Tìm các số nguyên x, biết:
a) 10⋮ (x - 1)
b) (x + 5) ⋮ (x - 2)
c) (3x + 8) ⋮ (x - 1)
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a) \(x.\left(x+7\right)=0\)
TH1:\(x=0\)
\(\Rightarrow0.\left(0+7\right)=0.7=0\)
\(TH2:x+7=0\)
\(\Rightarrow x=0-7\)
\(\Rightarrow x=-7\)
Vậy \(x\in\left\{-7;0\right\}\)
b) \(\left(x+12\right).\left(x-3\right)\)
\(TH1:x+12=0\)
\(\Rightarrow x=0-12\)
\(\Rightarrow x=-12\)
\(TH2:x-3=0\)
\(\Rightarrow x=0+3\)
\(\Rightarrow x=3\)
Vậy\(x\in\left\{-12;3\right\}\)
a) \(-45:5.\left(-3-2x\right)=3\)
\(-9.\left(-3-2x\right)=3\)
\(-3-2x=\left(3:-9\right)\)
\(-3-2x=\dfrac{-1}{3}\)
\(-2x=-3-\dfrac{1}{3}\)
-2x=\(\dfrac{-10}{3}\)
\(x=\dfrac{-10}{3}:-2\)
\(x=\dfrac{5}{3}\)
b)
3x - 28 = x + 36
<=> 3x - x = 36 + 28
<=> 2x = 64
<=> x = 32
Vậy x = 32
c)
(-12)2.x = 56 + 10.13.x
144.x = 56 + 130.x
144x – 130 x = 56
14x = 56
x = 56: 14
x = 4
Vậy x = 4
Bài 4: Thực hiện phép tính:
a, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + 0,25 \(\times\) \(\dfrac{7}{9}\) - \(\dfrac{7}{9}\)
= \(\dfrac{7}{9}\) (0,75 + 0,25 - 1)
= 0
b, \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): (\(\dfrac{3}{8}\) + \(\dfrac{1}{2}\).[\(\dfrac{9}{4}\) - 2.(19,5 -18,5)])
= \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): ( \(\dfrac{3}{8}\) + \(\dfrac{1}{2}\). [ \(\dfrac{9}{4}\)- 2])
= \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): ( \(\dfrac{3}{8}\) + \(\dfrac{9}{8}\)- 1)
= \(\dfrac{5}{12}\) - \(\dfrac{1}{12}\): ( \(\dfrac{1}{2}\))
= \(\dfrac{5}{12}\) - \(\dfrac{1}{6}\)
= \(\dfrac{1}{4}\)
c, [ 50,2 - 0,2 .(12,05 - 10,05)] - 37,38
= [ 50,2 - 0,2. 2 ] - 37,38
= [ 50,2 - 0,4] - 37,38
= 49,8 - 37,38
= 12,42
Bài 3:
B = \(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}+\dfrac{1}{7.11}+...+\dfrac{1}{97.100}\)
B = \(\dfrac{1}{3}\).( \(\dfrac{3}{1.4}\)+ \(\dfrac{3}{4.7}\)+ \(\dfrac{3}{7.11}\)+...+ \(\dfrac{3}{97.100}\))
B = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) +...+ \(\dfrac{1}{97}\) - \(\dfrac{1}{100}\))
B = \(\dfrac{1}{3}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{100}\))
B = \(\dfrac{1}{3}\). \(\dfrac{99}{100}\)
B = \(\dfrac{33}{100}\)
Số có hai chữ số có dạng: \(\overline{ab}\)
Khi thêm 12 vào bên trái số đó ta được số mới: \(\overline{12ab}\)
Theo bài ra ta có: \(\overline{12ab}\) : \(\overline{ab}\) = 26
(1200 + \(\overline{ab}\)): \(\overline{ab}\) = 26
1200: \(\overline{ab}\) + 1 = 26
1200: \(\overline{ab}\) = 26 - 1
1200: \(\overline{ab}\) = 25
\(\overline{ab}\) = 1200: 25
\(\overline{ab}\) = 48
Gọi số cần tìm là: ab
12ab = ab x 26
1200 + ab = ab x 26
1200 = ab x 25
⇒ ab = 1200 : 25 = 48
Vậy số cần tìm là: 48
Gọi số thập phân đúng cần tìm là \(x\) số còn lại là 149,96 - \(x\)
Số thập phân khi tính viết nhầm dấu phẩy là: \(x\) :10 = 0,1\(x\)
Theo bài ra ta có: 0,1\(x\) + 149,96 - \(x\) = 36,074
0,9\(x\) = 149,96 - 36,074
0,9\(x\) = 113,886
\(x\) = 126,54
Số còn lại là: 149,96 - 126,54 = 23,42
Kết luận:...
(-8,3)\(x\) + (-11,7).\(x\) = 0,2
\(x\).( -8,3 - 11,7) = 0,2
\(x\).(-20) = 0,2
\(x\) = 0,2: (-20)
\(x\) = -0,01
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
A = 2019 \(\times\) 2021 + 2023
A = (2018 + 1).(2022 -1) + 2023
A = 2018.2022 - 2018 + 2023 > 2018.2022 - 2022
Vậy A > B
Cách 1: Nhìn qua là biết A > B :))
Cách 2: Giải cụ thể:
A = 2019 x 2021 + 2023
= 2018 x 2021 + 2021 + 2023 = 2018 x 2021 + 4044
B = 2018 x 2022 - 2022
= 2018 x 2021 + 2018 - 2022 = 2018 x 2021 - 4
⇒ A > B và lớn hơn: 4044 + 4 = 4048
a) \(10⋮\left(x-1\right)\) (đkxđ \(x\ne1\))
\(\Rightarrow x-1\in\left\{-1;1;-2;2;-5;5;-10;10\right\}\)
\(\Rightarrow x\in\left\{0;2;-1;3;-4;6;-9;11\right\}\)
b) \(\left(x+5\right)⋮\left(x-2\right)\left(đkxđ,x\ne2\right)\)
\(\Rightarrow\left(x+5\right)-\left(x-2\right)⋮\left(x-2\right)\)
\(\Rightarrow x+5-x+2⋮\left(x-2\right)\)
\(\Rightarrow7⋮\left(x-2\right)\)
\(\Rightarrow x-2\in\left\{-1;1;-7;7\right\}\)
\(\Rightarrow x\in\left\{1;3;-5;9\right\}\)
c) \(\left(3x+8\right)⋮\left(x-1\right)\left(đkxd,x\ne1\right)\)
\(\Rightarrow\left(3x+8\right)-3\left(x-1\right)⋮\left(x-1\right)\)
\(\Rightarrow3x+8-3x+3⋮\left(x-1\right)\)
\(\Rightarrow11⋮\left(x-1\right)\)
\(\Rightarrow x-1\in\left\{-1;1;-11;11\right\}\)
\(\Rightarrow x\in\left\{0;2;-10;12\right\}\)
a) x∈{0;2;−1;3;−4;6;−9;11}
b) x∈{1;3;−5;9}
c) x ∈ {0;2;−10;12}