a) 187 - {[497 - ( 8 x X + 11) : X] : 3 - 78} = 150

=> {[497 - ( 8 x X + 11) : X] : 3 - 78} = 187 - 150

=> {[497 - (8 x X + 11) : X] : 3 - 78} = 37

=> [497 - (8 x X +11): X ] : 3 - 78 = 37

=> [497 - (8 x X + 11) : X] : 3 = 115

=> 497 - ( 8 x X + 11) : X = 345

=> (8 x X  + 11) : X = 497 - 345 = 152

=> 8X + 11 = 152X

=> 152X - 8X = 11

=> 144X = 11

=> X = 11/144

b) 19,96 + 4,19 - 24,15 : \(\left(x:\frac{1}{4}-\frac{1}{4}\right)=23,15\)

=> 19,96 + 4,19 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)=23,15\)

=> 24,15 - 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 23,15

=> 24,15 : \(\left(x\cdot4-\frac{1}{4}\right)\)= 1

=> \(x\cdot4-\frac{1}{4}=24,15\)

=> \(x\cdot4=24,15+\frac{1}{4}=24,4\)

=> x = 24,4 : 4 = 6,1

Còn câu c tương tự

\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{12}-\frac{10}{24}+\frac{14}{39}}\)

\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)

\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)

\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)

\(B=\frac{x}{y}-\frac{3}{2}\)

Thế x = 0, 5 = 1/2 ; y = 3 ta được :

\(B=\frac{\frac{1}{2}}{3}-\frac{3}{2}=\frac{1}{6}-\frac{9}{6}=-\frac{8}{6}=-\frac{4}{3}\)

Ta có:\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)

\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\left(\frac{2}{12}+\frac{10}{24}-\frac{14}{39}\right)}\)

\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{-\frac{2}{3}\left(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\right)}\)

\(B=\frac{x}{y}+\frac{1}{-\frac{2}{3}}\)(Do\(\frac{1}{4}+\frac{5}{8}-\frac{7}{13}\ne0\))

\(B=\frac{x}{y}-\frac{3}{2}\)

Thay x = 0,5; y = 3 vào B ta được:

\(B=\frac{0,5}{3}-\frac{3}{2}\)

\(B=\frac{1}{6}-\frac{3}{2}\)

\(B=\frac{1}{6}-\frac{9}{6}\)

\(B=-\frac{4}{3}\)

Vậy\(B=-\frac{4}{3}\)tại x = 0,5; y = 3

Linz

a)

\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)

\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow\frac{-10x-11}{12}=\frac{3x-60}{12}\)

\(\Leftrightarrow\frac{-10x-11-3x+60}{12}=0\)

\(\Leftrightarrow\frac{49-13x}{12}=0\)

\(\Rightarrow49-13x=0\)

\(\Rightarrow x=\frac{-49}{13}\)

b)

\(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)

\(\Leftrightarrow\frac{8x-3-6x+4}{4}=\frac{4x-2+x+3}{4}\)

\(\Leftrightarrow\frac{2x+1}{4}=\frac{5x+1}{4}\)

\(\Leftrightarrow\frac{2x+1-5x-1}{4}=0\)

\(\Leftrightarrow\frac{-3x}{4}=0\)

\(\Rightarrow-3x=0\)

\(\Rightarrow x=0\)

e)

\(\frac{3\cdot\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\cdot\left(x-7\right)\)

\(\Leftrightarrow\frac{45x-18-24-28x+60x-420}{12}=0\)

\(\Leftrightarrow\frac{77x-462}{12}=0\)

\(\Rightarrow77x-462=0\)

\(\Rightarrow x=\frac{462}{77}=6\)

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

⇔\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)

⇔\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)

⇔\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)

⇔\(9x-10=0\)

hay 9x=10

⇔\(x=\frac{10}{9}\)

Vậy: \(x=\frac{10}{9}\)

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

⇔\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)

⇔\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)

⇔\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)

⇔\(6x-3-5x+10-3x-21=0\)

⇔\(-2x-14=0\)

⇔\(-2x=14\)

hay x=-7

Vậy: x=-7

d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)

⇔\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

⇔\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

⇔\(6x-18+7x-35-13x-4=0\)

⇔\(-21\ne0\)

Vậy: x∈∅

e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)

⇔\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)

⇔\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)

⇔\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)

⇔\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)

⇔\(-12x+96=0\)

⇔\(-12x=-96\)

hay x=8

Vậy: x=8

1 cách khác nó phức tạp và khó hơn "n" lần :)) Cơ mà nó chẳng khác của cậu là mấy :v

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

\(\Leftrightarrow4+\frac{x}{1+\frac{1}{\frac{7}{3}}}=\frac{x}{4+\frac{1}{\frac{7}{2}}}\)

\(\Leftrightarrow4+\frac{x}{1+\frac{3}{7}}=\frac{x}{4+\frac{2}{7}}\)

\(\Leftrightarrow4+\frac{x}{\frac{10}{7}}=\frac{x}{\frac{30}{7}}\)

\(\Leftrightarrow4+x.\frac{7}{10}=x.\frac{7}{30}\)

\(\Leftrightarrow4+\frac{7x}{10}=\frac{7x}{30}\)

\(\Leftrightarrow120+21x=7x\)

\(\Leftrightarrow120=7x-21\)

\(\Leftrightarrow120=-14x\)

\(\Leftrightarrow-\frac{120}{14}=-\frac{60}{7}=x\)

\(\Rightarrow x=-\frac{60}{7}\)

Tuấn Huỳnh cách của a có khác gì cách của e đâu.chỉ một bên chọn MSC còn a thì chuyển vế thôi mà

bài của Tuấn Huỳnh khác với bài của cậu chứ zZz Cool Kid zZz

b) \(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+...+\frac{87}{1000}+\frac{99}{1000}\)

\(=\frac{1+13+25+...+85+97}{1000}=\frac{\left(97+1\right).\left[\left(97-1\right):12+1\right]:2}{1000}\)

\(=\frac{49.9}{1000}=\frac{441}{1000}.\) ( Đề bài sai nhé bạn tử số : 1; 13; 25; 37; 49 ; 61; 73; 85 ; 97. )

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

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

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

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

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

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

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)

= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)

= \(x^8.\frac{1}{10}.1.1.1.1\)

= \(x^8.\frac{1}{10}\)

Mk ko pik co dung ko nua

=4/10

\(\frac{x^8}{10}\)

a) Ta có: \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-5x+45-\frac{20x+1,5}{6}=0\)

\(\Leftrightarrow\frac{21x}{24}-\frac{120x}{24}+\frac{1080}{24}-\frac{4\left(20x+1,5\right)}{24}=0\)

\(\Leftrightarrow-99x+1080-4\left(20x+1,5\right)=0\)

\(\Leftrightarrow-99x+1080-80x-6=0\)

\(\Leftrightarrow1074-179x=0\)

\(\Leftrightarrow179x=1074\)

hay x=6

Vậy: x=6

b) Ta có: \(4\left(0,5-1,5x\right)=-\frac{5x-6}{3}\)

\(\Leftrightarrow2-6x=\frac{6-5x}{3}\)

\(\Leftrightarrow\frac{3\left(2-6x\right)}{3}-\frac{6-5x}{3}=0\)

\(\Leftrightarrow6-18x-6+5x=0\)

\(\Leftrightarrow-13x=0\)

mà -13≠0

nên x=0

Vậy: x=0

c) Ta có: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{6\left(x+4\right)}{30}+\frac{30\left(-x+4\right)}{30}-\frac{10x}{30}+\frac{15\left(x-2\right)}{30}=0\)

\(\Leftrightarrow6\left(x+4\right)+30\left(4-x\right)-10x+15\left(x-2\right)=0\)

\(\Leftrightarrow6x+24+120-30x-10x+15x-30=0\)

\(\Leftrightarrow-19x+114=0\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: x=6

d) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)

\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)

\(\Leftrightarrow-181x-362=0\)

\(\Leftrightarrow-181x=362\)

hay x=-2

Vậy: x=-2

e) Ta có: \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right)-\frac{1}{3}\left(x+2\right)\)

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

\(\Leftrightarrow\frac{3\left(x+3\right)}{12}-\frac{36}{12}+\frac{6\left(x+1\right)}{12}+\frac{4\left(x+2\right)}{12}=0\)

\(\Leftrightarrow3x+9-36+6x+6+4x+8=0\)

\(\Leftrightarrow13x-13=0\)

\(\Leftrightarrow13x=13\)

hay x=1

Vậy: x=1