Bài 1: Thu thập số liệu, tần số
Các câu hỏi tương tự
Bài 1:Thực hiện phép tính
a, 0,5-frac{5}{41}+frac{1}{2}-frac{36}{41}
b,(-frac{1}{3})2.frac{4}{11}+frac{7}{11}.(frac{1}{3})2
c,(-frac{2}{3}+frac{3}{7}):frac{4}{5}+(-frac{1}{3}+frac{4}{7}):frac{4}{5}
d,frac{48^2.8^5.100^9}{12^2.2^{15}.4^2}
e,(-2)3 -sqrt{0,36} - |-2,4|
f,|-frac{3}{4}|.sqrt{frac{16}{9}}+ 3sqrt{49}
Đọc tiếp
Bài 1:Thực hiện phép tính
a, 0,5-\(\frac{5}{41}\)+\(\frac{1}{2}\)-\(\frac{36}{41}\)
b,(-\(\frac{1}{3}\))2.\(\frac{4}{11}\)+\(\frac{7}{11}\).(\(\frac{1}{3}\))2
c,(-\(\frac{2}{3}\)+\(\frac{3}{7}\)):\(\frac{4}{5}\)+(-\(\frac{1}{3}\)+\(\frac{4}{7}\)):\(\frac{4}{5}\)
d,\(\frac{48^2.8^5.100^9}{12^2.2^{15}.4^2}\)
e,(-2)3 -\(\sqrt{0,36}\) - |-2,4|
f,|-\(\frac{3}{4}\)|.\(\sqrt{\frac{16}{9}}\)+ 3\(\sqrt{49}\)
tìm x , biết
a, \(\frac{4}{x+5}=\frac{3}{x-4}\)
b.5-\(\frac{2}{x}=\frac{3}{-7}\)
c, \(\frac{x+5}{6}-4=\frac{5}{2}\)
d,8-\(\frac{7}{x+4}=\frac{5}{2}\)
e,\(\frac{x}{7}=\frac{-15}{14}\)
f,2-\((2x+3)=7\)
cho tỉ lệ thức : frac{a}{b} frac{c}{d} CM:
1)frac{a+c}{a}frac{b+d}{b}
2)frac{2A+3B}{2A-3B} frac{2C+3D}{2C-3D}
3)frac{A.B}{C.D} frac{a^2-b^2}{c^2-d^2}
4)(frac{a+b}{c+d})2frac{a^2+b^2}{c^2+d^2}
ai giúp mk vs ạ ! mình đang cần gấp ạ!
Đọc tiếp
cho tỉ lệ thức : \(\frac{a}{b}\) =\(\frac{c}{d}\) CM:
1)\(\frac{a+c}{a}\)=\(\frac{b+d}{b}\)
2)\(\frac{2A+3B}{2A-3B}\)= \(\frac{2C+3D}{2C-3D}\)
3)\(\frac{A.B}{C.D}\)= \(\frac{a^2-b^2}{c^2-d^2}\)
4)(\(\frac{a+b}{c+d}\))2=\(\frac{a^2+b^2}{c^2+d^2}\)
ai giúp mk vs ạ ! mình đang cần gấp ạ!
rút gọn
a,\(\frac{8x^2y^3}{4xy^5}\)
b, \(\frac{-2(x^2y)^6}{(y^6x)^2}\)
c,\(\frac{(x+1)^2}{(x-1)\times(x+1)}\)
a) | x+y - frac{1}{100}| -| x| -| y| -| frac{1}{10}|
b) | x-frac{1}{2}y| +| y+frac{4}{5}|
Đọc tiếp
a) \(|\) x+y - \(\frac{1}{100}\)\(|\) = -\(|\) x\(|\) -\(|\) y\(|\) -\(|\) \(\frac{1}{10}\)\(|\)
b) \(|\) x-\(\frac{1}{2}\)y\(|\) +\(|\) y+\(\frac{4}{5}\)\(|\)
[TEX]frac{a}{c} frac{b}{d}[/TEX]
Rightarrow [TEX]adcb[/TEX]
Rightarrow [TEX]2ad2cb[/TEX]
Rightarrow [TEX]ad-cbcb-ad[/TEX]
Rightarrow[TEX]ab(ad-cb)ab(cb-ad)[/TEX]
Rightarrow[TEX]a^2bd- acb^2 acb^2-a^2bd[/TEX]
Rightarrow[TEX]16a^2b^2-36abcd+24a^2bd-24acb^2 16a^2b^2- 36abcd+24acb^2-24a^2bd[/TEX]
Rightarrow[TEX]( 4a^2 - 6ac)(4b^2 + 6bd) ( 4a^2 +6ac)(4b^2 - 6bd)[/TEX]
C2: [TEX]frac{a}{c} frac{b}{d}[/TEX]Rightarrow[TEX] (frac{a}{b})^2( frac{c}{d})^2 frac{ac}{bd}frac{-12ac}{-12bd} frac{(4a^2-6ac)-...
Đọc tiếp
\( [TEX]\frac{a}{c} = \frac{b}{d}[/TEX] \Rightarrow [TEX]ad=cb[/TEX] \Rightarrow [TEX]2ad=2cb[/TEX] \Rightarrow [TEX]ad-cb=cb-ad[/TEX] \Rightarrow[TEX]ab(ad-cb)=ab(cb-ad)[/TEX] \Rightarrow[TEX]a^2bd- acb^2= acb^2-a^2bd[/TEX] \Rightarrow[TEX]16a^2b^2-36abcd+24a^2bd-24acb^2 = 16a^2b^2- 36abcd+24acb^2-24a^2bd[/TEX] \Rightarrow[TEX]( 4a^2 - 6ac)(4b^2 + 6bd) =( 4a^2 +6ac)(4b^2 - 6bd)[/TEX] C2: [TEX]\frac{a}{c} = \frac{b}{d}[/TEX]\Rightarrow[TEX] (\frac{a}{b})^2=( \frac{c}{d})^2= \frac{ac}{bd}=\frac{-12ac}{-12bd}= \frac{(4a^2-6ac)-(4a^2+6ac)}{(4b^2- 6bd)- (4b^2+6bd)} [/TEX]\)
\(\frac{1}{8}\) . 102 = 2 n
tím a nguyên sao cho \(M=\frac{a^2+3}{a-1}\)nhận gt nguyên
(\(\frac{1}{9}\))^23 : (\(\frac{1}{3}\))^41