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Bài 3:
Số nghịch đảo của $x$ là: $\frac{1}{x}$.
Theo bài ra ta có:
$5.\frac{1}{x}=\frac{1}{2}$
$\frac{1}{x}=\frac{1}{2}:5=\frac{1}{10}$
$x=10$
Vậy $x=10$
Bài 2:
a)
\(\frac{7}{12}+\frac{x}{15}=\frac{1}{20}\)
\(\frac{x}{15}=\frac{1}{20}-\frac{7}{12}=\frac{-8}{15}\)
\(x=-8\)
b)
\(x=\frac{1}{2}+25\text{%}x=\frac{1}{2}+\frac{x}{4}\)
\(\frac{3}{4}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{3}{4}=\frac{2}{3}\)
c)
\(x+\frac{-7}{15}=-1\frac{1}{20}=\frac{-21}{20}\)
\(x=\frac{-21}{20}+\frac{7}{15}=\frac{-7}{12}\)
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\(-1\frac{5}{7}.15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{7}-\frac{4}{5}+\frac{1}{7}\right)\)
\(=\frac{-12}{7}.15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{-13}{35}\right)\)
\(=\left(\frac{-12}{7}+\frac{-2}{7}\right).15+39\)
\(=\left(-2\right).15+39\)
\(=\left(-30\right)+39\)
\(=9\)
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a/ \(\frac{2}{3}+\frac{4}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{82}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow82< x< 92\)
\(\Rightarrow x=\left\{83;84;85;86;87;88;89;90;91\right\}\)
b/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)
\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)
\(\Rightarrow-1\le x\le7\)
\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)
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bạn có thể giải thích rõ hơn không, tại sao không có X sao mà tính được
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= 15.(-12/7 -2/7) + (-105)(-2/15 +1/7)
=15.(-2) + (-105). 1/105
= -30-1
= -31
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a. 5x +2 4/5 = 15
=> 5x + 14/5 = 15
=> 5x = 15 - 14/5 = 61/5
=> x = 61/5 : 5
=> x = 61/25
a)5x + 2 4/5 =15
5x+14/5=15
5x=15- 14/5
5x=71/5
x=71/5 :5
x=71
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ra kết quả thui nhé!
a/ 120
b/ -100
c/ 430
d/ -98
e/ A=-32
f/ B=1/10
tik nhá
\(\frac{-105}{2}< x< \frac{20}{7}\)
\(\Rightarrow\frac{-735}{14}< x< \frac{40}{14}\)
\(\Rightarrow x\in\left\{\frac{-734}{14};\frac{-733}{14};...;\frac{39}{14}\right\}\)
Đề bạn thiếu \(x\inℤ\)
\(\frac{-105}{2}=-52.5;\frac{20}{7}\approx2.85\)
\(\Rightarrow-52.5< x< 2.85\left(x\inℤ\right)\)
\(\Rightarrow x\in\left\{-52;-51;-50;...-1;0;1;2\right\}\)