Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
a) 72x-1 = 343
=> 72x-1 = 73
=> 2x - 1 = 3 => 2x = 4 => x = 2
b) (7x - 11)3 = 25.32 + 200
=> (7x - 11)3 = 32.9 + 200
=> (7x - 11)3 = 488
xem kĩ lại đề này :vvv
c) 174 - (2x - 1)2 = 53
=> (2x - 1)2 = 174 - 53
=> (2x - 1)2 = 174 - 125 = 49
=> (2x - 1)2 = (\(\pm\)7)2
=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)
Bài 2 :
a) x5 = 32 => x5 = 25 => x = 2
b) (x + 2)3 = 27
=> (x + 2)3 = 33
=> x + 2 = 3 => x = 3 - 2 = 1
c) (x - 1)4 = 16
=> (x - 1)4 = 24
=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)
d) (x - 1)8 = (x - 1)6
=> (x - 1)8 - (x - 1)6 = 0
=> (x - 1)6 [(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)
+) x - 1 = 1 => x = 2 ( tm)
+) x - 1 = -1 => x = 0 ( tm)
Vậy x = 1,x = 2,x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1 tính giá trị biểu thức
( - 25 ) nhân ( -3 ) nhân x với x = 4
\(\left(-25\right).\left(-3\right).4\)
\(=\left(-25\right).4.\left(-3\right)\)
\(=-100.\left(-3\right)=300\)
( -1 ) nhân ( -4 ) nhân 5 nhân 8 nhân y với y =25
\(\left(-1\right).\left(-4\right).5.8.25\)
\(=4.5.8.25=4.25.5.8\)
\(=100.40=40000\)
( 2ab mũ 2 ) : c với a =4 ; b= -6 ; c =12
\(\left(2.4.\left(-6\right)\right)^2:12\)
\(=\left(-48\right)^2:12\)
\(=2304:12=192\)
[ ( -25 ) nhân ( - 27 ) nhân ( -x ) ] : y với x = 4 ; y = -9
\(\left[\left(-25\right).\left(-27\right).\left(-4\right)\right]:-9\)
\(=-2700:\left(-9\right)\)
\(=300\)
(a mũ 2 _ b mũ 2) : ( a + b ) nhân ( a _ b ) với a + 5 , b = -3
\(\left(5^2-\left(-3\right)^2\right):\left(5-3\right).\left(5+3\right)\)
\(=16:2.8\)
\(=8.8=64\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)
Đặt \(A=\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...+\frac{20}{53.55}\)
\(A=\frac{10.2}{11.13}-\frac{10.2}{13.15}-\frac{10.2}{15.17}-...-\frac{10.2}{53.55}\)
\(A=10\left(\frac{2}{11.13}-\frac{2}{13.15}-\frac{2}{15.17}-...-\frac{2}{53.55}\right)\)
\(A=10\left(\frac{1}{11}-\frac{1}{55}\right)\)
\(A=10.\frac{4}{55}\)
\(A=\frac{40}{55}\)\(A=10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{10}{3}.x+\frac{67}{4}=\frac{53}{4}\)
\(\frac{10}{3}.x=\frac{53}{4}-\frac{67}{4}\)
\(\frac{10}{3}.x=-\frac{7}{2}\)
\(x=-\frac{7}{2}:\frac{10}{3}\)
\(x=-\frac{21}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5.\left(x+35\right)=515\)
\(x+35=515:5\)
\(x+35=103\)
\(x=103-35\)
\(\Rightarrow x=68\)
\(\left(x+40\right).15=75.12\)
\(\Rightarrow\left(x+40\right)=5.12\)[ rút gọn theo t/c đẳng thức ]
\(x+40=60\)
\(x=20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(7x -11)3 - 200 = 25. 52
(7x - 11)3 - 200 = 32 . 25
(7x - 11)3 - 200 = 800
(7x - 11)3 = 800 + 200
(7x - 11)3 = 1000
(7x -11)3 = 103
7x - 11 = 10
7x = 10 + 11
7x = 21
x = 21: 7
x = 3
Vậy x=3
a, 7*(x-3) + 4 = 18
<=> 7*(x+3) = 14
<=> x+3=2
<=> x= -1
b, ( 7x - 11)3 =825
<=> 7x - 11=\(\sqrt[3]{825}\)
<=> x=\(\frac{\sqrt[3]{825}+11}{7}\) (loại)
$\dfrac{2}{3}$