{[27x11-(x+12)x8]x3-79}:5=100

[27x11-(x+12)x8]x3-79=100x5=500

[27x11-(x+12)x8]x3=500+79=579

27x11-(x+12)x8=579:3=193

27x11-(x+12)=193:8=24,125

x+12=27x11-24,125

x+12=297-24,125

x+12=272,875

x=272,875-12

x=260,875.

Vay x=260,875.

{[27x11-(x+12)x8]x3-79}:5=100

   [297 -(x+12)x8]x3-79    =100x5

   [297 -(x+12)x8]x3-79    =500

   [297 -(x+12)x8]x3         =500+79

   [297 -(x+12)x8]x3         =579

    297 -(x+12)x8              =579:3

    297 -(x+12)x8              =193

           (x+12)x8              =297-193

           (x+12)x8              =104

            x+12                   =104:8

            x+12                   =13

            x                         =13-12

            x                         =1

Thử lại: {[27x11-(1+12)x8]x3-79}:5=100

dấu " \(.\) " là dấu nhân

\(\left\{\left[27.11-\left(x+12\right).8\right].3-79\right\}:5=100\)

=>\(\left[297-8.x+12.8\right].3-79=100.5\)

=> \(\left[297-8.x+96\right].3-79=500\)

=> \(\left[297-8.x+96\right].3=500+79\)

=> \(\left[\left(297+96\right)-8.x\right].3=579\)

=> \(393-8.x=579:3\)

=> \(393-8.x=193\)

=> \(8.x=393-193\)

=> \(8.x=200=>x=200:8=25\)

Vậy \(x=25\)

Chúc bạn Hk tốt!!!!

bạn tìm câu hỏi này ở đâu

a) \(\dfrac{3}{4}=0.75\)

b) \(-2\dfrac{1}{2}=\dfrac{-5}{2}< \dfrac{-5}{3}\)

1: 243^5=(3^5)^5=3^25

3*27^8=3*(3^3)^8=3^25

=>243^5=3*27^8

6: 125^5=(5^3)^5=5^15

25^7=(5^2)^7=5^14

=>125^5>25^7(15>14)

5: 78^12-78^11=78^11(78-1)=78^11*77

78^11-78^10=78^10*77

mà 11>10

nên 78^12-78^11>78^11-78^10

Đề yêu cầu gì vậy em?

đây là so sánh hay quy đồng v 

\(\dfrac{3}{5}v\text{à}\dfrac{7}{3}\\ MSC:15\\ \dfrac{3}{5}=\dfrac{3\times3}{5\times3}=\dfrac{9}{15};\dfrac{7}{3}=\dfrac{7\times5}{5\times3}=\dfrac{35}{15}\\ 9< 35\\ \Rightarrow\dfrac{3}{5}< \dfrac{7}{5}\\ \dfrac{9}{10}v\text{à}\dfrac{5}{4}\\MSC:20\\ \dfrac{9\times2}{10\times2}=\dfrac{18}{20};\dfrac{5}{4}=\dfrac{5\times5}{4\times5}=\dfrac{25}{20}\\ 18< 25\\ \Rightarrow\dfrac{9}{10}< \dfrac{5}{4}\\ \dfrac{2}{7}v\text{à}\dfrac{5}{4}\\ MSC:28\\ \dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28};\dfrac{5}{4}=\dfrac{5\times7}{4\times7}=\dfrac{35}{28}\\ 8< 35\\ \dfrac{2}{7}< \dfrac{5}{4}\\ \dfrac{8}{9}v\text{à}\dfrac{11}{6}\\ MSC:18\\ \dfrac{8\times2}{9\times2}=\dfrac{16}{18};\dfrac{11}{6}=\dfrac{11\times3}{6\times3}=\dfrac{33}{18}\\ 16< 33\\ \Rightarrow\dfrac{8}{9}< \dfrac{11}{6}\)

Quy đồng hả ?

\(\dfrac{3}{5}và\dfrac{7}{3}=\dfrac{3}{5}=\dfrac{3\times3}{5\times3}=\dfrac{9}{15};\dfrac{7}{3}=\dfrac{7\times5}{3\times5}=\dfrac{35}{15}\)

\(\dfrac{9}{10}và\dfrac{5}{4}=\dfrac{9}{10}=\dfrac{9\times2}{10\times2}=\dfrac{18}{20};\dfrac{5}{4}=\dfrac{5\times5}{4\times5}=\dfrac{25}{20}\)

\(\dfrac{2}{7}và\dfrac{5}{4}=\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28};\dfrac{5}{4}=\dfrac{5\times7}{4\times7}=\dfrac{35}{28}\)

\(\dfrac{8}{9}và\dfrac{11}{6}=\dfrac{8}{9}=\dfrac{8\times2}{9\times2}=\dfrac{16}{18};\dfrac{11}{6}=\dfrac{11\times3}{6\times3}=\dfrac{33}{18}\)

\(\dfrac{6}{5}và\dfrac{13}{8}=\dfrac{6\times8}{5\times8}=\dfrac{48}{40};\dfrac{13}{8}=\dfrac{13\times5}{8\times5}=\dfrac{65}{40}\)

\(\dfrac{4}{5}và\dfrac{3}{2}=\dfrac{4}{5}=\dfrac{4\times2}{5\times2}=\dfrac{8}{10};\dfrac{3}{2}=\dfrac{3\times5}{2\times5}=\dfrac{15}{10}\)

\(\dfrac{5}{12}và\dfrac{7}{8}=\dfrac{5}{12}=\dfrac{5\times2}{12\times2}=\dfrac{10}{24};\dfrac{7}{8}=\dfrac{7\times3}{8\times3}=\dfrac{21}{24}\)

\(\dfrac{11}{4}và\dfrac{13}{6}=\dfrac{11}{4}=\dfrac{11\times3}{4\times3}=\dfrac{33}{12};\dfrac{13}{6}=\dfrac{13\times2}{6\times2}=\dfrac{26}{12}\)

A=1/2-1/3+1/3-1/4+...+1/149-1/150

=1/2-1/150

=37/75

B=3A=37/25

\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{149\cdot150}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{149}-\frac{1}{150}\)

\(A=\frac{1}{2}-\frac{1}{150}\)

\(A=\frac{75}{150}-\frac{1}{150}\)

\(A=\frac{74}{150}=\frac{37}{75}\)

Vậy \(A=\frac{37}{75}\)

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

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

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

\(=\left(x-5\right)\left(x^2-x-1\right)\)

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

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

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

\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

\(3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs