Tìm x \(\in\)N, biết:
6x . ( 7+72 +73+ .....+752) + 7=753
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c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
\(a,\Rightarrow x+2=-40\\ \Rightarrow x=-42\\ b,\Rightarrow6x-7-2x=5\\ \Rightarrow4x=12\Rightarrow x=3\\ c,\Rightarrow68-56-x=-2\\ \Rightarrow12-x=-2\\ \Rightarrow x=14\)
a) Xem lại đề.
b) 5. (x - 3)2 + 17 = 142
5. (x - 3)2 = 142 - 17
5. (x - 3)2 = 125
(x - 3)2 = 125 : 5
(x - 3)2 = 25
(x - 3)2 = 52 (cùng số mũ)
⇒ x - 3 = 5
x = 5 + 3 = 8
c) [(6x - 72) : 2 - 84] . 28 = 5628
[(6x - 72) : 2 - 84] = 5628 : 28
[(6x - 72) : 2 - 84] = 201
(6x - 72) : 2 = 201 + 84
(6x - 72) : 2 = 285
6x - 72 = 285. 2
6x - 72 = 570
6x = 570 + 72
6x = 642
x = 642 : 6
x = 107
a) \(3x-18.28:14=308\)
\(=>3x-36=308\)
\(=>3x=344\)\(\)
\(=>x=\frac{344}{3}\)
\(\)b)\(38+\left|x\right|=\left(-12\right)+65\)
\(=>38+\left|x\right|=53=>\left|x\right|=15=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
c)\(15+3.\left(x-1\right):5=30\)
\(=>15+3.\left(x-1\right)=150\)
\(=>3.\left(x-1\right)=135\)
\(x-1=45=>x=46\)
d) \(5.\left(7-3x\right)+7.\left(2+2x\right)=1\)
\(=>35-15x+14+14x=1\)
\(=>49-x=1=>x=48\)
e)\(3.\left(x-7\right)=21\)
\(=>x-7=7=>x=14\)
g)\(\left[\left(6x-72\right):2-84\right]:28=5628\)
\(\left(6x-72\right):2-84=157584\)
\(\left(6x-72\right):2=157668\)
\(6x-72=315336\)
\(6x=315408=>x=52568\)
h)\(123-5\left(x+4\right)=28\)
\(5.\left(x+4\right)=95\)
\(x+4=19=>x=15\)
p)\(14.\left(x-3\right)-138=73\)
\(14.\left(x-3\right)=211\)
\(x-3=\frac{211}{14}=>x=\frac{253}{14}\)
\(t\)
\(\left|x\right|-5=2\)
\(\left|x\right|=7=>\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
v)\(4x-72=\left|100-8.25\right|\)
\(4x-72=\left|-100\right|\)
\(4x-72=100=>4x=172=>x=43\)
cậu có thể tham khảo bài làm trên đây ạ, chúc cậu học tốt ^^
\(2x+1+\frac{1}{6}+1+\frac{1}{12}+..+1+\frac{1}{90}=10\)
=> 2x + 8 + \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=10\)
=> 2x + \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=10-8\)
\(2x+1-\frac{1}{10}=2\)
=> 2x + \(\frac{9}{10}=2\)
=> 2x = 2 - 9/10
=>2x = 11/10
=> x = 11/10 : 2
x = 11/20
Đặt \(S=7+7^2+7^3+...+7^{52}\)
\(\Rightarrow7S=7^2+7^3+7^4+...+7^{53}\)
\(\Rightarrow7S-S=6S=7^2+7^3+7^4+...+7^{53}-\left(7+7^2+7^3+...+7^{52}\right)\)
\(\Leftrightarrow6S=7^{53}-7\)
\(\Rightarrow S=\frac{7^{53}-7}{6}\)
Thay vào biểu thức ta có:
\(\frac{6x\left(7^{53}-7\right)}{6}+7=7^{53}.\)
\(\Leftrightarrow x\left(7^{53}-7\right)=7^{53}-7\)
\(\Rightarrow x=1\)
đặt A=\(7+7^2+...+7^{52}\)
7A =\(7^2+7^3+...+7^{53}\)
=>7A-A=\(\left(7^2+7^3+...+7^{53}\right)-\left(7+7^2+...+7^{52}\right)\)
6A=\(7^{53}+7\)
A=\(\frac{7^{53}+7}{6}\)
TA CÓ : 6X.\(\left(7+7^2+...+7^{52}\right)\)+7=\(7^{53}\)
(=)6X.\(\frac{7^{53}+7}{6}\)+7=\(7^{53}\)
(=)6X.\(\frac{7^{53}+7}{6}\)=\(7^{53}-7\)
(=)X.\(7.\left(7^{52}+1\right)\)=\(7.\left(7^{52}-1\right)\)
X=7.\(\left(7^{52}-1\right):[7.\left(7^{52}+1]\right)\)
X=\(\frac{7^{52}-1}{7^{52}+1}\)