e/(x+6)(x-1)(x²+5x+16)

Help me!!!

e: =>x=0 hoặc x=1

a:565-13.x=30

13.x=565-370

13.x=195

x=195:13

x=15

Nếu sai sót gì thì e xin lỗi ạ ^^

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

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

\(3x-1=2\)

\(3x=3\)

\(x=1\)

a: \(\left(x+y+z\right)^2-\left(y+z\right)^2\)

\(=\left(x+y+z-y-z\right)\left(x+y+z+y+z\right)\)

\(=x\left(x+2y+3z\right)\)

b: \(\left(x+3\right)^2+4\left(x+3\right)+4\)

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

\(=\left(x+5\right)\left(x+5\right)\)

c: \(25+10\left(x+1\right)+\left(x+1\right)^2\)

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

\(=\left(x+6\right)\left(x+6\right)\)

a: \(x-2-\frac{x^2-10}{x+2}\)

\(=\frac{\left(x-2\right)\left(x+2\right)-x^2+10}{x+2}\)

\(=\frac{x^2-4-x^2+10}{x+2}=\frac{6}{x+2}\)

b: \(\frac{x}{y^2-xy}-\frac{y}{xy-x^2}\)

\(=\frac{-x}{y\left(x-y\right)}+\frac{y}{x\left(x-y\right)}=\frac{-x^2+y^2}{xy\left(x-y\right)}=\frac{-\left(x-y\right)\left(x+y\right)}{xy\left(x-y\right)}\)

\(=\frac{-x-y}{xy}\)

c: \(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=-\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-2}+\frac{1}{x-3}-\frac{1}{x-3}+\frac{1}{x-4}-\frac{1}{x-4}+\frac{1}{x-5}\)

\(=\frac{1}{x-5}-\frac{1}{x}=\frac{x-\left(x-5\right)}{x\left(x-5\right)}=\frac{5}{x\left(x-5\right)}\)

1: \(\frac{2x+6}{3x^2-x}:\frac{x^2+3x}{1-3x}\)

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

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

2: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

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

3: \(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{1}{3x-2}-\frac{1}{3x+2}+\frac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

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

4: \(\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{x^2-1}\)

\(=\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+3\right)\left(x-1\right)+\left(2x-1\right)\left(x+1\right)+x+5}{\left(x-1\right)\left(x+1\right)}\)

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

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

6x^2

P = 4\(x^2\).\(x\) + (3³ + 2²)

P = 4\(x^3\) + ( 27 + 4)

P = 4\(x^3\) + 31

Thay \(x\) = 1 vào P ta có:

P = 4.1³ + 31 

P = 35

Thay \(x\) = 3 vào P ta có:

P = 4.3³ + 31

P = 4.27 + 31

P = 108 + 31

P = 139

b) Thay a=25, b=9 vào biểu thức D=1+2(a+b)-\(4^3\) ta có:

\(1+2.\left(25+9\right)-4^3\)

\(\Rightarrow3.34-64\)

\(\Rightarrow102-64\)

\(=38\)

Vậy giá trị của biểu thức D=1+2(a+b)- \(4^3\) khi a=25, b=9 là: 38