1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

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) x - 2 = -6

x = - 6 + 2

x = -4

2) -5x-(-3)=13

-5x+3=13

-5x=13-3=10

x=10:(-5)

x=-2

3) 15-(x-7)=21

15-x+7=21

15-x=21-7=14

x=15-14=1

4) 3x+17=2

3x=2-17=-15

x=-15:3=-5

5) 45-(x-9)=-35

45-x+9=-35

45-x=-35-9=-44

x=45-(-44)=89

6) -5+x=15

x=15-(-5)

x=20

7) 2x-(-17)=15

2x+17=15

2x=15-17=-2

x=-2:2=-1

ai giúp mk câu này với

8) \(\left|x-2\right|=3\\ \Rightarrow x-2=\pm3\)

+) x - 2 = 3

x = 5

+) x - 2 = -3

x = -1

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

Các câu 9; 10; 12 tương tự câu 8; còn câu 11 tương tự câu 3; 5

Chúc bạn học tốt!

tìm x biết:

(3x-1) [- 1/2x+5]=0

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

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !

3(3x-1/2)^3+1/9=0

a: \(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{3}{4}=3-\dfrac{1}{2}x-\dfrac{1}{2}-\dfrac{1}{3}x-\dfrac{2}{3}\)

=>13/12x=13/12

hay x=1

b: \(\Leftrightarrow\dfrac{3x-11}{11}-\dfrac{x}{3}=\dfrac{3x-5}{7}-\dfrac{5x-3}{9}\)

\(\Leftrightarrow\dfrac{3}{11}x-1-\dfrac{1}{3}x=\dfrac{3}{7}x-\dfrac{5}{7}-\dfrac{5}{9}x+\dfrac{1}{3}\)

\(\Leftrightarrow x\cdot\dfrac{46}{693}=\dfrac{13}{21}\)

hay x=429/46

b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)

=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)

=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)

c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)

=>2x=4

=>x=4/2=2

d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)

=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)

=>x=-2

e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)

=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)

=>\(\dfrac{7}{3}x=2,4\)

=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)

f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)

=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)

=>\(x=-\dfrac{2}{15}\cdot45=-6\)

g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)

=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)

=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)

=>x=10/2=5

h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)

=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)

=>4x=7

=>\(x=\dfrac{7}{4}\)