Tìm x 320 : ( x - 1 ) = ( 5^3 + 1 ) : 13
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a: Ta có: \(20:\left(x+1\right)=\left(5^2+1\right):13\)
\(\Leftrightarrow x+1=10\)
hay x=9
b: Ta có: \(320:\left(x-1\right)=2^2\cdot5^2-20\)
\(\Leftrightarrow x-1=4\)
hay x=5
1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)
\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)
\(x\times\left(8-25\right)=2\times29-33\)
\(x\times-17=25\)
\(x=-\dfrac{25}{17}\)
2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)
\(15\div\left(x+2\right)=\left(27+3\right)\div1\)
\(15\div\left(x+2\right)=30\div1\)
\(15\div\left(x+2\right)=30\)
\(x+2=\dfrac{1}{2}\)
\(x=-\dfrac{3}{2}\)
3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)
\(20\div\left(x+1\right)=\left(25+1\right)\div13\)
\(20\div\left(x+1\right)=26\div13\)
\(20\div\left(x+1\right)=2\)
\(x+1=20\div2\)
\(x+1=10\)
\(x=9\)
4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)
\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)
\(320\div\left(x-1\right)=100\div4+15\)
\(320\div\left(x-1\right)=25+15\)
\(320\div\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
5) \(240\div\left(x-5\right)=2^2\times5^2-20\)
\(240\div\left(x-5\right)=4\times25-20\)
\(240\div\left(x-5\right)=100-20\)
\(240\div\left(x-5\right)=80\)
\(x-5=30\)
\(x=35\)
6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)
\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)
\(70\div\left(x-3\right)=80\div4-10\)
\(70\div\left(x-3\right)=20-10\)
\(70\div\left(x-3\right)=10\)
\(x-3=7\)
\(x=10\)
2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
a. 8 + (x - 9) = 125 - 64
8 + (x - 9) = 61
x - 9 = 53
x = 62
b. 5 x (X + 7) - 10 = 8 x 5
5 x (X + 7) - 10 = 40
5 x (X + 7) = 50
X + 7 = 10
X = 3
=> 320 : (x - 1) = 100 : 4 + 15
=> 320 : (x - 1) = 25 + 15
=> 320 : (x - 1) = 40
=> x - 1 = 320 : 40
=> x - 1 = 8
=> x = 8 + 1
=> x = 9
Cái này dễ
320:(x-1)=(125-25):4+15
320:(x-1)=100:4+15
320:(x-1)=25+15
320:(x-1)=40
x-1=320:40
x-1=8
x=8+1
=>x=9
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
Ta thấy: Nếu thêm 11 vào số cần tìm sẽ được số mới chia hết cho 8 được thương hơn thương cũ 2 đơn vị và số mới cũng chia hết cho 12 được thương mới hơn thương cũ 1 đơn vị. Bài toán trở thành tìm một số chia hết cho 8 và 12, biết hiệu hai thương là 14
Gọi số mới là a
Ta có:
a/8 - a/12 = 14
3a/24 - 2a/24 = 14
a/24 = 14
a = 14 x 24
a = 336
Vậy: Số cần tìm là: 336 - 11= 325
\(\frac{1}{\left(x-1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+7\right)}+....+\frac{1}{\left(x+76\right).\left(x+80\right)}=-\frac{81}{320}\)
\(\Rightarrow\frac{4}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x+3\right)\left(x+7\right)}+...+\frac{4}{\left(x+76\right)\left(x+80\right)}=\frac{-81}{80}\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+7}+...+\frac{1}{x+76}-\frac{1}{x+80}=\frac{-81}{80}\)
\(\Rightarrow\frac{1}{x-1}-\frac{1}{x+90}=\frac{-81}{80}\)
Vì \(\frac{-81}{80}< 0\Rightarrow\frac{1}{x-1}< \frac{1}{x+90}\)
\(\Leftrightarrow x-1>x+90\)( luôn sai \(\forall x\in R\))
Vậy không tìm được x
ta có: \(320:\left(x-1\right)=\left(5^3+1\right):13\)
\(\Leftrightarrow320:\left(x-1\right)=\dfrac{126}{13}\)
\(\Leftrightarrow x-1=\dfrac{1391}{42}\)
hay \(x=\dfrac{1433}{42}\)