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8 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
\(B=(x+1)^2-2(2x-1)(1+x)+4x^2-4x+1\)
`= x^2 + 2x + 1 - 2(2x^2 + x - 1) + 4x^2 - 4x + 1`
`= 5x^2 - 2x + 2 - 4x^2 - 2x + 2`
`= x^2 - 4x + 4`
KV
8 tháng 8 2023
\(B=\left(x+1\right)^2-2\left(2x-1\right)\left(1+x\right)+4x^2-4x+1\)
\(=\left(x+1\right)^2-2\left(x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(x+1-2x+1\right)^2\)
\(=\left(2-x\right)^2\)
HG
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 4 2022
ĐKXĐ: \(x\ne\pm1\)
\(A=\left(\dfrac{\left(1+x\right)^2}{\left(1-x\right)\left(1+x\right)}-\dfrac{\left(1-x\right)^2}{\left(1-x\right)\left(1+x\right)}+\dfrac{4x^2}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)
\(=\left(\dfrac{x^2+2x+1-\left(x^2-2x+1\right)+4x^2}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x+1\right)}{x-1}\)
\(=\left(\dfrac{4x^2+4x}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{4\left(x+1\right)}{x-1}\)
\(=\dfrac{4x\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)}{4\left(x+1\right)}=-\dfrac{x}{x+1}\)
26 tháng 6 2021
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
26 tháng 6 2021
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
22 tháng 8 2021
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
11 tháng 1 2022
\(A=\left(\dfrac{1}{x^2-4x}+\dfrac{2}{16-x^2}+\dfrac{4}{4x+16}\right):\dfrac{1}{4x}\left(x\ne4;x\ne-4;x\ne0\right).\)
\(A=\left(\dfrac{1}{x\left(x-4\right)}+\dfrac{-2}{\left(x+4\right)\left(x-4\right)}+\dfrac{1}{x+4}\right).4x\).
\(A=\dfrac{x+4-2x+x^2-4x}{x\left(x-4\right)\left(x+4\right)}.4x.\)
\(A=\dfrac{x^2-5x+4}{\left(x-4\right)\left(x+4\right)}.4.\)
\(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}.4.\)
\(A=\dfrac{4\left(x-1\right)}{x+4}.\)
11 tháng 12 2020
Bài 1 :
\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)
\(=x^2-4x+4-x+9=x^2-5x+13\)
Bài 2 :
a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)
\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)
b, Thay x = -4 ta được :
\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)
16 tháng 8 2023
`@` `\text {Ans}`
`\downarrow`
`A= (2x - 3)^2 - (2x + 3)^2`
`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`
`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`
`= -6 * 4x`
`= -24x`
TD
0
A
3 tháng 8 2023
\(\text{ P = (2x-1).4x^2+2x+1+(x+1)x^2-x+1}\)
\(\text{P =}\) \(\text{[(2x-1) . 4x^2 ]}\)\(\text{[(x+1) .x^2]}\)
\(\text{P = }\) \(\text{8x^3 - 4x^2 + 2x^3 + 2x^2 + 2x + 1 + x^3 - x + 1}\)
\(\text{P =}\) \(\text{(8x^3 + 2x^3 + x^3) + (-4x^2 + 2x^2) + (2x - x) + (1 + 1)}\)
\(\text{P =}\) \(\text{11x^3 - 2x^2 + x + 2}\)
4x(x2 -x+1) -(4x-1)(x2-x)
= 4x3 - 4x2 + 4x - (4x3 - 4x2 - x2 + x)
= 4x3 - 4x2 + 4x - 4x3 + 4x2 + x2-x
=(4x3 - 4x3) - (4x2 - 4x2-x2)+ (4x-x)
= x2 + 3x
4x(x^2-x+1)-(4x-1)(x^2-x)
=4x^3-4x^2+4x-4x^3+4x^2+x^2-x
=x^2+3x