\(a,x+\frac{3}{4}=\frac{1}{4}\)

\(\Rightarrow x=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}\)

\(b,\frac{2}{3}x-\frac{3}{7}=\frac{1}{7}\)

\(\Rightarrow\frac{2}{3}x=\frac{1}{7}+\frac{3}{7}\)

\(\Rightarrow\frac{2}{3}x=\frac{4}{7}\)

\(\Rightarrow x=\frac{4}{7}:\frac{2}{3}=\frac{4}{7}\cdot\frac{3}{2}=\frac{6}{7}\)

\(c,1\frac{2}{3}x-50\%x=-1\frac{1}{6}\)

\(\Rightarrow\frac{5}{3}x-\frac{50}{100}x=-\frac{7}{6}\)

\(\Rightarrow\left[\frac{5}{3}-\frac{50}{100}\right]x=-\frac{7}{6}\)

\(\Rightarrow\left[\frac{5}{3}-\frac{1}{2}\right]x=-\frac{7}{6}\)

\(\Rightarrow\frac{7}{6}x=-\frac{7}{6}\Leftrightarrow x=-\frac{7}{6}:\frac{7}{6}=-\frac{7}{6}\cdot\frac{6}{7}=-1\)

a,x+\(\frac{3}{4}\)=\(\frac{1}{4}\)

=>x           =\(\frac{1}{4}\)-\(\frac{3}{4}\)

=> x          =\(\frac{-1}{2}\)   

Vậy x=\(\frac{-1}{2}\)

b,\(\frac{2}{3}\)x-\(\frac{3}{7}\)=\(\frac{1}{7}\)

=>\(\frac{2}{3}\)x          =\(\frac{1}{7}\)+\(\frac{3}{7}\)

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

=>         x            =\(\frac{4}{7}\) :\(\frac{2}{3}\)

=>         x            =\(\frac{4}{7}\).\(\frac{3}{2}\)

=>         x             =\(\frac{6}{7}\)

Vậy x=\(\frac{6}{7}\)

c,\(\frac{12}{3}\)x-50%x   =\(\frac{-11}{6}\)

=>\(\frac{12}{3}\)x-\(\frac{1}{2}\)x=\(\frac{-11}{6}\)

=>x(\(\frac{12}{3}\)-\(\frac{1}{2}\))=\(\frac{-11}{6}\)

=>x(\(\frac{24}{6}\)-\(\frac{3}{6}\))=\(\frac{-11}{6}\)

=>x\(\frac{21}{6}\)              =\(\frac{-11}{6}\)

=>x                           =\(\frac{-11}{6}\):\(\frac{21}{6}\)

=>x                            =\(\frac{-11}{6}\).\(\frac{6}{21}\)

=>x                            =\(\frac{-11}{21}\)

Vậy x=\(\frac{-11}{21}\)

Ý c, mk tưởng 1 2/3 và -1 1/6 là phân số nên mk giải sai ý đó

BÀI GIẢI LẠI :

1\(\frac{2}{3}\)x-50%x=-1\(\frac{1}{6}\)

\(\frac{5}{3}\)x-\(\frac{1}{2}\)x=\(\frac{-7}{6}\)

x(\(\frac{5}{3}\)-\(\frac{1}{2}\))=\(\frac{-7}{6}\)

x\(\frac{7}{6}\)             =\(\frac{-7}{6}\) 

x                       =\(\frac{-7}{6}\):\(\frac{7}{6}\)

x                        =-1

Vậy x=-1

xl vì bài giải trước nha

a) \(\left(x+2\right)\left(x+3\right)-\left(x+1\right)\left(x+7\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-8x-7=6\)

\(\Leftrightarrow-3x=7\)

\(\Leftrightarrow x=-\frac{7}{3}\)

b) \(\left(8x-3\right)\left(3x+2\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(\Leftrightarrow\left(8x-3\right)\left(9x^2+12x+4\right)-4x^2-23x-28=10x^2+3x-1-33\)

\(\Leftrightarrow72x^3+69x^2-4x-12-14x^2-26x+6=0\)

\(\Leftrightarrow72x^3+55x^2-30x-6=0\)

Nghiệm vô tỉ: \(x_1=-1,078...\) ; \(x_2=0,476...\) ; \(x_3=-0,162...\)

a) (x + 2)(x + 3) - (x + 1)(x + 7) = 6

=> x(x + 3) + 2(x + 3) - x(x + 7) - 1(x + 7) = 6

=> x² + 3x + 2x + 6 - x² - 7x - x - 7 = 6

=> x² + 5x + 6 - x² - 7x - x - 7 = 6

=> (x² - x²) + (5x - 7x - x) + (6 - 7) = 6

=> -3x - 1 = 6

=> -3x = 7

=> x = -7/3

b) (8x - 3)(3x + 2)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33

=> (8x - 3)(9x² + 12x + 4) - [4x(x + 4) + 7(x + 4)] = 2x(5x - 1) + 1(5x - 1) - 33

=> 8x(9x² + 12x + 4) - 3(9x² + 12x + 4) - (4x² + 16x + 7x + 28) = 10x² - 2x + 5x - 1 - 33

=> 72x³ + 96x² + 32x - 27x² - 36x - 12  - 4x² - 16x - 7x - 28 - 10x² + 2x - 5x + 1 + 33 = 0

=> 72x³ + (96x² - 27x² - 10x² - 4x²) + (32x - 36x  - 16x -  7x + 2x - 5x)  + (-12  - 28 + 1 +  33) = 0

=> 72x³ + 55x² - 30x - 6 = 0

=> x vô nghiệm

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

Đây là toán lớp 7 mà bạn.

BBạn có trả lời dc ko, ko thì thôi

a) |3x-2|=7

=> 3x-2=7 hoặc 3x-2=-7

=> 3x=9 hoặc 3x=-5

=> x=3 hoặc x=\(-\dfrac{5}{3}\)

a)\(2x\left(x+1\right)-3-2x=5\)

\(\Leftrightarrow2x^2+2x-3-2x=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4=\left(-2\right)^2=2^2\)

              \(\Rightarrow x=2;-2\)

b)\(2x\left(3x+1\right)+\left(4-2x\right)=7\)

\(\Leftrightarrow6x^2+2x+4-2x=7\)

\(\Leftrightarrow6x^2+4=7\)

\(\Leftrightarrow6x^2=3\)

\(\Leftrightarrow x^2=\frac{1}{2}=-\sqrt{\frac{1}{2}}=\sqrt{\frac{1}{2}}\)

c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x-1\right)^2=6\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2-2x+1\right)=6\)

\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)

​\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)​​

​\(\Leftrightarrow3x^2+15x=0\)

\(\Leftrightarrow3x\left(x+5\right)=0\)

         \(\Rightarrow\orbr{\begin{cases}3x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

a.   2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0

2x=- 4/5 hoặc 3x=1/2

x=-2/5 hoặc x=\(\dfrac{1}{6}\)

b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0

x=2/5 hoặc x=-\(\dfrac{4}{7}\)

d. x(1+5/8-12/16)=1

\(\dfrac{7}{8}\)x=1=> x=8/7

Chuyển vế->tìm x