Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a> x=13/12:2/3=13/8. vậy x=13/8
b>x=5/4 nhân 5/4=25/16. vậy...
c>x=7/4+1/2=9/4. vậy x=+-9/4
d>x=x=1-1/4=3/4. vậy...
a) \(0,5-\frac{2}{3}x=\frac{-7}{12}\)
\(\frac{1}{2}-\frac{2}{3}x=\frac{-7}{12}\)
\(\frac{2}{3}x=\frac{1}{2}-\frac{-7}{12}\)
\(\frac{2}{3}x=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{2}{3}\)
\(x=\frac{13}{8}\)
b) \(\frac{3}{4}-x:\frac{5}{4}=\frac{-1}{2}\)
\(x:\frac{5}{4}=\frac{3}{4}-\frac{-1}{2}\)
\(x:\frac{5}{4}=\frac{5}{4}\)
\(x=\frac{5}{4}.\frac{5}{4}\)
\(x=\frac{25}{16}\)
c) \(\left|x-\frac{1}{2}\right|-\frac{1}{4}=\frac{3}{2}\)
\(\left|x-\frac{1}{2}\right|=\frac{3}{2}+\frac{1}{4}\)
\(\left|x-\frac{1}{2}\right|=\frac{7}{4}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{7}{4}\\x-\frac{1}{2}=\frac{-7}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{-5}{4}\end{cases}}}\)
Vậy x = 9/4 hoặc x = -5/4
d) \(\left|1-x\right|=\frac{1}{4}\)
\(\Rightarrow\orbr{\begin{cases}1-x=\frac{1}{4}\\1-x=\frac{-1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{5}{4}\end{cases}}}\)
Vậy x = 3/4 hoặc 5/4
a, => x^3 < 0 ; x-3 > 0 hoặc x^3 > 0 ; x-3 < 0
=> 0 < x < 3
b, => x^4.(2x-8) < 0
=> x^4.(x-4) < 0
Vì x^4 >= 0
=> x-4 < 0
=> x < 4
c, Vì x-1 < x+12
=> x-1 < 0 ; x+12 >0
=> -12 < x < 1
d, => x-12 > 0 ; x-1 > 0 hoặc x-12 < 0 ; x-1 < 0
=> x >12 hoặc x < 1
Tk mk nha
Ta có: \(A=1-2+3-4+5-6+7-8+9\)
\(=(1+9)-(2+8)+(3+7)-(4+6)+5\)
\(=10-10+10-10+5\)
\(=5\)
Vậy \(A=5\)
B = 12 - 14 + 16 - 18 + ... + 2008 - 2010
B = -2 + (-2)+ (-2)+ (-2) + ...+ (-2)
B = -2 . 100
B = -200
S = 1 + 2 + 22 + 23 +24 + 25 +...+ 260 + 261 + 262 + 263
= ( 1 + 22) +( 2 + 23) + (24 + 26) + ( 25 + 27) +...+ (260 + 262) + ( 261 + 263)
=( 1 + 22) + 2 ( 1 + 22) + 24 (1 + 22) + 25 (1 +22)+...+ 260 ( 1 + 22) + 261( 1 + 22)
= ( 1 + 22)( 1 + 2 +24 + 25 +...+ 260)
= 5 ( 1 + 2 +24 + 25 +...+ 260)
Vậy S chia hết cho 5 vì có một thừa số là 5.
bai2
UCLN (n,n+2)=d
=>(n+2)-n chia hết cho d
2 chia het cho d
vay d thuoc uoc cua 2={1,2}
nếu n chia hết cho 2 uoc chung lon nhta (n,n+2) la 2
neu n ko chia het cho 2=> (n,n+2) nguyen to cung nhau
BCNN =n.(n+2) neu n le
BCNN=n.(n+2)/2
$a)\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}$
$=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}$
$=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}$
$=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}$
b) Đặt $B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{18.20}$
$=>\dfrac{1}{2}B=\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{18.20}$
$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}$
$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}$
$=>B=\dfrac{9}{10}$
c) $\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}$
$=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}$
$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}$
$=1-\dfrac{1}{9}=\dfrac{8}{9}$
d) Viết lại đề rõ ràng nha bạn.