1)Tìm x, biết
A) 8x +15-3x=(-400)
B) -32x +12x-5x=900
C) 3(x-1)-(x-5)=(-18)
D) 5-(x-2)-(x+3)=15
E) x+(x+1)+(x+2)+.....+2008+2009=2009
2)Bỏ dấu ngoặc các biểu thức sau
A) a-b.(c+d)
B) (a+b).(a-b)-(b-a).b
3)Biến đổi toorng sau thành dạng tích
A) ab-ac+ad
B) ac+ad-bc-bd
4)Cho a, b,c là các sống nguyên, biết ab-ac+bc-c. c=(-1).Chứng tỏ rằng a và b là 2 số đối. nhau
1a) 8x + 15 - 3x = -400 b) -32x + 12x - 5x = 900
=> 5x = -400 - 15 => -25x = 900
=> 5x = -415 => x = 900 : (-25)
=> x = -415 : 5 => x = -36
=> x = -83
c) 3(x - 1) - (x - 5) = -18
=> 3x - 3 - x + 5 = -18
=> 2x + 2 = -18
=> 2x = -18 - 2
=> 2x = -20
=> x = -10
d,e tự làm
a) \(8x+15-3x=-400\)
\(\Leftrightarrow8x-3x=-400-15\)
\(\Leftrightarrow5x=-415\)
\(\Leftrightarrow x=-415\div5\)
\(\Leftrightarrow x=-83\)
b) \(-32x+12x-5x=900\)
\(\Leftrightarrow-25x=900\)
\(\Leftrightarrow x=900\div\left(-25\right)\)
\(\Leftrightarrow x=-36\)
c) \(3\left(x-1\right)-\left(x-5\right)=-18\)
\(\Leftrightarrow3x-3-x+5=-18\)
\(\Leftrightarrow2x+2=-18\)
\(\Leftrightarrow2x=-18-2\)
\(\Leftrightarrow2x=-20\)
\(\Leftrightarrow x=-20\div2\)
\(\Leftrightarrow x=-10\)
d) \(5-\left(x-2\right)-\left(x+3\right)=15\)
\(\Leftrightarrow5-x+2-x-3=15\)
\(\Leftrightarrow4-2x=15\)
\(\Leftrightarrow2x=4-15\)
\(\Leftrightarrow2x=-11\)
\(\Leftrightarrow x=\frac{-11}{2}\)
e) \(x+\left(x+1\right)+\left(x+2\right)+...+2008+2009=2009\)
\(\Leftrightarrow x+\left(x+1\right)+\left(x+2\right)+...+2008=0\)
\(\Leftrightarrow\frac{x\left(2008+x\right)}{2}=0\)
\(\Leftrightarrow x\left(2008+x\right)=0\)
\(\Leftrightarrow2008x+x^2=0\)
\(\Rightarrow\)\(\hept{\begin{cases}2008x\\x^2\end{cases}}\)là hai số đối nhau.
Mà \(x^2\ge0\)\(\Rightarrow x^2=0\)\(\Leftrightarrow x=0\)
\(\Rightarrow x=0-2008\)
\(\Leftrightarrow x=-2008\)
2)a) \(a-b\left(c+d\right)=a-bc-bd\)
b) \(\left(a+b\right)\left(a-b\right)-b\left(b-a\right)\)
\(=a\left(a+b\right)-b\left(a+b\right)-b^2+ab\)
\(=a^2+ab-ab-b^2-b^2+ab\)
\(=a^2+\left(ab-ab\right)-\left(b^2+b^2\right)+ab\)
\(=a^2+0-2b^2+ab\)
\(=a^2-2b^2+ab\)
\(=a\left(a+b\right)-2b^2\)
3) a ) \(ab-ac+ad=a\left(b-c+d\right)\)
