a:

ĐKXĐ: x<>2

|2x-3|=1

=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào A, ta được:

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

b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)

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

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

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

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

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

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

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

Để P lớn nhất thì \(\dfrac{2}{x-2}\) max

=>x-2=1

=>x=3(nhận)

câu b)  \(\sqrt{-2}\) không xác định

a) \(A=4x-\sqrt{8}-\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)

\(=4x-\sqrt{8}-\frac{\sqrt{x^2\left(x+2\right)}}{\sqrt{x+2}}=4x-\sqrt{8}-x=3x-\sqrt{8}\)

b) \(x=\sqrt{-2}\) (không thỏa mãn)

Mọi người tk mình đi mình đang bị âm nè!!!!!!

Ai tk mình mình tk lại nha !!!

\(2\left(x-2\right)\left(x+3\right)-x^2+4=0\)

\(2\left(x^2+3x-2x-6\right)-x^2+4=0\)

\(2x^2+6x-4x-12-x^2+4=0\)

\(x^2+2x-8=0\)

\(x^2+4x-2x-8=0\)

\(x\left(x+4\right)-2\left(x+4\right)=0\)

\(\left(x+4\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+4=0\rightarrow x=\left(-4\right)\\x-2=0\rightarrow x=2\end{cases}}\)

3/ 

a/ \(2\left(x+1\right)^2-3\left(x-1\right)^2+\left(x+2\right)\left(5-x\right)\)

\(=2\left(x^2+2x+1\right)-3\left(x^2-2x+1\right)+\left(5x-x^2+10-2x\right)\)

\(=2x^2+4x+2-3x^2+6x-3+5x-x^2+10-2x\)

\(=-2x^2+13x+9\)

b/ \(\left(3x-1\right)^3+\left(3x-1\right)^3-6x^2+9\)

\(=2\left(3x-1\right)^3-6x^2+9\)

\(=2\left(\left(3x\right)^3-3\left(3x\right)^2\cdot1+3\cdot3x\cdot1-1\right)-6x^2+9\)

\(=2\left(27x^3-27x^2+9x-1\right)-6x^2+9\)

\(=54x^3-54x^2+18x-2-6x^2+9\)

\(=54x^3-60x^2+18x+7\)

Số hơi dài, nên dễ tính sai -,- tính mik hay cẩu thả có j sai ibbb ạ

2) 2.(x - 2).(x + 3) - x² + 4 = 0

<=> x² + 2x - 8 = 0

<=> (x - 2).(x + 4) = 0

        x - 2 = 0 hoặc x + 4 = 0

        x = 0 + 2         x = 0 - 4

        x = 2               x = -4

=> x = 2 hoặc x = -4

3) a) 2.(x + 1)² - 3.(x - 1)² + (x + 2).(5 - x)

= 2.(x² + 2x + 1) - 3.(x² - 2x + 1) + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + (x + 2).(5 - x)

= 2x² + 4x + 2 - 3x² + 6x - 3 + 3x - x² + 10

= (2x² - 3x² - x²) + (4x + 6x + 3x) + (2 - 3 + 10)

= -2x² + 13x + 9

b) (3x - 1)³ + (3x - 1)³ - 6x² + 9

= 2.(3x - 1)³ - 6x² + 9

= 2.(27x³ - 27x² + 9x - 1) - 6x² + 9

= 54x³ - 54x² + 18x - 2 - 6x² + 9

= 54x³ + (-54x² - 6x²) + 18x + (-2 + 9)

= 54x³ - 60x + 18x + 7

4) a) A = (x - 5).(x + 2) + 3.(x - 2).(x + 2) - (3x - 1)² + 5x²

A = (x - 5).(x + 2) + 3.(x - 2).(x + 3) - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - (9x² - 6x + 1) + 5x²

A = x² - 3x - 10 + 3x² - 12 - 9x² + 6x - 1 + 5x²

A = (x² + 3x² - 9x² + 5x²) + (-3x + 6x) + (-10 - 12 - 1)

A = 3x - 23 (1)

b) Thay x = 1/2 vào (1), ta có:

A = 3x - 23 = 3.(1/2) - 23

                   = 3/2 - 23

                   = -43/2

A khi x = 1/2 là -43/2

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

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

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

\(A=-x^2+7\)

Để A có giá trị bằng 3 thì :

\(-x^2+7=3\)

\(-x^2=-4\)

\(x^2=4\)

\(x\in\left\{\pm2\right\}\)

Vậy..........

a)

\(M=2+\sqrt{\left(2x\right)^2-2.2x.3+3^2}\)

\(\Rightarrow M=2+\sqrt{\left(2x-3\right)^2}\)

\(\Rightarrow M=2+2x-3\)

\(\Rightarrow M=2x-1\)

b)

(+) x=5/2

=> \(M=2.\frac{5}{2}-1=5-1=4\)

(+) x= - 1/5

=> \(M=2.\frac{\left(-1\right)}{5}-1=-\frac{2}{5}-1=-\frac{7}{5}\)