\(\left(x+2\right)-2=0\)

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

 (- (x - 3))/2 - 2 = 5(x + 2)/4

=> \(\dfrac{-\left(x-3\right)-4}{2}=\dfrac{5\left(x+2\right)}{4}\)

=> \(\dfrac{-2\left(x-3\right)-8}{4}=\dfrac{5\left(x+2\right)}{4}\)

=. -2x + 6 - 8 = 5x + 10

=> 7x = -12

=> x = -12/7

Các câu còn lại có cách làm tương tự là tính lần lượt trong ngoặc trước, quy đồng về cùng mẫu số để triệt tiêu mẫu và xử lý phần tử số có x như câu đầu tiên em nhé!

Chúc em học vui vẻ nha!

2) Ta có: \(\dfrac{2\left(2x+1\right)}{5}-\dfrac{6+x}{3}=\dfrac{5-4x}{15}\)

\(\Leftrightarrow\dfrac{6\left(2x+1\right)}{15}-\dfrac{5\left(6+x\right)}{15}=\dfrac{5-4x}{15}\)

\(\Leftrightarrow12x+6-30-5x-5+4x=0\)

\(\Leftrightarrow11x-29=0\)

\(\Leftrightarrow x=\dfrac{29}{11}\)

Vậy: \(S=\left\{\dfrac{29}{11}\right\}\)

123,45 + 23,56 +76,44 + 54,55

= (123,45 + 54,55) + (23,56 + 76,44)

= 178 + 100

= 278

456 x 45 + 456 x 10 + - 456 x 55

=456 x (45 +10-55)

=456 x 0

=0

2,5 x 8,5 + 7,5 x 8,5 

=8,5 x (2,5 +7,5)

= 8,5 x 10

= 85

86 x 5 + 86 x 4 + 86

= 86 x (5+4+1)

= 86 x10

860

25 x 8 x 17 x 50

= 50 x 4 x 17 x 50

= 200 x 17 x 50 

= 200 x 50 x 17

= 10000 x 17

= 170000

(12 x 2 + 12 x 4 - 12 x 6 ) : ( 2 + 4 + 6 +... + 12 +14)

=(12 x (2 + 4 - 6) ) : ( 2 + 4 + 6 + ... + 12 + 14 )
= (12 x 0 ) : (2 + 4 + 6 + ... + 12 + 14 )

=0 : ( 2 + 4 + 6 + ... + 12 + 14)

= 0

a: \(=3\cdot\dfrac{6+14-9}{42}\cdot\dfrac{14}{11}\)

\(=3\cdot\dfrac{11}{42}\cdot\dfrac{14}{11}=1\)

b: \(=\dfrac{19}{5}\cdot\dfrac{11}{10}+\dfrac{9}{5}\cdot\dfrac{7}{3}\)

\(=\dfrac{209}{50}+\dfrac{63}{15}=4.18+4.2=8,38\)

A. 5(x-11) + 4(x-3) =10
<=>5x -55 + 4x -12 =10
<=>9x -67 =10
<=>9x =10+67 =77
<=>x =77 : 9 =77/9
B.21 - 5 ( x-7) = 2-3(x-5)
<=>21-5x+35 = 2 - 3x +15
<=>-5x + 56 = -3x + 17
<=>-5x +3x=17-56=-39
<=> -2x =-39
<=> x = -39 : -2 =39/2
C.3(2x-1)-7(x+4)=8-9x
<=>6x -4 -7x-28=-9x+8
<=>-x + 9x = 8+4+28=40
<=> 8x = 40
<=> x = 40:8=5
D. l 4-x l =11
<=>\(\orbr{\begin{cases}4-x=11\\4-x=-11\end{cases}}\)<=> \(\orbr{\begin{cases}x=-7\\x=15\end{cases}}\)
E. 5-3 l x+2 l = -28
<=> 3 l x+2 L = -28 -5 =-33
<=> l x+2 l = -33 :3 =11
<=>\(\orbr{\begin{cases}x+2=11\\x+2=-11\end{cases}}\)<=> \(\orbr{\begin{cases}x=9\\-13\end{cases}}\)
 T I C K mk nhé!!!^_^

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)