a) \(x+5=20-\left(12-7\right)\)

\(\Rightarrow x+5=20-5\)

\(\Rightarrow x+5=15\)

\(\Rightarrow x=15-5\)

\(\Rightarrow x=10\)

b) \(15-\left(3+2x\right)=2^2\)

\(\Rightarrow3+2x=15-4\)

\(\Rightarrow3+2x=11\)

\(\Rightarrow2x=11-3\)

\(\Rightarrow2x=8\)

\(\Rightarrow x=\dfrac{8}{2}\)

\(\Rightarrow x=4\)

c) \(-11-\left(19-x\right)=50\)

\(\Rightarrow19-x=-11-50\)

\(\Rightarrow19-x=-61\)

\(\Rightarrow x=61+19\)

\(\Rightarrow x=80\)

d) \(159-\left(25-x\right)=43\)

\(\Rightarrow25-x=159-43\)

\(\Rightarrow25-x=116\)

\(\Rightarrow x=25-116\)

\(\Rightarrow x=-91\)

e) \(\left(79-x\right)-43=-\left(17-52\right)\)

\(\Rightarrow\left(79-x\right)-43=52-17\)

\(\Rightarrow79-x-43=35\)

\(\Rightarrow36-x=35\)

\(\Rightarrow x=1\)

f) \(\left(7+x\right)-\left(21-13\right)=32\)

\(\Rightarrow7+x-8=32\)

\(\Rightarrow x-1=32\)

\(\Rightarrow x=32+1\)

\(\Rightarrow x=33\)

g) \(-x+20=-15+8+13\)

\(\Rightarrow-x+20=6\)

\(\Rightarrow x=20-6\)

\(\Rightarrow x=14\)

h) \(-\left(-x+13-142\right)+18=55\)

\(\Rightarrow x-13+142+18=55\)

\(\Rightarrow x+147=55\)

\(\Rightarrow x=55-147\)

\(\Rightarrow x=-92\)

a) 7(x - 5) + 2 = 51

\(\Leftrightarrow\) 7(x - 5) = 51 - 2

\(\Leftrightarrow\) 7(x - 5) = 49

\(\Leftrightarrow\) x - 5 = 49 : 7

\(\Leftrightarrow\) x - 5 = 7

\(\Leftrightarrow\) x = 7 + 5

\(\Leftrightarrow\) x = 12.

Vậy x = 12.

k) 2412 : (3x + 147) = |-38| + (-26)

\(\Leftrightarrow\) 2412 : (3x + 147)= 38 + (-26)

\(\Leftrightarrow\) 2412 : (3x + 147)= 12

\(\Leftrightarrow\) 3x + 147 = 2412 : 12

\(\Leftrightarrow\) 3x + 147 = 201

\(\Leftrightarrow\) 3x = 201 - 147

\(\Leftrightarrow\) 3x = 54

\(\Leftrightarrow\) x = 54 : 3

\(\Leftrightarrow\) x = 18.

Vậy x = 18.

I) 4824 : (4x + 137) = |-59| + (-35)

\(\Leftrightarrow\) 4824 :(4x + 137) = 59 + (-35)

\(\Leftrightarrow\) 4824 :(4x + 137) = 24

\(\Leftrightarrow\) 4x + 137 = 4824 : 24

\(\Leftrightarrow\) 4x + 137 = 201

\(\Leftrightarrow\) 4x = 201 - 137

\(\Leftrightarrow\) 4x = 64

\(\Leftrightarrow\) x = 64 : 4

\(\Leftrightarrow\) x = 16.

Vậy x = 16.

c) |-123| - 5(x - 3) = (-28) + 66

\(\Leftrightarrow\) 123 - 5(x - 3) = 38

\(\Leftrightarrow\) 5(x - 3) = 123 - 38

\(\Leftrightarrow\) 5(x - 3) = 85

\(\Leftrightarrow\) x - 3 = 85 : 5

\(\Leftrightarrow\) x - 3 = 17

\(\Leftrightarrow\) x = 17 + 3

\(\Leftrightarrow\) x = 20.

Vậy x = 20.

m) 7x - 4 . 6 = 2058

\(\Leftrightarrow\) 7x - 24 = 2058

\(\Leftrightarrow\) 7x = 2058 + 24

\(\Leftrightarrow\) 7x = 2082

\(\Leftrightarrow\) x = 2082 : 7

\(\Leftrightarrow\) x = \(\frac{2058}{7}\).

Vậy x = \(\frac{2058}{7}\).

Phần b) ra phân số nên mình để mai làm và cả những bài còn lại nữa.

Chúc bạn học tốt !!!

---------------NHANH NHA MK ĐANG CẦN GẤP--------------

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Bài 5 :

S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399

S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)

S = 1 + 0 + 0 + ... + 0 + (- 401)

S = 1 - 401

S = - 400

Bài 5

A= 1+3-5-7+9+11-13-15+...-397-399

A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)

A= -8 -8 -...-8

A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)

A= -400

Bài 6: 

b) Ta có: x(3x+9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: S={0;-3}

c) Ta có: 52-(x+21)=19-x

\(\Leftrightarrow52-x-21-19+x=0\)

\(\Leftrightarrow12=0\)(vô lý)

Vậy: \(x\in\varnothing\)

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

Cần bổ sung điều kiện \(x;y\inℤ\)

a) \(\left(x-7\right)\left(xy+1\right)=9=1\cdot9=3\cdot3=\left(-1\right)\cdot\left(-9\right)=\left(-3\right)\cdot\left(-3\right)\)

Xét bảng :

Vì x,y thuộc Z nên ta có (x;y)={(8;1),(16;0),(-2;1),(4;-1)

b) \(\left(x+5\right)\left(3x-12\right)>0\)

TH1 : \(\hept{\begin{cases}x+5>0\\3x-12>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-5\\x>4\end{cases}\Leftrightarrow x>4}}\)

TH2 : \(\hept{\begin{cases}x+5< 0\\3x-12< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -5\\x< 4\end{cases}\Leftrightarrow x< -5}}\)

Vậy....

c) \(\left|x+3\right|< 3\)

Mặt khác : \(0< \left|x+3\right|< 3\)

\(\Rightarrow\left|x+3\right|\in\left\{1;2\right\}\)

\(\Rightarrow\left(x+3\right)\in\left\{\pm1;\pm2\right\}\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)

Vậy....