\(A=\dfrac{9x^2-4}{4x^2-1+\left(2x+1\right)\left(x-1\right)}\)
a) Rút gọn A.
b) x bằng mấy để A > 0
c) Tìm x thuộc Z để A thuộc Z.
\(A=\dfrac{9x^2-4}{4x^2-1+\left(2x+1\right)\left(x-1\right)}\)
a) Rút gọn A.
b) x bằng mấy để A > 0
c) Tìm x thuộc Z để A thuộc Z.
\(A=\left[\left(\dfrac{1}{x}+1\right)\cdot\dfrac{2}{x^3+3x^2+3x+1}+\dfrac{1}{1+2x+x^2}\cdot\left(1+\dfrac{1}{x^2}\right)\right]:\dfrac{x-1}{x^3}\)
a) Rút gọn
b) x bằng mấy để A = 3?
c) x thuộc Z, x bằng mấy để A thuộc Z?
Cho biểu thức:
A\(=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)
a/ Rút gọn A
b/ Tìm x ∈ Z để A nguyên
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 \(x\in\left\{3;1;4;0;5;8;-4\right\}\)
A= \(\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x-1}{x+3}\right):\dfrac{3x+12}{x^3-1}\)
a/ rút gọn A
b/ tìm x thuộc Z để A nguyên
c/ tính A vs x = -2, x = -3
d/ tìm x để A = 1
Cho biểu thức A = \(\left(\dfrac{2}{x+2}-\dfrac{1}{x-3}+\dfrac{5-x}{x^2-x-6}\right)\left(x-\dfrac{6}{x-1}\right)\)
a, Rút gọn biểu thức A.
b, Tìm x để A<0
c, Tìm các số tự nhiên x thoả mãn \(A^2-\left|A\right|=6\)
a) Ta có: \(A=\left(\dfrac{2}{x+2}-\dfrac{1}{x-3}+\dfrac{5-x}{x^2-x-6}\right)\cdot\left(x-\dfrac{6}{x-1}\right)\)
\(=\left(\dfrac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{x+2}{\left(x-3\right)\left(x+2\right)}+\dfrac{5-x}{\left(x-3\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-1\right)-6}{x-1}\)
\(=\dfrac{2x-6-x-2+5-x}{\left(x+2\right)\left(x-3\right)}\cdot\dfrac{x^2-x-6}{x-1}\)
\(=\dfrac{-3}{x-1}\)
A=\(\dfrac{4x^2+\left(2x+3\right)\left(x+1\right)-9}{9x^2-4}\)
a) Rút gọn A
b) Tìm các số nguyên x để A đạt giá trị nguyên
a, \(A=\dfrac{4x^2+2x^2+5x+3-9}{9x^2-4}=\dfrac{6x^2+5x-6}{9x^2-4}=\dfrac{\left(3x-2\right)\left(2x+3\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{2x+3}{3x+2}\)
b, Ta có \(6x+9⋮3x+2\Leftrightarrow2\left(3x+2\right)+5⋮3x+2\Rightarrow3x+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
3x+2 | 1 | -1 | 5 | -5 |
x | loại | -1 | 1 | loại |
A ) \(Q=\left(\frac{2x-x^2}{2x^2+8}-\frac{2x^2}{x^3-2x^2+4x-8}\right)\left(\frac{2}{x^2}+\frac{1-x}{x}\right)\)
a, RÚT GỌN Q
b, TÌM x THUỘC TẬP HỢP Z ĐỂ Q THUỘC TẬP HỢP Z
vào thống kê xem link nhé:
Câu hỏi của Kim Trân Ni - Toán lớp 8 - Học toán với OnlineMath
Cho biểu thức A=\(\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\)
và B=\(\dfrac{x^2+x-2}{x^3-1}\)
a Rút gọn biểu thức M=A.B
b Tìm x thuộc Z để M thuộc Z
c Tìm GTLN của biểu thức N=\(A^{-1}-B\)
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
B1: A=\(\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\)
a) Rút gọn
b) Tìm x thuộc Z để A nguyên
c) Tính A với x=-2; x=-3
d) Tìm x dể A=1
B2: Phân tích thành nhân tử
a) x2-2xy-4+y2
b) x2-4x+3
c) 9x2(x-y)-x+y
B3: Rút gọn
a) (x-2)3-(x+2)3-(x-1)(x2+x+1)
b) (5x+3y)(5x-3y)+(4x-3y)2
B4: P(x)=x4+x3+mx2-3x+5
a) Khi m=4, thực hiện phép chia P(x) cho x2-x+1
b) Tìm m để P(x)⋮(x-1)