2x(x+3) -- x(2x-1)
Đề : thực hiện phép tính
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Ta có: \(\dfrac{2x+3}{1-x^2}+\dfrac{2x+1}{x^2-2x+1}\)
\(=\dfrac{-2x-3}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x+1}{\left(x-1\right)^2}\)
\(=\dfrac{\left(-2x-3\right)\left(x-1\right)}{\left(x-1\right)^2\cdot\left(x+1\right)}+\dfrac{\left(2x+1\right)\left(x+1\right)}{\left(x+1\right)\cdot\left(x-1\right)^2}\)
\(=\dfrac{-2x^2+2x-3x+3+2x^2+2x+x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{2x+4}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
thực ra mình cũng cố rồi nhưng mà IQ có hạn nên nghĩ mãi ko ra, thế nên mới phải cầu cứu mấy bạn giỏi hơn đấy =)
a, \(\left(2x-1\right)\left(3-2x\right)=6x-4x^2-3+2x=-4x^2+8x-3\)
b, \(\left(x+2\right)+\left(1+x\right)\left(1-x\right)=x+2+1-x^2=-x^2+x+3\)
1.(2x+3).(x-5)+2x(3-x)+x-10
=2x^2 -10x+3x-15+6x-2x^2+x-10
=2x-25
2.(-x-2)3+(2x-4).(x2+2x+4)-x2.(x-6)
=-x^3+6x^2-12x-8+2x^3+4x^2+8x-4x^2+8x-16-x^3+6x^2
ĐKXĐ: \(x\ne0;x\ne\pm1\)
\(\dfrac{3}{2x^2+2x}+\dfrac{2x-1}{x^2-1}-\dfrac{2}{x}\)
\(=\dfrac{3}{2x\left(x+1\right)}+\dfrac{2x-1}{\left(x-1\right)\left(x+1\right)}-\dfrac{4}{2x}\)
\(=\dfrac{3\left(x-1\right)}{2x\left(x-1\right)\left(x+1\right)}+\dfrac{2x\left(2x-1\right)}{2x\left(x-1\right)\left(x+1\right)}-\dfrac{4\left(x^2-1\right)}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{3x-3+4x^2-2x-4x^2+4}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{2x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{2x\left(x-1\right)}\)
\(=\dfrac{1}{2x^2-2x}\)
b: \(\dfrac{xy}{2x-y}-\dfrac{2x^2}{y-2x}=\dfrac{xy}{2x-y}+\dfrac{2x^2}{2x-y}=\dfrac{xy+2x^2}{2x-y}\)
b: \(\dfrac{3x^2-x}{x-1}+\dfrac{x+2}{1-x}+\dfrac{3-2x^2}{x-1}\)
\(=\dfrac{3x^2-x-x-2+3-2x^2}{x-1}\)
\(=\dfrac{x^2-2x+1}{x-1}=x-1\)
Bạn cần viết lại đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo). Viết như thế này nhìn khó đọc quá.
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
a) \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b) \(\left(x+3\right)^2+\left(x-1\right)\left(3+2x\right)=x^2+6x+9+3x+2x^2-3-2x\)
\(=3x^2+7x+6\)
a: \(2x\left(x+5\right)-2x^2=2x^2+10x-2x^2=10x\)
b: \(\left(x+3\right)^2+\left(2x+3\right)\left(x-1\right)\)
\(=x^2+6x+9+2x^2-2x+3x-3\)
\(=3x^2+7x+6\)
\(2x\left(x+3\right)-x\left(2x-1\right)\)
\(=2x^2+6x-2x^2+x\)
\(=7x\)
\(2x\left(x+3\right)-x\left(2x-1\right)\)
\(=2x^2+6x-2x^2+x=7x\)