b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

ĐKXĐ : \(x\ne0;x\ne\pm1\)

a) Bạn ghi lại rõ đề.

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

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

c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)

Không tồn tại Min P \(\forall x\inℝ\)

a) đK: \(x\ne0;2\)

B = \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{3x-4}{-4}=\dfrac{4-3x}{4}\) \(\dfrac{x-4+2x}{x\left(x-2\right)}:\dfrac{\left(x-2\right)\left(x+2\right)-x^2}{x\left(x-2\right)}\)

= \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{4-3x}{4}\)

b) Thay x = -2 (TMDK) vào B, ta có:

\(B=\dfrac{4-3.\left(-2\right)}{4}=\dfrac{4+6}{4}=\dfrac{5}{2}\)

c) Để \(\left|B\right|-2x=5\)

<=> \(\left|\dfrac{4-3x}{4}\right|-2x=5\)

TH1: \(x\le\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)

PT <=> \(\dfrac{4-3x}{4}-2x=5\)

<=> \(\dfrac{4-3x-8x}{4}=5\)

<=> \(4-11x=20\)

<=> x = \(\dfrac{-16}{11}\) (Tm)

TH2: \(x>\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)

PT <=> \(\dfrac{3x-4}{4}-2x=5\)

<=> \(\dfrac{3x-4-8x}{4}=5\)

<=> \(-5x-4=20\)

<=> \(x=\dfrac{-24}{5}\left(l\right)\)

d) Xét (2-x)B = \(\dfrac{\left(2-x\right)\left(4-3x\right)}{4}\)  = \(\dfrac{3x^2-10x+8}{4}\)

= \(\dfrac{3\left(x-\dfrac{5}{3}\right)^2-\dfrac{1}{3}}{4}\)

Mà \(3\left(x-\dfrac{5}{3}\right)^2\ge\) 0

=> (2-x)B \(\ge\dfrac{\dfrac{-1}{3}}{4}=\dfrac{-1}{12}\)

Dấu "=" <=> x = \(\dfrac{5}{3}\left(tm\right)\)

e) Số nguyên âm lớn nhất là -1

Để B = -1

<=> \(\dfrac{4-3x}{4}=-1\)

<=> 4 - 3x = -4
<=> \(x=\dfrac{8}{3}\left(tm\right)\)

g) 

TH1: \(x\le\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)

BDT <=> \(\dfrac{4-3x}{4}< 2x-4\)

<=> \(4-3x< 8x-16\)

<=> \(x>\dfrac{20}{11}\left(l\right)\)

TH2: \(x>\dfrac{4}{3}\)

<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)

BDT <=> \(\dfrac{3x-4}{4}< 2x-4\)

<=> \(3x-4< 8x-16\)

<=> x > \(\dfrac{12}{5}\)

KHDK: \(x>\dfrac{12}{5}\)

1/

( a + b )³ + ( a - b )³ - 6ab² < đã sửa >

= a³ + 3a²b + 3ab² + b³ + a³ - 3a²b + 3ab² - b³ - 6ab²

= 2a³ 

2/

A = x² + y² - 2x - 4y + 6 = ( x² - 2x + 1 ) + ( y² - 4y + 4 ) + 1 = ( x - 1 )² + ( y - 2 )² + 1 ≥ 1 ∀ x, y

Dấu "=" xảy ra khi x = 1 ; y = 2

=> MinA = 1 <=> x = 1 ; y = 2

B = 2x² + 8x + 10 = 2( x² + 4x + 4 ) + 2 = 2( x + 2 )² + 2 ≥ 2 ∀ x

Dấu "=" xảy ra khi x = -2

=> MinB = 2 <=> x = -2

C = 25x² + 3y² - 10x + 11 = ( 25x² - 10x + 1 ) + 3y² + 10 = ( 5x - 1 )² + 3y² + 10 ≥ 10 ∀ x, y

Dấu "=" xảy ra khi x = 1/5 ; y = 0

=> MinC = 10 <=> x = 1/5 ; y = 0

D = ( x - 3 )² + ( x - 11 )²

Đặt t = x - 7

D = ( t + 4 )² + ( t - 4 )²

    = t² + 8t + 16 + t² - 8t + 16

    = t² + 32 ≥ 32 ∀ t

Dấu "=" xảy ra khi t = 0

=> x - 7 = 0 => x = 7

=> MinD = 32 <=> x = 7

Cảm ơn bn nhiều nhé!

Bài 1:

\(\left(a+b\right)^3+\left(a-b\right)^3-6ab^2\)

\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)-6ab^2\)

\(=2a\left(a^2+3b^2\right)-6ab^2\)

\(=2a^3+6ab^2-6ab^2\)

\(=2a^3\)

Bài 2:

\(A=x^2+y^2-2x-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu"=" xảy ra khi \(\hept{\begin{cases}x-1=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Vậy...

\(B=2x^2+8x+10\)

\(=2\left(x^2+4x+4\right)+2\)

\(=2\left(x+2\right)^2+2\ge2\forall x\)

Dấu"="xảy ra khi \(x+2=0\Leftrightarrow x=-2\)

Vậy...

\(P=x^2+4xy+4y^2-4xy-4y^2+2x+3\)

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

a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)

\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)

\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)

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