tĩm,ybiet

B=x^2-2xy+2y^2+2y+1

5( x+2 ) . ( x-2 ) -12 .( 6 - 8x )² +17 

=5.x^2-2^2-12*6^2

a) \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)

\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)

\(=x^2+2x+1-x^2+2x-1-3x^2+3\)

\(=4x+3\)

b) \(5\left(x+2\right)\left(x-2\right)-\frac{1}{2}\left(6-8x\right)+17\)

\(=5\left(x^2-4\right)-3+4x+17\)

\(=5x^2-20-3+4x+17\)

\(=5x^2-6+4x\)

ĐKXĐ : \(\left\{{}\begin{matrix}4x^2-1\ne0\\8x^3+1\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm\dfrac{1}{2}\)

\(P=\dfrac{2x^5-x^4-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

\(=\dfrac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(=\dfrac{x^4-1}{2x+1}+\dfrac{2}{2x+1}=\dfrac{x^4+1}{2x+1}\)

https://sg.docworkspace.com/l/sIM-LioBEocfloAY

a) Ta có: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)

\(=x^2+2x+1-x^2+2x-1-3\left(x^2-1\right)\)

\(=4x-3x^2+3\)

\(=-3x^2+4x+3\)

b) Ta có: \(5\left(x+2\right)\left(x-2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\)

\(=5\left(x^2-4\right)-\dfrac{1}{2}\left(64x^2-96x+36\right)+17\)

\(=5x^2-20-32x^2+48x-16+17\)

\(=-27x^2+48x-19\)

a) P = 2x(-3x + 2) - (x + 2)² + 8x² - 1

= -6x² + 4x - x² - 4x - 4 + 8x² - 1

= (-6x² - x² + 8x²) + (4x - 4x) + (-4 - 1)

= x² - 5

b) Thay x = 3 vào P, ta được:

P = 3² - 5

= 4

c) Để P = -1 thì x² - 5 = -1

x² = -1 + 5

x² = 4

x = 2 hoặc x = -2

Vậy x = 2; x = -2 thì P = -1

\(a,P=2x\left(-3x+2\right)-\left(x+2\right)^2+8x^2-1\)

\(=-6x^2+4x-\left(x^2+4x+4\right)+8x^2-1\)

\(=-6x^2+4x-x^2-4x-4+8x^2-1\)

\(=\left(-6x^2-x^2+8x^2\right) +\left(4x-4x\right)+\left(-4-1\right)\)

\(=x^2-5\)

Vậy \(P=x^2-5\).

\(b,\) Ta có: \(P=x^2-5\)

Thay \(x=3\) vào \(P\), ta được:

\(P=3^2-5=9-5=4\)

Vậy \(P=4\) khi \(x=3\).

\(c,\) Có: \(P=-1\)

\(\Leftrightarrow x^2-5=-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(P=-1\) khi \(x\in\left\{2;-2\right\}\).

#\(Toru\)

\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

\(=\dfrac{\left(1-x^2\right)\left(1+x^2\right)+2x\left(x^2-1\right)}{4x^2-1}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(=\dfrac{\left(1-x^2\right)\left(1+x^2-2x\right)}{4x^2-1}+\dfrac{2}{2x+1}\)

\(=\dfrac{\left(1-x^2\right)\left(x^2-2x+1\right)+4x-2}{4x^2-1}\)

TKS bạn

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: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

\(=3x-y-y-x+2x^2-2x\)

\(=2x^2-2y\)

https://olm.vn/hoi-dap/question/1027904.html

tk nhé 

^_^

\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{ }\)

\(P=\frac{x^4\left(2x-1\right)-\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(P=\frac{\left(x^4-1\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}+\frac{2}{2x+1}\)

\(P=\frac{x^4-1}{2x+1}+\frac{2}{2x+1}\)

\(P=\frac{x^4+1}{2x+1}\)

Vậy \(P=\frac{x^4+1}{2x+1}\)