2, 2(x+1)-1=3-(1-2x)

2x+2-1=2-1+2x

2x-2x=2-1-2+1

0x=0

Vậy không tồn tại giá trị của x thỏa mãn đề bài

3, (3x+5)(2x-7)=0

\(\orbr{\begin{cases}3x+5=0\\2x-7=0\end{cases}}\)

\(\orbr{\begin{cases}3x=0-5=-5\\2x=0+7=7\end{cases}}\)

\(\orbr{\begin{cases}x=\left(-5\right):3\\x=7:2=3,5\end{cases}}\)Vô lí

Vậy x=3,5

P = 3x/(x(x + 5)) > 0

<=> 3/(x + 5) > 0

<=> 3 = 0 (vô lý)

=> vô nghiệm

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

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

\(\Leftrightarrow\left(x-1\right)^2-\left(1-x\right)\left(1+x\right)-\left(1-x\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2-\left(1-x\right)\left(1+x-x-3\right)=0\)

\(\Leftrightarrow x^2-2x+1+2\left(1-x\right)=0\)

\(\Leftrightarrow x^2-2x+1+2-2x=0\)

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

\(\Leftrightarrow x^2-x-3x+3=0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

1) (2x^2 + 1)(x^2 - 2x - 1)

= 2x^4 - 4x^3 - 2x^2 + x^2 - 2x - 1

= 2x^4 - 4x^3 - x^2 - 2x - 1

2) (x^2 - x^4)/(x^2 - 1 + 1)

= (x^2.(1 - x^2))/(x^2 - 1 + 1)

= (x^2.(1 + x)(1 - x))/x^2

= (1 + x)(1 - x)

3) (3x + y)^3 + x^3 - 3x^2 + 3x + 1

Thay x = 1,1; y = -0,7 vào biểu thức, ta có:

= [3.1,1 + (-0,7)]^3 + 1,1^3 - 3.1,1^2 + 3.1,1 + 1

= 19,577

a) thay x = -3 vào biểu thức, ta có: 

\(A=\frac{\left(-3\right)^2+2.\left(-3\right)}{\left(-3\right)+1}=-\frac{3}{2}\)

b) M = A.B

\(M=\left(-\frac{3}{2}\right)\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)\)

\(M=-\frac{3\left(\frac{x+2}{x-2}-\frac{x-2}{x+2}+\frac{16}{4-x^2}\right)}{2}\)

\(M=-\frac{3.\frac{8}{x+2}}{2}\)

\(M=-\frac{\frac{24}{x+2}}{2}\)

\(M=-\frac{24}{2\left(x+2\right)}\)

\(M=-\frac{12}{x+2}\)