p(x)= (x+2)+(x+5)+(x+7)+...+(x+152)-154
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CM
9 tháng 4 2019
a ) | x | − 3 4 = 5 3 = > | x | = 29 12 = > x = 29 12 x = − 29 12
b) x − 3 2 = 1 = > x − 3 2 = 1 x − 3 2 = − 1 = > x = 5 2 x = 1 2
c) x − 3 2 + 11 4 = 15 4 = > x − 3 2 = 1 = > x = 5 2 x = 1 2
d) 17 4 − 5 6 − x = − 7 4 = > 5 6 − x = 6 = > x = − 31 6 x = 41 6
e)
x + 1 5 2 − 9 25 = 0 = > x + 1 5 2 = 9 25 = > x + 1 5 = 3 5 x + 1 5 = − 3 5 = > x = 2 5 x = − 4 5
f) x + 1 5 2 + 17 25 = 26 25 = > x + 1 5 2 = 9 25 = > x = 2 5 x = − 4 5
VN
1
T
20 tháng 2 2020
\(\frac{x+143}{157}+\frac{x+146}{154}=\frac{x+149}{151}+\frac{x+152}{148}\)
\(\Leftrightarrow\frac{x+143}{157}+1+\frac{x+146}{154}+1=\frac{x+149}{151}+1+\frac{x+152}{148}+1\)
\(\Leftrightarrow\frac{x+300}{157}+\frac{x+300}{154}=\frac{x+300}{151}+\frac{x+300}{148}\)
\(\Leftrightarrow\left(x+300\right)\left(\frac{1}{157}+\frac{1}{154}-\frac{1}{151}-\frac{1}{148}\right)=0\)
có \(\frac{1}{157}+\frac{1}{154}+\frac{1}{151}+\frac{1}{148}\ne0\)
\(\Leftrightarrow x+300=0\)
\(\Leftrightarrow x=-300\)
CM
9 tháng 7 2019
a) 1 ≤ x ≤ 6 ⇒ x ∈ - 3 ; - 2 ; - 1 ; 0 ; 1
b) - 7 ≤ x ≤ - 3 ⇒ x ∈ - 3 ; - 2 ; - 1 ; 0 ; 1
CM
21 tháng 10 2018
a) – 3 < x < 2 => x ∈ {-2;-1;0;1}
b) − 4 21 < x 21 < 2 21 => x ∈ {-3;-2;-1;0;1}
c) 1 ≤ x ≤ 6 => x ∈ {-3;-2;-1;0;1}
d) -7 ≤ x ≤ -3 => x ∈ {-3;-2;-1;0;1}
2 tháng 7 2021
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
DK
2
10 tháng 2 2022
a: \(\Leftrightarrow\left(2x-5\right)^3=225-4\cdot100=125\)
=>2x-5=5
=>2x=10
hay x=5
b: \(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)
hay \(x\in\left\{2;5\right\}\)
c: =>x-5-2x-7=-8
=>-x-12=-8
=>x+12=8
hay x=-4
TV
3 tháng 3 2020
a, -154+(x-9-18)=40
x-27=40-(-154)
x-27=194
x=194+27
x=221
b,\(\left|9-x\right|\)=64+(-7)
\(\left|9-x\right|\)=57
TH1:
9-x=57
x=9-57
x=-48
TH2:
9-x=-57
x=9-(-57)
x=66
TD
6 tháng 4 2020
a, -154+(x-9-18)=40
x-27=40-(-154)
x-27=194
x=194+27
x=221
b,|9−x||9−x|=64+(-7)
|9−x||9−x|=57
TH1:
9-x=57
x=9-57
x=-48
TH2:
9-x=-57
x=9-(-57)
x=66