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320 : ( x - 1 ) = ( 53 - 52 ) : 4 + 15
<=> 320 : ( x - 1 ) = ( 125 - 25 ) : 4 + 15
<=> 320 : ( x - 1 ) = 100 : 4 + 15
<=> 320 : ( x - 1 ) = 40
<=> x - 1 = 8
<=> x = 9
240 : ( x - 5 ) = 22 . 52 - 20
<=> 240 : ( x - 5 ) = 4 . 25 - 20
<=> 240 : ( x - 5 ) = 80
<=> x - 5 = 3
<=> x = 8
70 : ( x - 3 ) = ( 34 - 1 ) : 4 - 16
<=> 70 : ( x - 3 ) = 80 : 4 - 16
<=> 70 : ( x - 3 ) = 4
<=> x - 3 = 35/2
<=> x = 41/2
a) 320: (x-1) = (53 - 52):4 + 15 b) 240: (x-5) = 22.52 - 20 c) 70:(x-3) = (34 - 1):4 - 16
320: (x-1) = (125- 25): 4 + 15 240: (x-5) = 4.25 - 20 70:(x-3) = (81-1): 4 - 16
320: (x-1) = 100:4 + 15 240: (x-5) = 100 - 20 70: (x-3) = 80: 4 - 16
320: (x-1) = 25 + 15 240: (x-5) = 80 70: (x-3) = 20 -16
320: (x-1) = 40 (x-5) = 240:80 70: (x-3) = 4
(x-1) = 320:40 (x-5) = 3 (x-3) = 70:4
(x-1) = 80 x = 5+3 (x-3) = 17,5
x = 80+1 x = 8 x = 17,5 : 3
x = 81 x = 5,83

Trả lời
a)
\(x^2:\frac{16}{11}=\frac{11}{4}\)
\(\Leftrightarrow x^2=\frac{11}{4}\cdot\frac{16}{11}\)
\(\Leftrightarrow x^2=\frac{16}{4}\)
\(\Leftrightarrow x^2=\left(\frac{4}{2}\right)^2\)
\(\Leftrightarrow x=\frac{4}{2}\)
Vậy x=\(\frac{4}{2}\)
b) (bạn thiếu nhóm \(\frac{1}{10\cdot13}\))
Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\)
\(\Rightarrow3A=3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\right)\)
\(\Rightarrow3A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\)
\(\Rightarrow3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}\)
\(\Rightarrow3A=1-\frac{1}{19}\Leftrightarrow3A=\frac{18}{19}\)
\(\Rightarrow A=\frac{18}{19}:3\Leftrightarrow A=\frac{6}{19}\)

a) x2=16
<=> x = \(\sqrt{16}\)
<=> x = 4
b) x - 12 = 4
<=> x = 4 + 1
<=> x = 5
a)\(x^2=16\)
\(\Leftrightarrow x^2=4^2\)
\(\Leftrightarrow x=4\)
vậy...
b)\(x-\left(1\right)^2=4\)
\(\Leftrightarrow x-\left(1\right)^2=\left(\pm2\right)^2\)
\(\Leftrightarrow x-1=\pm2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
vậy...
k mik nhé

a) \(\frac{4}{7}=\frac{12}{21}=\frac{28}{49}=\frac{52}{91}\)
b) \(\frac{4}{5}=\frac{12}{15}=\frac{16}{20}=\frac{8\cdot\left(16-15\right)}{10}\)
=> x,y,y phù hợp vs từng vị trí
hok tốt

Để \(\left(n+8\right)⋮\left(n+5\right)\) thì
\(\left(n+8\right)-\left(n+5\right)⋮\left(n+5\right)\)
\(\Rightarrow\)\(3⋮\left(n+5\right)\)
\(\Rightarrow\)\(\left(n+5\right)\inƯ\left(3\right)\)
\(\Rightarrow\)\(\left(n+5\right)\in\left(1;-1;3;-3\right)\)
\(\Rightarrow\)\(n\in\left(-4;-6;-2;-8\right)\)
Để \(\left(16-3n\right)⋮\left(n+4\right)\) thì
\(\left(16-3n\right)+\left(n+4\right)⋮\left(n+4\right)\)
\(\Rightarrow\)\(\left(16-3n\right)+3\left(n+4\right)⋮\left(n+4\right)\)
\(\Rightarrow\)\(16-3n+3n+12⋮\left(n+4\right)\)
\(\Rightarrow\)\(28⋮\left(n+4\right)\)
\(\Rightarrow\)\(\left(n+4\right)\inƯ\left(28\right)\)
\(\Rightarrow\)\(\left(n+4\right)\in\left\{\pm1;\pm2;\pm4;\pm7;\pm14;\pm28\right\}\)
\(\Rightarrow\)\(n\in\left\{-3;-4;-2;-6;0;-8;3;-11;10;-18;24;-32\right\}\)
x4=16
=>x=16:4
=>x=4
x*4=16
x=16:4
x=4