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{ [ 27 * 11 - ( x + 12 ) * 8 ] * 3 - 79 } : 5 = 100
[ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 100 * 5
[ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 500
[ 297 - ( x + 12 ) * 8 ] * 3 = 500 + 79
[ 297 - ( x + 12 ) * 8 ] * 3 = 579
297 - ( x + 12 ) * 8 = 579 : 3
297 - ( x + 12 ) * 8 = 193
( x + 12 ) * 8 = 297 - 193
( x + 12 ) * 8 = 104
x + 12 = 104 : 8
x + 12 = 13
x = 13 - 12
x = 1.^^
a) 12 : x + 5 = 33
12 : x = 33 - 5
12 : x = 28
x = 12 : 28
x = 1/4
Vậy x = 1/4
b) 5*(x + 19) = 170 - 50
5*(x+19) = 120
x + 19 = 120 : 5
x + 19 = 24
x = 24 - 19
x = 5
Vậy x = 5
c) 11*(x - 6) - 11 = 4*x
11*(x-7) = 4x
11x - 77 = 4x
11x - 4x = 77
7x = 77
x = 11
Vậy x = 11
\(\left\{\left[27\times11-\left(x+12\right)\times8\right]\times3-79\right\}\div5=100\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=100\times5\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=500\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3=500+79\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3=579\)
\(27\times11-\left(x+12\right)\times8=579:3\)
\(27\times11-\left(x+12\right)\times8=193\)
\(297-\left(x+12\right)\times8=193\)
\(\left(x+12\right)\times8=297-193\)
\(\left(x+12\right)\times8=104\)
\(x+12=104:8\)
\(x+12=13\)
\(x=13-12\)
\(x=1\)
\(k\)\(dung\)\(cho\)\(minh\)\(nhe!\)
Bài 1:
\(A=\frac{5}{3.6}+\frac{5}{6.9}+....+\frac{5}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{3}{3.6}+\frac{3}{6.9}+....+\frac{3}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)
Bái 2:
\(B=\frac{2}{3.7}+\frac{2}{7.11}+...+\frac{2}{99.103}\)
\(\Rightarrow2B=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{99.103}\)
\(\Rightarrow2B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}\div2=\frac{50}{309}\)
Bài 1:
Ta có:
\(\frac{5}{n.\left(n+3\right)}=\frac{5}{3}.\frac{3}{n.\left(n+3\right)}=\frac{5}{3}.\frac{\left(n+3\right)-n}{n.\left(n+3\right)}=\frac{5}{3}.\left[\frac{n+3}{n.\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right]\)\(=\frac{5}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+...+\frac{5}{96.99}=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\right)\)
\(\left(2.x-4\right).\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2.x-4=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
\(\orbr{\begin{cases}2X-4=0\\X-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}X=2\\X=1\end{cases}}\)
Bạn phải sửa dấu ngoặc như thế này mới đúng này.
{ [ 27 x 11 - ( x +12) x 8 ] x 3 - 79 } = 100
= [ 297 - ( x +12) x 8 ] x 3 - 79 = 100
= [ 297 - ( x + 12) x 8] x 3 = 100 + 79
= [ 297 - ( x + 12) x 8] x 3 = 179
= [ 297 - ( x + 12) x 8] = 79 : 3
mình làm đến đây rùi mới nhận ra là thiếu đề hoặc đề sai