2x=3`

`=>x=3/2`

Vậy `x=3/2` là nghiệm của đa thức `q(x)`

Cho `Q(x)=0`

`=>3-2x=0`

`=>2x=3`

`=>x=3/2`

Vậy nghiệm của `Q(x)` là `x=3/2`

a) \(\dfrac{3}{71}\)   và   \(\dfrac{27}{211}\)

\(\dfrac{3}{71}=\dfrac{633}{14981}\)

\(\dfrac{27}{111}=\dfrac{1917}{14981}\)

Vì \(633< 1917\)   nên \(\dfrac{3}{71}< \dfrac{27}{211}\)

b) \(\dfrac{-19}{35}\)   và   \(\dfrac{-13}{21}\)   và   \(\dfrac{-17}{42}\)

\(\dfrac{-19}{35}=\dfrac{-16758}{30870}\)

\(\dfrac{-13}{21}=\dfrac{-19110}{30870}\)

\(\dfrac{-17}{42}=\dfrac{-12495}{30870}\) 

Vì: \(-12495>-16758>-19110\)

Nên:\(\dfrac{-17}{42}>\dfrac{-19}{35}>\dfrac{-13}{21}\)

Ai bt thì giúp mk đc ko ạ :(( Mk gấp lắm rồi

`3^2 .2^5 .(2/3)^2`

`= 9 . 32 . 4/9`

`= (9.4)/9 .32`

`= 1/4 .32 = 8 =2^3`

`(1/3)^2 . 1/3 .9 `

`= 1/9 . 1/3 .9`

`=  1.1/3 ` \(=\left(\dfrac{1}{\sqrt{3}}\right)^2\) 

`9.3^3 .1/81 .3^2 `

`= 9. 3^3/3^4 .3^2`

`= 9.1/3 .9 = 9/3 .3 = 3.3 =9 =3^2`

`4.2^5 :(2^3 .1/16)`

`= 4. 32 : (8/16)`

`= 4.32:1/2 = 4.32.2 = 256 = 2^8`

1/

\(=3^2.3^3.\dfrac{1}{3^4}.3^2=3^3\)

2/

\(=2^2.2^5:\left(2^3.\dfrac{1}{2^4}\right)=2^7.2=2^8\)

Vì `\hat{AOB}` và `\hat{BOC}` kề bù

 `=>\hat{AOB}+\hat{BOC}=180^o`

   `=>72^o +\hat{BOC}=180^o`

   `=>\hat{BOC}=108^o`

Ta có `2` góc kề bù `+` lại ` = 180^0`

Mà : \(\widehat{AOB}=72^0\)

`=>` \(\widehat{BOC}=180^0-72^0\)

`=108^0`

Vậy....

9.3³.\(\dfrac{1}{81}\).3² = 3².3³.\(\dfrac{1}{3^4}\)= 3¹

4.2⁵:(2³ . \(\dfrac{1}{16}\)) = 2².2⁵:(2³.\(\dfrac{1}{2^4}\) ) = 2⁶

3².2⁵.(\(\dfrac{2}{3}\))² = 2⁷

(\(\dfrac{1}{3}\))² . \(\dfrac{1}{3}\).9 = \(\dfrac{1}{3^3}\) . 3² = \(\dfrac{1}{3}\) = 3^-1

`9.3^3 . 1/81 .3^2=3^2 .3^3 . 1/[3^4].3^2=3^3`

`4.2^5:(2^3 .1/16)=2^2 .2^5 :(2^3 .1/[2^4])=2^7 :1/2=2^7 .2=2^8`

`3^2 . 2^5 . (2/3)^2 =3^2 .2^5 . [2^2]/[3^2]=2^7`

`(1/3)^2 . 1/3 .9=[1^2]/[3^2] . 1/3 .3^2=1/3=(1/\sqrt{3})^2`

`(1 + 1/2 - 1/4)^2 . ( 2 + 3/7)`

` = ( 4/4 + 2/4 - 1/4)^2 . ( 14/7 + 3/7)`

` = (5/4)^2 . 17/7`

` = 25/16 . 17/7`

` = 425/112`

__________________________________

`4 : ( 1/2 - 1/3)^2`

` = 4 : ( 3/6 - 2/6)^2`

` = 4 : (1/6)^2`

` = 4 : 1/36`

` = 4 xx 36`

`= 144`

\(\left(1+\dfrac{1}{2}-\dfrac{1}{4}\right)^2\times\left(2+\dfrac{3}{7}\right)\)

\(=\left(\dfrac{4}{4}+\dfrac{2}{4}-\dfrac{1}{4}\right)\times\left(\dfrac{14}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{5}{4}\times\dfrac{17}{7}\)

\(=\dfrac{85}{28}\)

\(4:\left(\dfrac{1}{2}-\dfrac{1}{3}\right)^3\)

\(=4:\left(\dfrac{3}{6}-\dfrac{2}{6}\right)^3\)

\(=4:\left(\dfrac{1}{6}\right)^3\)

\(=864\)

Bài 1:

A B C M D E

a/ Xét tg vuông MBD và tg vuông MAD có

MD chung

MB=MA (gt)

\(\Rightarrow\Delta MBD=\Delta MAD\) (Hai tg vuông có 2 cạnh góc vuông bằng nhau)

b/

Xét tg ABC và tg BAE có

AB chung

AE=BC (gt)

tg MAD = tg MBD (cmt) \(\Rightarrow\widehat{ABC}=\widehat{BAE}\)

\(\Rightarrow\Delta ABC=\Delta BAE\) (c.g.c)

Bài 2

A B C D

a/

Xét tg ABC và tg ADC có

AC chung

\(\widehat{BAC}=\widehat{DCB}\) (góc so le trong)

\(\widehat{ACB}=\widehat{DAC}\) (góc so le trong)

\(\Rightarrow\Delta ABC=\Delta ADC\) (g.c.g)

b/

Nối BC. Chứng mhinh tương tự câu a có \(\Delta ADB=\Delta CBD\) (g.c.g)