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`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/
\(=2^2.2^5:\left(2^3.\dfrac{1}{2^4}\right)=2^7.2=2^8\)
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
`(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/ 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/
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)
Lời giải:
a.
ĐKĐB $\Leftrightarrow \frac{-7}{12}\leq x\leq \frac{1}{4}$
Vì $x$ nguyên nên $x=0$
b.
ĐKĐB $\Leftrightarrow \frac{-1}{12}\leq x\leq \frac{1}{8}$
Vì $x$ nguyên nên $x=0$
\(a)\)
\(VT=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{3}{4}=\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{9}{12}=-\dfrac{7}{12}\)
\(VP=\dfrac{1}{24}-\left(\dfrac{1}{8}-\dfrac{1}{3}\right)=\dfrac{1}{24}-\dfrac{1}{8}+\dfrac{1}{3}=\dfrac{1}{24}-\dfrac{3}{24}+\dfrac{8}{24}=\dfrac{1}{4}\)
Ta có: \(\dfrac{1}{4}=\dfrac{1.3}{4.3}=\dfrac{3}{12}\Rightarrow-\dfrac{7}{12}\le x\le\dfrac{3}{12}\)
Mà \(x\inℤ\Rightarrow x=-7;-6;-5;-4;...;1;2;3\)
\(b)\)
\(VT=\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{3}{12}=-\dfrac{1}{12}\)
\(VP=\dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)=\dfrac{1}{48}-\dfrac{1}{16}+\dfrac{1}{6}=\dfrac{1}{48}-\dfrac{3}{48}+\dfrac{8}{48}=\dfrac{1}{8}\)
Ta có: \(-\dfrac{1}{12}=\dfrac{\left(-1\right).2}{12.2}=-\dfrac{2}{24};\dfrac{1}{8}=\dfrac{1.3}{8.3}=\dfrac{3}{24}\Rightarrow-\dfrac{2}{24}< x< \dfrac{3}{24}\)
Mà số nguyên duy nhất lớn hơn \(-\dfrac{2}{24}\) và bé hơn \(\dfrac{3}{24}\) là \(0\Rightarrow x=0\)
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