Tìm x,y sao cho:
a,81x = 325 - 144y
b,5^x + 99=10^y
c,2^x+1 x 3^y = 12^y
d,1y96 = (3x + 12)^2
5, Tìm x, y ϵ Z, sao cho:
a) y = \(\dfrac{6x-4}{2x+3}\) b) \(\dfrac{1}{x}-\dfrac{y}{2}=\dfrac{1}{4}\)
c) xy-3x+2y=5 d) (3x-5)(2x+1)=12
a) Để y nguyên thì \(6x-4⋮2x+3\)
\(\Leftrightarrow-13⋮2x+3\)
\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)
hay \(x\in\left\{-1;-2;5;-8\right\}\)
Bài 6. Tìm x Z, sao cho:
a. (x – 3) – 12 = – 25
f. 121 – (35 – x) = 2.52
c. 72 – (84 – 9x) : 7 = 69
g. 2(17 + x) – (400 – 325) = – 31
d. (x + 7).(x – 5) =0
h. (x – 10).(x2 – 9) = 0
e. (x + 8).(x2 + 1) = 0
i. 17 – { – x + [– x – (– x)]} = – 16
a: \(\Leftrightarrow x-3=-13\)
hay x=-10
8) Tìm x, y ϵ Z, sao cho:
a) 3|x-5|+|y+4| = 5
b) 2|3x|+|y+3| = 10
12) Tìm x, y ϵ Z, sao cho:
a) \(\dfrac{x}{2}\) - \(\dfrac{1}{y}\)= \(\dfrac{1}{3}\)
b) \(\dfrac{4}{x}\) + \(\dfrac{y}{2}\) = \(\dfrac{-1}{4}\)
1. Tính:
D = 3/5.7 + 3/7.9 + 3/9.11 + ... + 3/53.55
2. Tìm x sao cho:
A = x+2/x-5 là số nguyên dương
3. Chứng tỏ rằng:
S = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^20 nhỏ hơn 1
A = 1/10 + 1/11 + 1/12 + ... + 1/99 + 1/100 lớn hơn 1
Giúp mình với. Hu hu
1:
Ta có: \(D=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{53\cdot55}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{53\cdot55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{11}{55}-\dfrac{1}{55}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{2}{11}=\dfrac{3}{11}\)
2) Để A là số nguyên dương thì
\(\left\{{}\begin{matrix}x+2⋮x-5\\x-5>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5+7⋮x-5\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7⋮x-5\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-5\inƯ\left(7\right)\\x>5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\in\left\{1;-1;7;-7\right\}\\x>5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;4;12;-2\right\}\\x>5\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{6;12\right\}\)
Giải:
1)D=3/5.7+3/7.9+3/9.11+...+3/53.55
D=3/2.(2/5.7+2/7.9+2/9.11+...+2/53.55)
D=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+...+1/53-1/55)
D=3/2.(1/5-1/55)
D=3/2.2/11
D=3/11
Bài 1: tìm đạo hàm của các hàm số sau
1. y=6x2 -\(\dfrac{4}{x}\)+1
2. y=\(\dfrac{2x+1}{-x+1}\)
3. y= \(\sqrt{x^2-3x+4}\)
4. y=\(\dfrac{\left(x^2-1\right)\left(x+3\right)}{x-4}\)
5. y=\(\dfrac{1}{2x^2-3x+5}\)
6. y=(x+1)\(\sqrt{x^2-1}\)
1.
\(y'=12x+\dfrac{4}{x^2}\)
2.
\(y'=\dfrac{3}{\left(-x+1\right)^2}\)
3.
\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)
4.
\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)
\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)
5.
\(y'=-\dfrac{4x-3}{\left(2x^2-3x+5\right)^2}\)
6.
\(y'=\sqrt{x^2-1}+\dfrac{x\left(x+1\right)}{\sqrt{x^2-1}}\)
1)a)2x-13=25+6x
b)12-x=3x+6
c)40-(25-2x)=x
d)|x-3|=5
e)|x-3|+(x+2)+(x+1)=12
2)tính
a)(-1).(-2).(-3).(-4).(-5).[(-3)-(-5)]
b)1-2+3-4+5-6+...-98+99
3)tim x
a)(x-1).(y-2)=5
b)x.(y-3)=12
Bài 3: Tính giá trị của biểu thức
a) A = 2x(
1
2 x 2 + y) – x(x 2 + y) + xy(x 3 – 1) tại x = 10; y = –
1
10
b) B = 3x 2 (x 2 – 5) + x(–3x 3 + 4x) + 6x 2 tại x = –5
Tìm các số tự nhiên x và y , sao cho :
a) (2x+1)(y-3)=10 b) (3x-2)(2y-3)=1
c) (x+1)(2y-1)=12 d) x+6=y(x-1)
e) x-3=y(x+2)