(a) Điều kiện : \(x\ne-1.\)

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

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

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

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

\(=x^2-x.\)

Vậy : Với mọi \(x\ne-1\) thì \(P=x^2-x.\)

(b) Ta có : \(P=x^2-x\)

\(=\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2\)

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

Vậy : \(MinP=-\dfrac{1}{4}.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=\dfrac{1}{2}.\)

cứuuuuuuuuuuu

a: \(P=\dfrac{2x^2-1-x^2+1+3x}{x\left(x+1\right)}=\dfrac{x^2+3x}{x\left(x+1\right)}=\dfrac{x+3}{x+1}\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

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 \(x\in\left\{2;-2\right\}\).

#\(Toru\)

a: Ta có: |x+4|=1

=>x+4=1 hoặc x+4=-1

=>x=-3(loại) hoặc x=-5

Khi x=-5 thì \(A=\dfrac{\left(-5\right)^2-5}{3\left(-5+3\right)}=\dfrac{20}{3\cdot\left(-2\right)}=\dfrac{-10}{3}\)

b: \(B=\dfrac{x-1+x+1-3+x}{\left(x-1\right)\left(x+1\right)}=\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x+1}\)

\(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne-\frac{1}{2}\end{cases}}\)

a) \(A=\left(\frac{1}{x-1}+\frac{x}{x^3-1}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)

\(\Leftrightarrow A=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)

\(\Leftrightarrow A=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)

\(\Leftrightarrow A=\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow A=\frac{x+1}{x-1}\)

b) Thay \(x=\frac{1}{2}\)vào A, ta được :

\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3\)