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Câu 1 : ( 2+x)2 =49
(2+x)2= 72
2+x = 7
=> x = 7 - 2
=>x=5
câu 2: 10 +2x = 45:43
10+2x = 42
10+2x = 16
2x = 16-10
2x = 6
=>x= 6 :2
=>x =3
câu 3: [ ( 4x +28) . 3+55 ] :5 = 35
(4x +28).3+55 = 35 . 5
(4x +28).3+55 = 175
( 4x+28).3 = 175 - 55
(4x+28).3 = 120
4x+28 = 120 :3
4x+28 = 40
4x = 40 - 28
4x = 12
x = 12:4
x = 3
**** cho mình nha!!!
Tự ghi đề nhé!
a. \(\frac{-3}{4}:x=\frac{-11}{36}+\frac{1}{2}\)
= 7/36
x = 7/36 : -3/4 = -7/27
k) \(2x-49=5.3^2\)
\(2x-49=45\)
\(2x=49+45\)
\(2x=94\)
\(x=47\)
l) \(3^2.\left(x+14\right)-5^2=5.2^2\)
\(9.\left(x+14\right)-25=20\)
\(9.\left(x+14\right)=45\)
\(x+14=5\)
\(x=-9\)
m) \(6x+x=5^{11}:5^9+3^1\)
\(7x=5^{11-9}+3\)
\(7x=5^2+3\)
\(7x=28\)
\(x=4\)
n) \(7x-x=5^{21}:5^{19}+3.2^2-\left(-7^{-0}\right)\)
\(6x=5^{21-19}+12-1\)
\(6x=5^2+11\)
\(6x=36\)
\(x=6\)
o) \(7x-2x=6^{17}:6^{15}+44:11\)
\(5x=6^{17-15}+4\)
\(5x=6^2+4\)
\(5x=40\)
\(x=8\)
o)7x-x=521:519+3.22-70
=> 6x = 5^2 + 12 -1
=> 6x = 36
=> x = 36/6 = 6
Kết quả 6
Học tốt
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
32x -1 = 243
32x -1 = 35
2x-1=5
2x=5+1
2x=6
x=6:2
x=3
(2 x-1)3 = 27
(2 x-1)3 =33
2x-1=3
2x=3+1
2x=4
x=4:2
x=2
\(x^3-\frac{1}{49}x=0\)
\(\Rightarrow x^2.x-\frac{1}{49}.x=0\)
\(\Rightarrow x\left(x^2-\frac{1}{49}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-\frac{1}{49}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{49}\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)
Vậy x = 0 hoặc x = \(\frac{1}{7}\)
a) 32x+2 = 9x
32.(x+1) = 9x
9x+1 = 9x
=> x + 1 = x
=> x + 1 - x =0 ( vô lí)
=> không tìm được x
b) x20 = x
=> x20 - x = 0
x.(x19 -1) = 0
=> x = 0
x19 - 1 = 0 => x19 = 1 => x = 1
KL:...
phần c lm tương tự như phần b nha bn
\(C=3^{10}+3^{11}+...+3^{50}\)
\(\Rightarrow3C=3^{11}+3^{12}+...+3^{51}\)
\(\Rightarrow3C-C=\left(3^{11}+...+3^{51}\right)-\left(3^{10}+...+5^{50}\right)\)
\(\Rightarrow2C=3^{51}-3^{10}\)
\(\Rightarrow C=\frac{3^{51}-3^{10}}{2}\)
a) => 2x2 = 50 => x2 = 25 => x = 5 hoặc x = -5
b) => \(|-9-x|=17\)
=> -9 - x = 17 hoặc -17
=> x = -26 hoặc x = 9
a, \(\left(2x-1\right)^2:9=49\)
\(\left(2x-1\right)^2=441\)
\(\Rightarrow\orbr{\begin{cases}2x-1=441\\2x-1=-441\end{cases}\Rightarrow\orbr{\begin{cases}x=221\\x=-220\end{cases}}}\)
b, \(3^x+3^{x+2}=810\)
\(3^x+3^x.3^2=810\)
\(3^x\left(1+3^2\right)=810\)
\(3^x.10=810\)
\(3^x=81=3^4\)
\(\Rightarrow x=4\)
\(\hept{\begin{cases}80⋮x\\56⋮x\end{cases}}\Rightarrow x\inƯC\left(80;56\right)\)
\(80=2^4.5\)
\(56=2^3.7\)
\(ƯCLN\left(80;56\right)=2^3=8\)
\(\RightarrowƯC\left(80;56\right)=Ư\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Mà \(x\ge3\)
\(\Rightarrow x\in\left\{4;8\right\}\)
\(=>\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\left[{}\begin{matrix}2x=7+3\\2x=-7+3\end{matrix}\right.\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\left[{}\begin{matrix}x=10:2\\x=-4:2\end{matrix}\right.\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
( 2x - 3)2 = 49
( 2x - 3)2 = 72
\(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=7+3\\2x=-7+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=10:2\\x=-4:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)