5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

<=> (x-1-3x-3)(x-1+3x+3) = 0

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)

a: \(\Leftrightarrow5x^2-45x-7x+63-\left(5x-2\right)\left(x-3\right)=2x^2-8x-2x^2-3x+2x+3\)

\(\Leftrightarrow5x^2-52x+63-\left(5x-2\right)\left(x-3\right)=-9x+3\)

\(\Leftrightarrow5x^2-52x+63-5x^2+15x+2x-6=-9x+3\)

=>-37x+57=-9x+3

=>28x=-54

hay x=-27/14

b: \(\Leftrightarrow-x^2+19x+3x-30+x^2-5x-24=\left(5x^2-1\right)\left(x+3\right)-5x^3-15x^2\)

\(\Leftrightarrow17x-54=5x^3+15x^2-x-3-5x^3-15x^2\)

=>18x=51

hay x=17/6

a/ pt đãcho tương đương với

6x\(^2\)+ 21x -2x-7-6x+5x-6x+5= 16

<=>18x=18

=> x=1

b/ pt đã cho tương đương với

10x\(^2\)+9x-10x\(^2\)-15x+2x+3= 8

<=> -4x=5

<=.> x=-\(\frac{5}{4}\)

c/ pt đã cho tương đương với

21x-15x\(^2\)-35+25x+15x\(^2\)-10x+6x-4-2=0

<=>42x=41

<=> x= \(\frac{41}{42}\)

d/ pt đã cho tương đương với

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

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

<=> 8x\(^2\)+6x-5x=0

<=>8x\(^2\)+16x-10x-5x=0

<=> (x+2)2x-5(x+2)=0

<=> (x+2)(2x-5)=0

<=>x+2=0 hoặc 2x+5=0

=> x=-2 hoặc x= -\(\frac{5}{2}\)

a) (3x - 1)(2x + 7) - (x + 1)(6x - 5) = 16

6x² + 21x - 2x - 7 - 6x² + 5x - 6x + 5 = 16

(6x² - 6x²) + (21x - 2x + 5x - 6x) + (-7 + 5) = 16

18x - 2 = 16

18x = 18

x = 1

Vậy x = 1

b) (10x + 9)x - (5x - 1)(2x + 3) = 8

10x² + 9x - 10x² - 15x + 2x + 3 = 8

(10x² - 10x²) + (9x - 15x + 2x) + 3 = 8

-4x + 3 = 8

-4x = 5

x = \(\frac{-5}{4}\)

Vậy x = \(\frac{-5}{4}\)

c) x(x + 1)(x + 6) - x³ = 5x

(x² + x)(x + 6) - x³ = 5x

x³ + 7x² + 6x - x³ = 5x

7x² + 6x = 5x

x(7x + 6) = 5x

=> 7x + 6 = 5

7x = -1

x = \(\frac{-1}{7}\)

Vậy x = \(\frac{-1}{7}\)

d) (3x - 5)(7 - 5x) + (5x + 2)(3x - 2) - 2 = 0

21x - 15x² - 35 + 25x + 15x² - 10x + 6x - 4 - 2 = 0

(-15x² + 15x²) + (21x + 25x - 10x + 6x) + (-35 - 4 - 2) = 0

42x - 41 = 0

42x = 41

x = \(\frac{41}{42}\)

Vậy x = \(\frac{41}{42}\)

cái này có ngoặc ko vậy

bài này là nhân đa vs đa hay là nhân đa vs đơn

\(A=x^5-5x^4+5x^3-5x^2+5x-1\)

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

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

\(=3\)

\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

\(g,\left(x-1\right)^2-\left(x+2\right)^2\)

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

\(=-6x-3\)

a, \(\frac{5}{7}.3x-\frac{8}{5}=\frac{9}{35}\)

=> \(\frac{15}{7}x=\frac{9}{35}+\frac{8}{5}\)=> \(\frac{15}{7}x=\frac{9}{35}+\frac{56}{35}\)

=> \(\frac{15}{7}x=\frac{65}{35}=\frac{13}{7}\)=> \(x=\frac{13}{7}:\frac{15}{7}=\frac{13}{15}\)

vậy \(x=\frac{13}{15}\)

b, \(\frac{2}{9}.5x+\frac{1}{2}-\frac{1}{18}=\frac{5}{36}\)

=> \(\frac{10}{9}x+\frac{1}{2}=\frac{5}{36}+\frac{1}{18}\)=\(\frac{5}{36}+\frac{2}{36}=\frac{7}{36}\)

=> \(\frac{10}{9}x=\frac{7}{36}-\frac{1}{2}\)=\(\frac{7}{36}-\frac{18}{36}\)=\(\frac{-11}{36}\)=> \(x=\frac{-11}{36}:\frac{10}{9}\)=\(\frac{-11}{36}.\frac{9}{10}\)=\(\frac{-11}{40}\)

vậy x=\(\frac{-11}{40}\)

a, ta có : \(\frac{5}{7}.\frac{3x-8}{5}=\frac{9}{35}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}:\frac{5}{7}\Leftrightarrow\frac{3x-8}{5}=\frac{9}{35}.\frac{7}{5}\)

\(\Leftrightarrow\frac{3x-8}{5}=\frac{9}{5.5}\Leftrightarrow3x-8=\frac{9}{25}.5\Leftrightarrow3x-8=\frac{9.5}{25}\)

\(\Leftrightarrow3x-8=\frac{9}{5}\Leftrightarrow3x-8=\frac{9}{5}+8\Leftrightarrow3x=\frac{9+8.5}{5}\)

\(\Leftrightarrow3x=\frac{49}{5}\Leftrightarrow x=\frac{49}{5}:3\Leftrightarrow x=\frac{49}{5}.\frac{1}{3}=\frac{49}{15}\)

~ Vậy, ta tìm được \(x=\frac{49}{15}\)

b, Ta có : \(\frac{2}{9}.\frac{5x+1}{2}-\frac{1}{18}=\frac{5}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5}{36}+\frac{1}{18}\)

\(\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{5+2.1}{36}\Leftrightarrow\frac{2}{9}.\frac{5x+1}{2}=\frac{7}{36}\)

\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}:\frac{2}{9}\Leftrightarrow\frac{5x+1}{2}=\frac{7}{36}.\frac{9}{2}\Leftrightarrow\frac{5x+1}{2}=\frac{7.9}{4.9.2}\)

\(\Leftrightarrow\frac{5x+1}{2}=\frac{7}{8}\Leftrightarrow5x+1=\frac{7}{8}.2\Leftrightarrow5x+1=\frac{7.2}{8}\)

\(\Leftrightarrow5x+1=\frac{7}{4}\Leftrightarrow5x=\frac{7}{4}-1\Leftrightarrow5x=\frac{7-1.4}{4}\)

\(\Leftrightarrow5x=\frac{3}{4}\Leftrightarrow x=\frac{3}{4}:5\Leftrightarrow x=\frac{3}{4}.\frac{1}{5}\Leftrightarrow x=\frac{3}{20}\)

~ Vậy, ta tìm được \(x=\frac{3}{20}\)

mk và CTV làm 2 đề khác nhau, vậy đề nào mới là đề đúng của bn vậy