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2
27 tháng 4 2022
`1 / 3 + 2 / 3 : x = -7`
`2 / 3 : x = -7 - 1 / 3`
`2 / 3 : x = -22 / 3`
`x = 2 / 3 : -22 / 3`
`x = -1 / 11`
PK
2
H9
HT.Phong (9A5)
CTVHS
12 tháng 1
\(\dfrac{-28}{4}< x\le\dfrac{-21}{7}\)
\(\Rightarrow-7< x\le-3\)
Nếu x ∈ Z thì:
\(x\in\left\{-6;-5;-4;-3\right\}\)
SV
Sinh Viên NEU
CTVVIP
12 tháng 1
\(\dfrac{-28}{4}< x\le\dfrac{-21}{7}\)
\(\Leftrightarrow-4< x\le-3\)
Nếu x nguyên thì x = -3
7 tháng 7 2021
a)=1/2 . 8/15 - 3/4.47/9
=4/15 - 47/12
=-73/20
b)=2-1/3 . -21/20
=2+7/20
=47/20
VN
1
S
12 tháng 4 2023
\(\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot...\cdot\dfrac{400}{399}\)
\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot...\cdot\dfrac{20\cdot20}{19\cdot21}\)
\(=\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot20\cdot20}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}\)
\(=\dfrac{2\cdot3\cdot4\cdot...\cdot20}{1\cdot2\cdot3\cdot...\cdot19}\cdot\dfrac{2\cdot3\cdot4\cdot...\cdot20}{3\cdot4\cdot5\cdot...\cdot21}\)
\(=20\cdot\dfrac{2}{21}\)
\(=\dfrac{40}{21}\)
PN
1
21 tháng 2 2023
\(\dfrac{4}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{2}{5}\cdot\dfrac{-3}{7}=\dfrac{15}{4}\)
\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)-\dfrac{-6}{35}=\dfrac{15}{4}\)
\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)+\dfrac{6}{35}=\dfrac{15}{4}\)
\(\Rightarrow\dfrac{2}{5}\left(\dfrac{2}{3}x-\dfrac{9}{5}\right)=\dfrac{15}{4}-\dfrac{6}{35}=\dfrac{501}{140}\)
\(\Rightarrow\dfrac{2}{3}x-\dfrac{9}{5}=\dfrac{501}{140}:\dfrac{2}{5}=\dfrac{501\cdot5}{28\cdot5\cdot2}=\dfrac{501}{56}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{501}{56}+\dfrac{9}{5}=\dfrac{2001}{280}\)
\(\Rightarrow x=\dfrac{2001}{280}:\dfrac{2}{3}=\dfrac{2001\cdot3}{280\cdot2}=\dfrac{6003}{560}\)
31 tháng 12 2021
a) (x-3)(y+2) = 7
=> (x−3)∈Ư(7);(y+2)∈Ư(7)(x−3)∈Ư(7);(y+2)∈Ư(7)
=> (x−3)∈{−7;−1;1;7}(x−3)∈{−7;−1;1;7}
(y+2)∈{−7;−1;1;7}
ta có bảng sau :
vạy có 4 cạp
xin like
7 tháng 7 2021
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
22 tháng 6 2019
Bài 2
\(a,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-5+1\right)\left(x-5-1\right)=0\)
\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)
\(\Rightarrow x\in\left\{4;5;6\right\}\)
\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)
\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)
\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)
Mà \(\frac{15}{2}\notin n\)
\(\Rightarrow x\in\left\{7;8\right\}\)
T
22 tháng 6 2019
#)Giải :
Bài 1 :
a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)
Bài 2 :
a) \(\left(x-5\right)^2=\left(x-5\right)^6\)
\(\Leftrightarrow x^4-625=x^6-15625\)
\(\Leftrightarrow x^6-x^4=15000\)
\(\Leftrightarrow x^6-x^4=5^6-5^4\)
\(\Leftrightarrow x=5\)
b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow2x-15=1\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
(x-1)^2=9
=> (x-1)^2 = 3^2
=> x-1= 3
=> x = 3 + 1
=> x =4
\(\left(x-1\right)^2=9\)
\(\Leftrightarrow\left(x-1\right)^2=3^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3+1\\x=-1+3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=2\end{cases}}}\)
Vậy \(x=4\)hoặc \(x=2\)
\(\left(x-3\right)^3+15=7\)
\(\Leftrightarrow\left(x-3\right)^3=7-15\)
\(\Leftrightarrow\left(x-3\right)^3=-8\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow x-3=-2\)
\(\Leftrightarrow x=-2+3\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)