\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)

\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)

A = ( -4/5 + 4/3 ) + (-5/4 + 14/5) - 7/3

  =   8/15 + 31/20 - 7/3

  =  25/12 - 7/3

  = -1/4

B = 8/3 x 2/5 x 3/8 x 10x 19/92

   = 16/15 x 15/4 x 19/92

   = 4x19/92

   = 19/23

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

\(\left(x-2\right)\left(x+1\right)=x^2-4\)

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)

\(2x^3+3x^2-32x=48\)

\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)

\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)

\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)

giúp mình

giup minh

\(\dfrac{5}{2}\times\dfrac{3}{7}-\dfrac{3}{4}\div\dfrac{7}{3}=\dfrac{5}{2}\times\dfrac{3}{7}-\dfrac{3}{4}\times\dfrac{3}{7}=\dfrac{5}{2}\times\dfrac{3}{7}-\dfrac{3}{7}\times\dfrac{3}{4}\) 

                           \(=\dfrac{3}{7}\times\left(\dfrac{5}{2}-\dfrac{3}{4}\right)=\dfrac{3}{7}\times\dfrac{7}{4}=\dfrac{3}{4}\) 

\(\dfrac{5}{4}\div\dfrac{15}{12}-\dfrac{5}{4}\div\dfrac{1}{2}=\dfrac{5}{4}\times\dfrac{12}{15}-\dfrac{5}{4}\times2\)

                              \(=\dfrac{5}{4}\times\left(\dfrac{12}{15}-2\right)\) 

                              \(=\dfrac{5}{4}\times\dfrac{-6}{5}=\dfrac{-3}{2}\) 

\(\dfrac{1}{26}+\dfrac{7}{30}+\dfrac{12}{26}+\dfrac{8}{3}=\left(\dfrac{1}{26}+\dfrac{12}{26}\right)+\left(\dfrac{7}{30}+\dfrac{8}{3}\right)\)

                                 \(=\dfrac{1}{2}+\dfrac{29}{10}=\dfrac{17}{5}\)

\(x+50-2\times3=54\)

\(x+50-2=54\div3\)

\(x+50-2=18\)

\(x+50=18+2\)

\(x+50=20\)

\(x=20-50\)

\(x=-30\)

Hình như sai do lớp 3 chưa học âm

X + 50 - 2x3 = 54

X + 50 = 54 + 6

X + 50 = 60

X = 60 - 50

X = 10

x+50-2x3=54

x+50-6=54

x+50=54+6

x+50=50

x=60-50

x=10

\(x:\left[\frac{8}{5}.\left(\frac{2}{3}\right)^2-\frac{2}{5}\right]=\frac{15}{7}+\frac{6}{5}.\left[\left(\frac{15}{7}\right)^2-\frac{50}{49}\right]\)

\(\Rightarrow x:\frac{14}{45}=\frac{15}{7}+\frac{6}{5}.\frac{25}{7}\)

\(\Rightarrow x:\frac{14}{45}=\frac{15}{7}+\frac{30}{7}\)

\(\Rightarrow x:\frac{14}{45}=\frac{45}{7}\)

\(\Rightarrow x=\frac{45}{7}.\frac{14}{45}\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

Chúc bạn học tốt!

bài 2 tìm x

a,106- ( x+ 7) =9

x+7 = 106 - 9

x+7 = 107

x= 107 - 7

x=100

b, 2 x ( x+ 4) + 5 =65

2 x (x+4) = 65 - 5

2 x (x+4) = 60

x+4 = 60:2

x+4= 30

x= 30 - 4

x=26

c, (16x x -32) x 45=0

16 x X - 32 = 0: 45

16 x X - 32 =0

16 x X = 0 + 32

16 x X = 32

X= 32:16

X=2

d, x+4 x x = 100 : 5

X + 4 x X = 20

(1+4) x X = 20

5 x X  = 20

X= 20:5

X=4

Bài 1:

a, 24 x 66 - 24 + 36 x 24

= 24 x (66-1+36)

= 24 x 101

= 24 x 100 + 24

= 2400+24

=2424

b, 29 x 173 - 29 x 72 - 29

= 29 x (173 - 72 -1)

= 29 x 100

= 2900

c, 1+5+10+15+...+55+60

= (5+60) x 6 + 1 (có 6 cặp)

= 65 x 6 + 1

= 390 +1 =391

d, D= 1 x 2 + 2 x 3 + 3 x 4 +...+18 x 19 + 19 x 20

3D= 1 x 2 x 3+ 2 x 3 x (4-1) + 3 x 4 x (5-2) +...+ 18 x 19 x (20 - 17) + 19 x 20 x (21-18)

3D= 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 +.... + 19 x 20 x 21 - 18 x 19 x 20

3D= 19 x 20 x 21 =7980

=> D= 7980:3=2660

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

\(a,=\left(2x^3-x^2+x+4x^2-2x+2-x+1\right):\left(2x^2-x+1\right)\\ =\left[x\left(2x^2-x+1\right)+2\left(2x^2-x+1\right)-x+1\right]:\left(2x^2-x+1\right)\\ =x+2\left(\text{dư }-x+1\right)\\ b,=\left[x^2\left(2x-5\right)+3\left(2x-5\right)\right]:\left(2x-5\right)\\ =x^2+3\)