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V )x+(x+1)+(x+2)+....+(x+30)=1240
31 . x + (1 + 2 + 3 + 4 +...+ 29 + 30) = 1240
31 . x + 31.15 = 1240
31 . x = 1240 - 31.15
31 . x = 775
x = 775 : 31
x = 25
c) 10 + 2x = 2(32 - 1)
10 + 2x = 16
2x = 6
x = 3
d) 2x - 138 = 23.32
2x - 138 = 72
2x = 72 + 138
2x = 210
x = 210/2 = 105
a) 1032 : (x : 4) = 6
x : 4 = 1032 : 6 = 172
X = 172 x 4 = 688
b) 123 - 5(x + 4) = 38
5(x + 4) = 123 - 38 = 85
x + 4 = 85/5 = 17
x = 17 - 4 = 13
Bài 1 :
a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\)
= \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\)
b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\)
= \(10+45-455+750=350\)
c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\)
= \(\left(-33\right)+\left(-15\right)+20-5-57=-90\)
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 \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
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\(a)\) Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
a) \(\left(x-2\right).\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)
b) \(\left(3x+9\right).\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-9\\3x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
c) (31 - 2x)3 =27
(31 - 2x)3 = 33
=> 31 - 2x = 3
2x = 31 - 3
2x = 28
x = 14
a. \(\left(x-2\right).\left(2x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(x=2\)hoặc \(x=\frac{1}{2}\)
b.\(\left(3x+9\right).\left(1-3x\right)=0\Leftrightarrow\orbr{\begin{cases}3x+9=0\\1-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}}\)
Vậy \(x=-3\)hoặc \(x=\frac{1}{3}\)
c.\(\left(31-2x\right)^3=-27\)
\(\Leftrightarrow\left(31-2x\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow31-2x=-3\)
\(2x=34\)
\(x=17\)
d.\(\left(x-2\right).\left(7-x\right)=0\Leftrightarrow\orbr{\begin{cases}x-2=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}}\)
Vậy \(x=2\)hoặc \(x=7\)
e.\(\left(x-5\right)^5=32\)
\(\Leftrightarrow\left(x-5\right)^5=2^5\)
\(\Leftrightarrow x-5=2\Leftrightarrow x=7\)
f.\(\left(2-x\right)^4=81\)
\(\Leftrightarrow\left(2-x\right)^4=3^4\)
\(2-x=3\Leftrightarrow x=-1\)
g.\(\left|x-7\right|< 3\Leftrightarrow-3< x-7< 3\Leftrightarrow4< x< 10\)
bài 2 :
a) 123-5.(x+4)=38
5.(x+4)=123-38=85
x+4=85:5=17
x=17-4=13
b) 10 + 2x=2.(32-1)
10+2x =16
2x=16-10=6
x=6:2=3
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