\(\text{a)}350:\left[19-\left(7x-3\right)\right]=350\)

\(\Rightarrow19-\left(7x-3\right)=1\)

\(\Rightarrow19-7x+3=1\)

\(\Rightarrow-7x=-21\)

\(\Rightarrow x=3\)

\(\text{b)}7.\left[109-\left(x-9\right)\right]-420=0\)

\(\Rightarrow7\left(109-x+9\right)=420\)

\(\Rightarrow118-x=60\)

\(\Rightarrow x=58\)

\(\text{c)}252-2\left\{200-4\left[90-7.\left(19-9\right)\right]\right\}.x=4\)

\(\Rightarrow2\left\{200-4\left[90-7.10\right]\right\}.x=250\)

\(\Rightarrow\left[200-4\left(90-70\right)\right].x=125\)

\(\Rightarrow\left[200-4.20\right].x=125\)

\(\Rightarrow120x=125\)

\(\Rightarrow x=\frac{125}{120}=\frac{25}{24}\)

\(\text{Xin điểm ạ}\)

350:[19-(7x-3)]=350

[19-(7x-3)]=350:350

[19-(7x-3)]=1

(7x-3)=19-1

(7x-3)=18

7x=18+3

7x=21

x=21:7

x=3

vậy x=3

7.[109-(x-19)]-420=0

7.[109-(x-19)]=0+420

7.[109-(x-19)]=420

[109-(x-19)]=420:7

[109-(x-19)]=60

(x-19)=109-60

(x-19)=49

x=49+19

x=58

vậy x=58

254-2{200-4.[90-7.(19-9)]}.x=4

254-2{200-4.[90-7.10]}.x=4

254-2{200-4.[90-70]}.x=4

254-2{200-4.20}.x=4

254-2{200-80}.x=4

254-2.120.x=4

254-240.x=4

240.x=254-4

240.x=250

x=250:240

x=25/24

vậy x=25/24

2:

a: =>(x-9)(x-1)=0

=>x=9 hoặc x=1

b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0

=>(x+4)(x^2-4x+16+x-16)=0

=>(x+4)(x^2-3x)=0

=>x(x-3)(x+4)=0

=>x=0;x=3;x=-4

 bài 2 :

a: =>(x-9)(x-1)=0

=>x=9 hoặc x=1

b: =>(x+4)(x^2-4x+16)+(x+4)(x-16)=0

=>(x+4)(x^2-4x+16+x-16)=0

=>(x+4)(x^2-3x)=0

=>x(x-3)(x+4)=0

=>x=0;x=3;x=-4

a, 7x - 3 = 3x + 1

7x - 3x = 1 + 3

4x = 4

x = 4 : 4

x = 1

b, 2x - 5 - 9x = 9

2x - 9x = 9 + 5

-7x = 14

x = 14 : (-7)

x = -2

c, 4 - (x + 3) = 7 - (2x + 6)

4 - x - x = 7 - 2x - 6

-x - x + 2x = 7 - 6 - 4

0x = -3

Vì 0 nhân cho số nào cũng bằng 0 nên không có giá trị nào của x thỏa mãn

7x-3=3x+1

7x-3x=3+1

4x=4

x=4/4

x=1

vậy x =1

b , 2x-5-9x=9

2x-9x-5=9

-7x=9+5

-7x=14

x=14/(-7)

x=-2

vậy x=-2

a) 150 : [17 - (2x - 3)] = 350

=> 17 - (2x - 3) = \(\frac{150}{350}=\frac{3}{7}\)

=> 2x - 3 = \(17-\frac{3}{7}=\frac{116}{7}\)

=> 2x = \(\frac{116}{7}+3\)

=> 2x = \(\frac{137}{7}\)

=> x = \(\frac{137}{7}:2=\frac{137}{14}\)

b) 4 . [105 - (x - 9)] - 486 = 0

=> 4. [105 - (x - 9)] = 486

=> 105 - (x - 9) = 243/2

=> x - 9 = \(105-\frac{243}{2}=-\frac{33}{2}\)

=> x = \(-\frac{33}{2}+9=-\frac{15}{2}\)

c) 1000 - 2{600 - 4.[198 - 7.(21 - 7)]}.x = 200

=> 1000 - 2{600 - 4.[198 - 7.14]}.x = 200

=> 1000 - 2{600 - 4.[198 - 98].x = 200

=> 1000 - 2{600 - 4.100}.x = 200

=> 1000 - 2{600 -400}.x = 200

=> 1000 - 2.200.x = 200

=> 2.200.x = 800

=> 400.x = 800

=> x = 2

Câu 2 :

a) 350 : [19 - (7x - 3)]  = 350

=> 19 - (7x - 3) = 1

=> 7x - 3 = 18

=> 7x = 21

=> x = 3

b) 7.[109 - (x - 9)] - 420 = 0

=> 7.[109 - (x - 9)] = 420

=> 109 - (x - 9) = 60

=> x - 9 = 49

=> x = 49 + 9 = 58

c) 254 - 2{200 - 4.[90 - 7.(19 - 9)]}.x = 54

=> 254 - 2{200 - 4.[90 - 7.10]}.x = 54

=> 254 - 2{200 - 4[90 - 70]} . x= 54

=> 254 - 2{200 - 4.20} . x = 54

=> 254 - 2{200 - 80}.x = 54

=> 254 - 2.120 . x=  54

=> 2.120.x = 200

=> 240.x = 200

=> \(x=\frac{5}{6}\)

2) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}\sqrt{25\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.\sqrt{25}.\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9\left(x+2\right)}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9}.\sqrt{x+2}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)

\(\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)

\(-\sqrt{x+2}=-9\)

\(x+2=81\)

\(\Rightarrow x=79\)

3) \(\sqrt{x^2-4x+4}=7x-1\)

\(\sqrt{x^2-2.x.2+2^2}=7x-1\)

\(\sqrt{\left(x-2\right)^2}=7x-1\)

\(x-2=7x-1\)

\(-2=7x-1-x\)

\(-2+1=7x-x\)

\(-1=6x\)

\(-\frac{1}{6}=x\)

\(\Rightarrow x=-\frac{1}{6}\)

Đề yêu cầu gì em?

\(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

\(x^2-7x+12=\left(x-2\right)\left(x-5\right)\)

\(x^2+x-12=\left(x-5\right)\left(x+6\right)\)

\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

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

\(4x^2-7x-2=4\left(x-2\right)\left(x+\dfrac{1}{4}\right)\)

\(4x^2+15x+9=4\left(x+\dfrac{3}{4}\right)\left(x+3\right)\)

\(\frac{9x-13}{16}=\frac{7y-11}{19}=\frac{z-5}{3}\)

\(\frac{\left(9x-13\right)\cdot\frac{5}{9}}{16\cdot\frac{5}{9}}=\frac{7y-11}{19}=\frac{3\left(z-5\right)}{3\cdot3}\)

\(\frac{5x-\frac{65}{9}}{\frac{80}{9}}=\frac{7y-11}{19}=\frac{3z-15}{9}\)

Áp dụng tc dãy tỉ số bằng nhau, ta được:

\(\frac{5x-\frac{65}{9}}{\frac{80}{9}}=\frac{7y-11}{19}=\frac{3z-15}{9}=\frac{5x-\frac{65}{9}-7y+11+3z-15}{\frac{80}{9}-19+9}=\frac{\left(5x-7y+3z\right)-\left(\frac{65}{9}-11-15\right)}{\frac{80}{9}-\frac{171}{9}+\frac{81}{9}}\)

\(=\frac{9-\left(\frac{65}{9}-\frac{99}{9}-\frac{135}{9}\right)}{-\frac{10}{9}}=\frac{9+\frac{169}{9}}{-\frac{10}{9}}=-\frac{88}{9}:-\frac{10}{9}=\frac{44}{5}\)

=>x=[(44/5*16)+13]/9=769/45

=>y=[(44/5*19)+11]/7=891/35

=>z=(44/5*3)+5=157/5

hình như sai hay sao, số lớn quá