Tìm x, biết: 3 x 7 + 1 : ( - 4 ) = - 1 28
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a) Ta có : ( x + 1 ).( 3 - x ) > 0
Th1 : \(\hept{\begin{cases}x+1>0\\3-x>0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x>3\end{cases}\Rightarrow}x>3}\)
Th2 : \(\hept{\begin{cases}x+1< 0\\3-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x< 3\end{cases}\Rightarrow}x< -1}\)
A/ \(\frac{4}{5}+\frac{1}{5}x=\left(-\frac{1}{28}\right)\)
\(\frac{1}{5}x=\left(-\frac{1}{28}\right)-\frac{4}{5}\)
\(\frac{1}{5}x=\left(-\frac{5}{140}\right)-\frac{112}{140}\)
\(\frac{1}{5}x=-\frac{117}{140}\)
\(x=\left(-\frac{117}{140}\right):\frac{1}{5}\)
\(x=\left(-\frac{117}{28}\right)\)
B/ \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\left(-\frac{1}{28}\right).\left(-4\right)\)
\(\frac{3x}{7}+1=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1\)
\(\frac{3x}{7}=\frac{1}{7}-\frac{7}{7}\)
\(\frac{3x}{7}=-\frac{6}{7}\)
\(\Rightarrow3x=\left(-6\right)\)(theo cách l8)
\(\Rightarrow x=\left(-6\right):3=\left(-2\right)\)
a. \(\frac{4}{5}+\frac{1}{5}x=\frac{1}{28}\)
\(\Leftrightarrow\frac{1}{5}x=\frac{1}{28}-\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{5}x=\frac{5}{140}-\frac{112}{140}\)
\(\Leftrightarrow\frac{1}{5}x=\frac{-107}{140}\)
\(\Leftrightarrow x=\frac{-107}{140}\div\frac{1}{5}\)
\(\Leftrightarrow x=\frac{-107}{140}\times5\)
\(\Leftrightarrow x=\frac{-107}{28}=\)
b. \(\left(\frac{3x}{7}+1\right)\div\left(-4\right)=\frac{-1}{28}\)
\(\Leftrightarrow\left(\frac{3x}{7}+\frac{7}{7}\right)=\frac{-1}{28}\times\left(-4\right)\)
\(\Leftrightarrow\frac{3x+1}{7}=\frac{1}{7}\)
\(\Leftrightarrow3x+1=1\)
\(\Leftrightarrow3x=1-1\)
\(\Leftrightarrow3x=0\Leftrightarrow x=0\)
Lớp 4?
a) \(\dfrac{x+1}{4}=\dfrac{36}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=144\)
\(\Rightarrow\left[{}\begin{matrix}x+1=12\\x+1=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=-13\end{matrix}\right.\)
Vậy: \(x\in\left\{11;-13\right\}\)
b) \(\dfrac{x}{7}=\dfrac{55x-4}{28}\)
\(\Rightarrow4x=55x-4\)
\(\Rightarrow-51x=-4\)
\(\Rightarrow x=\dfrac{4}{51}\)
Vậy: \(x=\dfrac{4}{51}\)
a) \(\dfrac{x + 1}{4} = \dfrac{36}{x + 1} \)
\(\Rightarrow\) \(( x + 1 )( x + 1 ) = 36 . 4 \)
\(\Rightarrow ( x + 1 )^2 = 144 \)
\(\Rightarrow ( x + 1 )^2 = 12^2 = ( -12 )^2 \)
\(\Rightarrow\) \(x + 1 ∈ \) { \(12 ; -12 \) }
\(\Rightarrow \) \(x \) \(∈ \) { \(11 ; -13 \) }
Vậy \(x ∈ \) { \(11 ; -13 \) }
(x+1) + (x+4) + (x+7) + .............+ (x+28) = 195
x+1 + x+4 + x+7 +.............+ x+28=195
Có số số x là:
(28-1) : 3+ 1 = 10
x x 10 + (1 + 4 + 7 + ...+ 28) = 195
x x 10 + 145 = 195
x x 10 = 195 - 145
x x 10 = 50
x = 5
=>(x+x+...+x)+(1+4+7+...+28)=195
có:(28-1):3+1=10(số x)
=>1+4+7+...+28=10x(10+1):2=55
=>10x+55=195
=>10x=195-55=140
=>x=140:10
=>x=14
tick nhé
(x + 1 ) + (x + 4 ) + ( x + 7 ) + ... + ( x + 28) = 195
=> 10x + (1 + 4 + 7 +...+ 28) = 195
Áp dụng công thức tính dãy số ta có
\(1+4+7+...+28=\frac{\left[\left(28-1\right):3+1\right].\left(28+1\right)}{2}=\frac{10.29}{2}=29.5=145\)
=> 10x + 145 = 195
=> 10x = 50
=> x = 5
(x + 1) + (x + 4) + (x + 7) + .... + (x + 28) = 195
<=>(x+x+x+...+x)+(1+4+7+...+28)=195
<=>10x+145=195
<=>10x=50
<=>x=5
Số các số hạng là:
(28 - 1) : 3 + 1 = 10 (số)
Ta có: (x + 1) + (x + 4) + (x + 7) + ... + (x + 28) = 195
=> 10x + (1 + 4 + 7 + ... + 28) = 195
=> 10x + 195 = 195
=> 10x = 0
=> x = 0
Vậy x = 0
(x+1)+(x+4)+(x+7)+.........+(x+28)
= (x nhân 10 ) + (1+4+....+28)
= x nhân 10 + 145 = 195
=> x nhân 10 = 195 - 145
x nhân 10 = 50
x = 50 : 10
x = 5