mình làm bài cho bạn rồi bạn tích cho mình đi

a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

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

b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

c) Đổi đề: \(a^2x+a^2y-7x-7y\)

\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)

d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)

e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

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

g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)

i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)

a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)

b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)

e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)

g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)

h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)

i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)

\(\dfrac{x-3}{2+x}< 1\\ \Leftrightarrow\dfrac{x-3}{2+x}-1< 0\\ \Leftrightarrow\dfrac{x-3}{2+x}-\dfrac{2+x}{2+x}< 0\\ \Leftrightarrow\dfrac{x-3-2-x}{2+x}< 0\\ \Leftrightarrow\dfrac{-5}{2+x}< 0\)

Vì \(-5< 0\)

\(\Rightarrow2+x>0\\ \Rightarrow x>-2\)

Vậy \(x>-2\)

\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\) (ĐK: \(x\ne0,x\ne-1\))

\(\Leftrightarrow\dfrac{360\left(x+1\right)}{x\left(x+1\right)}-\dfrac{400x}{x\left(x+1\right)}=\dfrac{x\left(x+1\right)}{x\left(x+1\right)}\)

\(\Leftrightarrow360\left(x+1\right)-400x=x\left(x+1\right)\)

\(\Leftrightarrow360x+360-400x=x^2+x\)

\(\Leftrightarrow-40x+360=x^2+x\)

\(\Leftrightarrow x^2+40x+x-360=0\)

\(\Leftrightarrow x^2+41x-360=0\)

\(\Rightarrow\Delta=41^2-4\cdot1\cdot\left(-360\right)=3121>0\)

\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{-41+\sqrt{3121}}{2\cdot1}\approx7\left(tm\right)\\x_2=\dfrac{-41-\sqrt{3121}}{2\cdot1}\approx-48\left(tm\right)\end{matrix}\right.\)

\(\dfrac{360}{x}-\dfrac{400}{x+1}=1\)

Điều kiện: \(x\ne0;x\ne-1\)

PT \(\Leftrightarrow\dfrac{360\left(x+1\right)-400x}{x\left(x+1\right)}=1\)

\(\Rightarrow-40x+360=x\left(x+1\right)\)

\(\Leftrightarrow-40x+360=x^2+x\)

\(\Leftrightarrow x^2+41x-360=0\)

\(\Leftrightarrow x^2+2.\dfrac{41}{2}.x+\dfrac{1681}{4}=\dfrac{3121}{4}\)

\(\Leftrightarrow\left(x+\dfrac{41}{2}\right)^2=\left(\dfrac{\sqrt{3121}}{2}\right)^2\)

\(\Leftrightarrow x+\dfrac{41}{2}=\dfrac{\sqrt{3121}}{2}\) hoặc \(x+\dfrac{41}{2}=-\dfrac{\sqrt{3121}}{2}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\) hoặc \(x=-\dfrac{\sqrt{3121}}{2}-\dfrac{41}{2}\)

Vậy...

a, để pt trên là pt bậc nhất khi m khác 2 

b, Ta có \(2x+5=x+7-1\Leftrightarrow x=1\)

Thay x = 1 vào pt (1) ta được 

\(2\left(m-2\right)+3=m-5\Leftrightarrow2m-1=m-5\Leftrightarrow m=-4\)

Lưu Anh Đức: không được đặt dấu ''='' giữa 2 vế ở biểu thức cuối, 0x không thể bằng 1

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

\(\Leftrightarrow\frac{2}{x}-2=\frac{4}{2x}-1\)

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

\(\Leftrightarrow0x=1\)

Vậy phương trình vô nghiệm

tích cho mình đi

có 4 nghiệm ,đó là x=0;2;2-(căn 2);2+(căn 2)