Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(19,96+4,19-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15-24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
\(24,15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24,15\)
\(4x=24,4\)
\(x=6,1\)
\(\text{Bài này cx đơn giản thôi!}\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=23.15-19.96\)
\(4.19-24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=4.19-3.19\)
\(24.15:\left(x:\frac{1}{4}-\frac{1}{4}\right)=1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15:1\)
\(x:\frac{1}{4}-\frac{1}{4}=24.15\)
\(x:\frac{1}{4}=24.15+\frac{1}{4}\)
\(x:\frac{1}{4}=24.4\)
\(x=24.4.\frac{1}{4}\)
\(x=6.1\)
a) 187 - {[497 - ( 8 x X + 11) : X] : 3 - 78} = 150
=> {[497 - ( 8 x X + 11) : X] : 3 - 78} = 187 - 150
=> {[497 - (8 x X + 11) : X] : 3 - 78} = 37
=> [497 - (8 x X +11): X ] : 3 - 78 = 37
=> [497 - (8 x X + 11) : X] : 3 = 115
=> 497 - ( 8 x X + 11) : X = 345
=> (8 x X + 11) : X = 497 - 345 = 152
=> 8X + 11 = 152X
=> 152X - 8X = 11
=> 144X = 11
=> X = 11/144
b) 19,96 + 4,19 - 24,15 : \(\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)
=> 19,96 + 4,19 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)=23,15\)
=> 24,15 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 23,15
=> 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 1
=> \(x\cdot4-\frac{1}{4}=24,15\)
=> \(x\cdot4=24,15+\frac{1}{4}=24,4\)
=> x = 24,4 : 4 = 6,1
Còn câu c tương tự
*19,96+4,19-24,15:(1/4-1/4) *252/x=84/97
=19,96+4,19-24,15:0 x=252x97:84
=19,96+4,19-0 x=291
=24,15-0 * x-2/255=114/153
=24,15 x-2/255=38/51
x=38/51+2/255
x=64/85
a ) ko rõ đề
b )\(\frac{252}{x}=\frac{84}{97}\)
=> 252 . 97 = x . 84
=> x . 84 = 24 444
x = 24 444 : 84
x = 291
Vậy x = 291
c ) x - \(\frac{2}{255}=\frac{114}{153}\)
x = 114/153 + 2/255
x = 64/85
Vậy x = 64/85
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
Bài 1:
\(B=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{4}+\frac{3}{8}-\frac{5}{12}}+\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{8}}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)\(=\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{6}}{\frac{1}{2}\left(\frac{1}{2}+\frac{3}{4}-\frac{5}{6}\right)}+\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{8}\right)}{\frac{1}{4}+\frac{1}{5}-\frac{1}{8}}\)
\(=\frac{1}{\frac{1}{2}}+3\) \(=2+3\) \(=5\)
Vậy B=5
Bài 2:
a) x3 - 36x = 0
=> x(x2-36)=0
=> x(x2+6x-6x-36)=0
=> x[x(x+6)-6(x+6) ]=0
=> x(x+6)(x-6)=0
\(\Rightarrow\orbr{\begin{cases}^{x=0}x+6=0\\x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}^{x=0}x=-6\\x=6\end{cases}}\)
Vậy x=0; x=-6; x=6
b) (x - y = 4 => x=4+y)
x−3y−2 =32
=>2(x-3) = 3(y-2)
=>2x-6= 3y-6
=>2x-3y=0
=>2(4+y)-3y=0
=>8+2y-3y=0
=>8-y=0
=>y=8 (thỏa mãn)
Do đó x=4+y=4+8=12 (thỏa mãn)
Vậy x=12 và y =8
B= 1/2 + 3/4 - 5/6/1/2(1.2 + 3/4 - 5/6) + 3(1/4+ 1/5 - 1/8)/ 1/4 1/5 - 1/8
B= 1/ 1/2 + 3
B= 2+3
B=5
B2:
a) x^3 - 36x = 0
x(x^2 - 36) = 0
=> x=0 hoặc x^2-36=0
=> x= 0 hoặc x^2=36
=> x=0 hoặc x= +- 6