Cho biểu thức:
A = (x + 1) + (2x + 4) + (3x +7) + (4x +10) +...+ (50x + a)
a) Tìm a
b) Tìm x để A=6275
K
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Những câu hỏi liên quan
25 tháng 1 2021
ĐKXĐ: \(x\notin\left\{-1;2;-2\right\}\)
a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
\(=\left(\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{\left(x-2\right)\left(x+2\right)}{3x\left(x+2\right)}\)
\(=\dfrac{x^3+3x^2+3x+1-2x^2-4x+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x-2}{3x}\)
\(=\dfrac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3x}{x-2}\)
\(=\dfrac{3x}{x-2}\)
b) Để A nguyên thì \(3x⋮x-2\)
\(\Leftrightarrow3x-6+6⋮x-2\)
mà \(3x-6⋮x-2\)
nên \(6⋮x-2\)
\(\Leftrightarrow x-2\inƯ\left(6\right)\)
\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{3;1;4;0;5;8;-4\right\}\)
Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)
TD
0
27 tháng 2 2022
a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)
\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)
\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)
b: Ta có: |2x-1|=5
=>2x-1=5 hoặc 2x-1=-5
=>x=-2
Thay x=-2 vào A, ta được:
\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)
c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)
=>x+3=24x+12
=>24x+12=x+3
=>23x=-9
hay x=-9/23
d: Để A<0 thì x+3<0
hay x<-3
LL
1
24 tháng 10 2021
a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)
\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)
\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)
\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)
b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)
\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)
\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)
\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)
6 tháng 11 2021
a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)
\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)
\(=-17x+18\)
HT
4
10 tháng 5 2022
a: Khi x=2/3 thì \(A=\dfrac{\dfrac{2}{3}-2}{\dfrac{2}{3}}=\dfrac{-4}{3}\cdot\dfrac{3}{2}=-2\)
b: \(B=\dfrac{4x}{x+1}-\dfrac{x}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x^2-4x-x^2-x+2x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x}{x+1}\)