Bài 8: Quy tắc dấu ngoặc

a. \(-\left(a+b-c\right)+\left(a-b-c\right)\)

\(=-a-b+c+a-b-c\)

\(=\left(-a+a\right)+\left(c-c\right)+\left[-b+\left(-b\right)\right]\)

\(=-2b\left(đpcm\right)\)

b. \(a.\left(b-c\right)+a\left(d+c\right)\)

\(=a.\left[\left(b-c\right)+\left(d+c\right)\right]\)

\(=a.\left(b-c+d+c\right)\)

\(=a.\left[\left(b+d\right)+\left(-c+c\right)\right]\)

\(=a.\left(b+d\right)\left(đpcm\right)\)

1,Tìm x:

\(\left(x+8\right)+\left(x-4\right)=x-20\)

\(\Leftrightarrow x+8+x-4=x-20\)

\(\Leftrightarrow x+x-x=-20+4-8\)

\(\Leftrightarrow x=-24\)

Vậy...

Vi x-3(x+1)=x+(-3)(x+1)=x+(-3x)+(-3.1)=x+(-3x)+(-3)=x-3x-3

Ở đây mình dùng tính chất phân phối tách :

x-3×(x+1)thành

x-3x-3×1 nhưng thầy giáo bạn viết gọn 3×1 thành 3 vì 3×1 vẫn bằng chính nó

\(4\left(x-5\right)-\left(x+7\right)=-19\)

\(\Leftrightarrow4x-20-x-7=-19\)

\(\Leftrightarrow3x-27=-19\)

\(\Leftrightarrow3x=8\)

\(\Leftrightarrow x=\dfrac{8}{3}\)

Vậy ....

a) 25-5x+3-4=0

=> 24-5x=0

=> 5x=24

=> x=\(\dfrac{24}{5}\)

b) 259-51/4x+3/=100.3+10=310

=> -51/4x+3/=310-259=51

=> /4x+3/=-51 (1)

Vi /4x+3/≥0 (2)

Tu (1)(2) => pt vo nghiem

c) 7x-21-15+5x=6

=> 12x-36-6=0

=> 12x-42=0

=> 12x=42

=> x=\(\dfrac{42}{12}=\dfrac{7}{2}\)

a/ \(-3⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(-3\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=3\\x-2=-1\\x-2=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\\x=1\\x=-1\end{matrix}\right.\)

Vậy ..

b/ \(-5⋮x-4\)

\(\Leftrightarrow x-4\inƯ\left(-5\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=1\\x-4=5\\x-4=-1\\x-4=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\\x=3\\x=-1\end{matrix}\right.\)

Vậy ..

c/ \(2x-9⋮x-5\)

Mà \(x-5⋮x-5\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9⋮x-5\\2x-10⋮x-5\end{matrix}\right.\)

\(\Leftrightarrow19⋮x-5\)

\(\Leftrightarrow x-5\inƯ\left(19\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=1\\x-5=19\\x-5=-1\\x-5=-19\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=24\\x=4\\x=-14\end{matrix}\right.\)

Vậy ..

h: =>7x-21-15+5x=11x-5

=>12x-36=11x-5

hay x=31

l: (x+3)(x-4)=0

=>x+3=0 hoặc x-4=0

=>x=-3 hoặc x=4

i: =>5x-30-2x-6=12

=>3x-36=12

hay x=16

m: =>119+27x=8x52

=>27x=297

hay x=11

1/3x+2/3(x-1)=4/15:4/5-5/7

<=>x/3+(2x-2)/3=1/3-5/7

<=>(x+2x-2)/3=1/3-5/7

<=>7(x+2x-2)/21=(7-15)/21

<=>7x+14x-14=-8

<=>7x+14x=6

<=>21x=6

<=>x=6/21

-5 - (24 - x ) = 12 + (42-65)

\(\Leftrightarrow\) -5-24-x = 12+42-65

\(\Leftrightarrow\) x - 29 = -11

\(\Leftrightarrow\) x = -11+29

\(\Leftrightarrow\) x = 18

a,-5 - (24 - x ) = 12+(42-65 )

-> -5 - (24-x) = 12+ ( -23)

-> -5 - (24-x) = -11

-> - (24-x) = -11 - 5

->- (24-x) = -16

->24-x = 16

-> x = 18

Vậy x = 18

-5 - (24 - x ) = 12+(42-65 )

-> -5 - (24-x) = 12+ ( -23)

-> -5 - (24-x) = -11

-> - (24-x) = -11 - 5

->- (24-x) = -16

->24-x = 16

-> x = 18

97+72:[60-(18-12)²]

=97+72:(60-6²)

^{=97+72:(60-36)}

^=97+72:24

⁼⁹⁷⁺³

⁼¹⁰⁰

97+72:[60-(18-12)²

=97+72:[60-6²]

=97+72:24

=97+3

=100