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Bài 1:
a) \((a-b)(a+b)=a^2-b^2\) (theo hằng đẳng thức đáng nhớ)
b) \((8x^3y^3-12y^3-12x^3y^5):(2x^3y^2)=\frac{8x^3y^3}{2x^3y^2}-\frac{12y^3}{2x^3y^2}-\frac{12x^3y^5}{2x^3y^2}\)
\(=4y-\frac{6y}{x^3}-6y^3=4y-6x^{-3}y-6y^3\)
c)
\((x^3+1):(x^2-x+1)=\frac{x^3+1}{x^2-x+1}=\frac{(x+1)(x^2-x+1)}{x^2-x+1}=x+1\)
Bài 2:
a)
\(6x^2y-18xy^2=6xy(x-3y)\)
b)
\(x^3+x^2-4x-4=(x^3+x^2)-(4x+4)=x^2(x+1)-4(x+1)\)
\(=(x+1)(x^2-4)=(x+1)(x^2-2^2)=(x+1)(x-2)(x+2)\)
a,\(\left|9+x\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}9+x=2x\\9x+x=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9=x\\9=-3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-3\end{matrix}\right.\)
Vậy...
Trường hợp 2 chưa chắc chắn lắm!!!
a) \(\left|9+x\right|=2x\)
Xét trường hợp 1:
\(9+x=2x\)
\(\Leftrightarrow9+x-2x=0\)
\(\Leftrightarrow9-x=0\)
\(\Leftrightarrow x=9\)
Xét trường hợp 2:
\(9+x=-2x\)
\(\Leftrightarrow9+x-\left(-2x\right)=0\)
\(\Leftrightarrow9+x+2x=0\)
\(\Leftrightarrow9+3x=0\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-9:3\)
\(\Leftrightarrow x=-3\)
Vậy x=9 hoặc x=-3
b) \(\left|x+6\right|-9=2x\)
\(\Leftrightarrow\left|x+6\right|=2x+9\)
Xét trường hợp 1:
\(x+6=2x+9\)
\(\Leftrightarrow x+6-\left(2x+9\right)=0\)
\(\Leftrightarrow x+6-2x-9=0\)
\(\Leftrightarrow-3-x=0\)
\(\Leftrightarrow x=-3\)
Xét trường hợp 2:
\(x+6=-\left(2x+9\right)\)
\(\Leftrightarrow x+6-\left[-\left(2x+9\right)\right]=0\)
\(\Leftrightarrow x+6+\left(2x+9\right)=0\)
\(\Leftrightarrow x+6+2x+9=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
\(\Leftrightarrow x=-15:3\)
\(\Leftrightarrow x=-5\)
Vậy x=-3 hoặc x=-5
Bài 2:
a: (x+1)(3-x)=0
=>x+1=0 hoặc 3-x=0
=>x=-1 hoặc x=3
b: (x-2)(2x-1)=0
=>x-2=0 hoặc 2x-1=0
=>x=2 hoặc x=1/2
c: (3x+9)(1-3x)=0
=>1-3x=0 hoặc 3x+9=0
=>x=1/3 hoặc x=-3
d: (x2+1)(81-x2)=0
=>(9+x)(9-x)=0
=>x=-9 hoặc x=9
a) 5x +3=2x-8 <=>5x-2x=-8-3 <=>3x=-11 <=> x=\(\dfrac{-11}{3}\)
b)6x-3(x+2)=5x+3<=> (6-3-5)x-9=0 <=> x=\(\dfrac{-9}{2}\)
c) (3x-9)(5x+10)=0<=> \(\left[{}\begin{matrix}3x-9=0\\5x+10=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d)8x(x+2)+16(x+2)=0<=>(x+2)(8x+16)=0<=>\(\left[{}\begin{matrix}x=-2\\x=-2\end{matrix}\right.\)
e)x2 -12x+35=0 <=>\(\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
a) \(\left[\left(4x+28\right).3+55\right]:5=35\)
\(\Leftrightarrow\left(4x+28\right).3+55=35.5\)
\(\Leftrightarrow\left(4x+28\right).3+55=175\)
\(\Leftrightarrow\left(4x+28\right).3=175-55\)
\(\Leftrightarrow\left(4x+28\right).3=120\)
\(\Leftrightarrow4x+28=120:3\)
\(\Leftrightarrow4x+28=40\)
\(\Leftrightarrow4x=40-28\)
\(\Leftrightarrow4x=12\)
\(\Leftrightarrow x=12:4\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
b) \(\left(12x-4^3\right).8^3=4.8^4\)
\(\Leftrightarrow12x-4^3=4.8^4:8^3\)
\(\Leftrightarrow12x-4^3=4.8^{4-3}\)
\(\Leftrightarrow12x-4^3=4.8\)
\(\Leftrightarrow12x-4^3=32\)
\(\Leftrightarrow12x-64=32\)
\(\Leftrightarrow12x=32+64\)
\(\Leftrightarrow12x=96\)
\(\Leftrightarrow x=96:12\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
c) \(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=8.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x-5\right)=720:40\)
\(\Leftrightarrow41-\left(2x-5\right)=18\)
\(\Leftrightarrow2x-5=41-18\)
\(\Leftrightarrow2x-5=23\)
\(\Leftrightarrow2x=23+5\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=28:2\)
\(\Leftrightarrow x=14\)
Vậy \(x=14\)
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