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(x+1)+(x+3)+(x+5)+....+(x+99)=0
=> (x+x+x...+x) + (1+3+5+...+99) = 0
=> 50x + 2500 = 0
=> 50x = -2500
=> x = -50
Vậy x = -50
a, \(390-\left(x-7\right)=13^2:12\)
\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)
\(x-7=390-\dfrac{169}{12}\)
\(x-7=\dfrac{4511}{12}\)
\(x=\dfrac{4511}{12}+7\)
\(x=\dfrac{4595}{12}\)
Vậy ...
b, \(\left(x-35.2^2\right):7=3^3-24\)
\(\left(x-35.4\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(\Leftrightarrow\left(x-140\right)=3.7\)
\(\Leftrightarrow x-140=21\)
\(\Leftrightarrow x=161\)
Vậy .....
c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)
\(x-3-\left(16.3-24\right):2:6=3\)
\(x-3-\left(48-24\right):2:6=3\)
\(x-3-24:2:6=3\)
\(x-3-2=3\)
\(x=3+2+3\)
\(x=8\)
Vậy ......
d) \(4x-5=5+5^2+5^3+.....+5^{99}\)
Đặt :
\(A=5+5^2+.........+5^{99}\)
\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)
\(\Leftrightarrow4A=5^{100}-5\)
\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)
\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)
Đến đây thì sao nữa nhỉ ?
e) \(\left(2x-1\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy ....
a)64:2mũ5×30×4
= 64 : 32 x 30 x 4
= 240
b)3 mũ 2× 5 - 2 mũ 2×7+2 mũ 0 × 5
= 9 x 5 - 4 x 7 + 1 x 5
= 45 - 28 + 5
= 22
c)2 mũ 3-5 mũ 3÷5 mũ 2 + 12×2 mũ 2
= 8 - 125 : 25 + 12 x 4
= 8 - 5 + 48
= 51
d)2[(7-3 mũ 3÷3 mũ 2) chia 2 mũ 2 + 99]-100
= 2[( 7 - 27 : 9) : 4 + 99] - 100
= 2[4 : 4 + 99] - 100
= 2. 100 - 100
= 200 - 100
= 100
e)4[(3 + 3^7:3^4)chia 10 + 97]-300
= 4[( 3 + 3^3) : 10 + 97] - 300
= 4[ 30 : 10 + 97 ] - 300
= 4. 100 - 300
= 400 - 300
= 100
f)2^2 x 5 [(5 mũ 2 cộng 2 mũ 3) chia 11 - 2] - 3^2 x 2
= 4 x 5 [ (25 + 8 ) : 11 - 2] - 9 x 2
= 20 [ 33 : 11 - 2] - 18
= 20. 1 - 18
= 20 - 18
= 2
gọi biểu thức trên là A , ta có :
\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+\dfrac{5}{3^5}-...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}\\ 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ \Rightarrow A+3A=\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)+\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\\ \Rightarrow4A\cdot3=12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)
từ đó ta được :
\(16A=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Rightarrow A=\dfrac{\dfrac{3-101}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\ \Rightarrow A=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}\)
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) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
Bài làm:
a) \(a=2+2^3+2^5+...+2^{99}+2^{101}\)
\(\Rightarrow4a=2^3+2^5+2^7+...+2^{101}+2^{103}\)
\(\Rightarrow4a-a=\left(2^3+2^5+2^7+...+2^{103}\right)-\left(2+2^3+2^5+...+2^{101}\right)\)
\(\Leftrightarrow3a=2^{103}-2\)
\(\Rightarrow a=\frac{2^{103}-2}{3}\)
Vậy \(a=\frac{2^{103}-2}{3}\)
b) \(b=1-5^3+5^6-5^9+...+5^{96}-5^{99}\)
\(\Rightarrow125b=5^3-5^6+5^9-5^{12}+...+5^{99}-5^{102}\)
\(\Rightarrow125b+b=\left(5^3-5^6+5^9-5^{12}+...+5^{99}-5^{102}\right)+\left(1-5^3+5^6-5^9+...+5^{96}-5^{99}\right)\)
\(\Leftrightarrow126b=1-5^{102}\)
\(\Rightarrow b=\frac{1-5^{102}}{126}\)
Vậy \(b=\frac{1-5^{102}}{126}\)
Học tốt!!!!
1 + 3 + 5 + ...... + 99
99 + 97 + 95 + .......+ 1
= 100 + 100 + .... + 100
có 25 số 100
= 100 . 25
= 2500
2500= ( x- 2)2
= 2500 + 2 = x2
= 2502 = x2
=2502 = x.x
chỉ giúp được đến đây thui nha
phần còn lại tự mà tìm hiểu