Tại \(a=\dfrac{5}{9}\); \(b=\dfrac{5}{2}\), ta có:

\(\dfrac{3}{4}\cdot\dfrac{5}{9}+\left(\dfrac{5}{2}-\dfrac{1}{2}\right)\)

\(=\dfrac{3\cdot5}{4\cdot3\cdot3}+\dfrac{4}{2}\)

\(=\dfrac{5}{12}+2\) hoặc \(\dfrac{29}{12}\)

Thay a = 5/9 và b = 5/2 vào A ta được:

A = 3/4 × 5/9 + (5/2 - 1/2)

= 5/12 + 2

= 29/12

Thay a = 5/9 và b = 5/2 vào A ta được:

A = 3/4 × 5/9 + (5/2 - 1/2)

= 5/12 + 2

= 29/12

\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)

\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)

\(\Rightarrow P=4\)

\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)

\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)

\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)

\(A=\left(2\times2^2-3\times2-5\right)\left(2^2-3\right)=\left(8-6-5\right)\left(4-3\right)=-3\times1=-3\)

quá đơn giản

a: 

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b: \(A=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)

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

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

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

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

Khi x=2005 thì \(A=\dfrac{-1}{2\cdot\left(2005-1\right)}=-\dfrac{1}{4008}\)

Vì x=1 không thỏa mãn ĐKXĐ

nên khi x=1 thì A không có giá trị

c: Để A=-1002 thì \(\dfrac{-1}{2\left(x-1\right)}=-1002\)

=>\(2\left(x-1\right)=\dfrac{1}{1002}\)

=>\(x-1=\dfrac{1}{2004}\)

=>\(x=\dfrac{1}{2004}+1=\dfrac{2005}{2004}\left(nhận\right)\)

Bài 2

a) ĐKXĐ: x - 10 0 và x + 10 0

*) x - 10 0

x 10

*) x + 10 0

x 10

Vậy ĐKXĐ: x -10; x 10

b) P = [(5x + 2)(x + 10) + (5x - 2)(x - 10)]/[(x - 10)(x + 10)] . (x - 10)/(x² + 4)

= (5x² + 50x + 2x + 20 + 5x² - 50x - 2x + 20)/[(x + 10)(x² + 4)]

= (10x² + 40)/[(x + 10)(x² + 4)]

= 10(x² + 4)/[(x + 10)(x² + 4)]

= 10/(x + 10)

c) Khi x = 2/5 ta có:

P = 10.(2/5 + 10)

= 4 + 100

= 104

giúp mik vs 

Bài 1

b) x⁵ - x³ - x² + 1

= (x⁵ - x³) - (x² - 1)

= x³.(x² - 1) - (x² - 1)

= (x² - 1)(x³ - 1)

= (x - 1)(x + 1)(x - 1)(x² + x + 1)

= (x + 1)(x - 1)²(x² + x + 1)

Bạn rút gọn ra bao nhiêu rồi mình làm luôn phần c cho.

mình rút gọn đc \(\frac{9x-18}{\left(x-3\right)\left(x+3\right)}\)

\(A=\frac{9x}{x^2-9}-\frac{3}{x-3}-\frac{3}{x+3}\)

\(=\frac{9x}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{9x-3x-9-3x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{3x}{x^2-9}\)

Mk ra vầy

\(A=\left(2\times2^2-3\times2-5\right)\left(2-2^2-3\right)=\left(8-6-5\right)\left(2-4-3\right)=\left(-3\right)\times\left(-5\right)=15\)

Bạn phân ra 2 trường hợp

1) x=2

2) x= -1

các bạn nhớ ghi rõ các làm nhé

1. x( x - 3 ) + y( y - 3 ) + 2xy - 35

= x² - 3x + y² - 3y + 2xy - 35

= ( x² + 2xy + y² ) - ( 3x + 3y ) - 35

= ( x + y )² - 3( x + y ) - 35

= 5² - 3.5 - 35

= 25 - 15 - 35 = -25

2. 4x² + y² + 8x - 4xy - 4y + 100

= ( 4x² - 4xy + y² + 8x - 4y + 4 ) + 96

= [ ( 4x² - 4xy + y² ) + ( 8x - 4y ) + 4 ] + 96

= [ ( 2x - y )² + 2.( 2x - y ).2 + 2² ] + 96

= ( 2x - y + 2 )² + 96

= ( 4 + 2 )² + 96

= 6² + 96 = 36 + 96 = 132