Theo đề bài ta có :

a3 =b8 =c5 Và 2a + 3b - c = 50

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

a\3 =b\8 =c\5 =2a+3b−c\6+24−5 =50\25 =2

⇒a=2.3=6

b=2.8=16

c=2.5=10

5x = 8y = 20z => 5x40 =8y40 =20z40 =>x8 =y5 =z2

áp dụng tính chất dãy tỉ số bằng nhau , và x - y - z = 3. ta có

x8 =y5 =z2 =x−y−z8−5−2 =31 =3

x8 =3=>x=8.3

=> x = 24

y5 =3=>y=5.3

=> y=15

z2 =3=>z=2.3

=> z=6

a) x : 1/6 = 3
    x         = 3 x 1/6
    x         = 1/2
b) 3/7 + x = 2/7 + 1
    3/7 + x = 9/7
             x = 9/7 - 3/7
             x = 6/7

\(x:\dfrac{1}{6}=3\)

\(x=\dfrac{1}{6}\times3=\dfrac{3}{6}=\dfrac{1}{2}\)

\(\dfrac{3}{7}+x=\dfrac{2}{7}+1\)

\(\dfrac{3}{7}+x=\dfrac{2}{7}+\dfrac{1}{1}=\dfrac{2}{7}+\dfrac{7}{7}=\dfrac{9}{7}\)

\(x=\dfrac{9}{7}-\dfrac{3}{7}=\dfrac{6}{7}\)

\(a.x:\dfrac{1}{6}=3\)

\(x=3\times\dfrac{1}{6}=\dfrac{1}{2}\)

\(b.\dfrac{3}{7}+x=\dfrac{2}{7}+1\)

\(x=1+\dfrac{2}{7}-\dfrac{3}{7}=\dfrac{6}{7}\)

a) \(8x^3-27y^6\)

\(=\left(2x\right)^3-\left(3y^2\right)^3\)

\(=\left(2x-3y^2\right)\left[\left(2x\right)^2+2x.3y+\left(3y\right)^2\right]\)

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

b) \(a^3b^3c^3-1\)

\(=\left(abc\right)^3-1^3\)

\(=\left(abc-1\right)\left(a^2b^2c^2+abc+1\right)\)

c) \(64x^3+\dfrac{1}{8}y^3\)

\(=\left(4x\right)^3+\left(\dfrac{1}{2}y\right)^3\)

\(=\left(4x+\dfrac{1}{2}y\right)\left[\left(4x\right)^2+4x.\dfrac{1}{2}y+\left(\dfrac{1}{2}y\right)^2\right]\)

\(=\left(4x+\dfrac{1}{2}y\right)\left(4x^2+2xy+\dfrac{1}{4}y^2\right)\)

d) \(125+y^3\)

\(=5^3+y^3\)

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

e) \(a^6-b^6\)

\(=\left(a^3\right)^2-\left(b^3\right)^2\)

\(=\left(a^3-b^3\right)\left(a^3+b^3\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)

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

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

\(=\left[2x-3\left(3x+5\right)\right]\left[2x+3\left(3x+5\right)\right]\)

\(=\left(2x-9x-15\right)\left(2x+9x+15\right)\)

\(=\left(-7x-15\right)\left(11x+15\right)\)

3:

a: =>x=0 hoặc x+5=0

=>x=0 hoặc x=-5

b: =>x^2=4

=>x=2 hoặc x=-2

c: =>(x-5)(2x+1+x+6)=0

=>(x-5)(3x+7)=0

=>x=5 hoặc x=-7/3

1.

a. 2x - 6 > 0 

\(\Leftrightarrow\)  2x  > 6

\(\Leftrightarrow\)    x  > 3

S = \(\left\{x\uparrow x>3\right\}\) 

b. -3x + 9 > 0

\(\Leftrightarrow\)  - 3x   > - 9 

\(\Leftrightarrow\)      x < 3

S = \(\left\{x\uparrow x< 3\right\}\) 

c. 3(x - 1) + 5 > (x - 1) + 3

\(\Leftrightarrow\) 3x - 3 + 5 > x - 1 + 3

\(\Leftrightarrow\) 3x - 3 + 5 - x + 1 - 3 > 0

\(\Leftrightarrow\) 2x > 0 

\(\Leftrightarrow\)   x > 0

S = \(\left\{x\uparrow x>0\right\}\) 

d. \(\dfrac{x}{3}-\dfrac{1}{2}>\dfrac{x}{6}\) 

\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3}{6}>\dfrac{x}{6}\)

\(\Leftrightarrow2x-3>x\)

\(\Leftrightarrow2x-3-x>0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

\(S=\left\{x\uparrow x>3\right\}\)

2.

a. 

Ta có: a > b

3a > 3b (nhân cả 2 vế cho 3)

3a + 7 > 3b + 7 (cộng cả 2 vế cho 7)

b. Ta có: a > b

a > b (nhân cả 2 vế cho 1)

a + 3 > b + 3 (cộng cả 2 vế cho 3) (1)

Ta có; 3 > 1

b + 3 > b + 1 (nhân cả 2 vế cho 1b) (2)

Từ (1) và (2) \(\Rightarrow\) a + 3 > b + 1 

c.

5a - 1 + 1 > 5b - 1 + 1 (cộng cả 2 vế cho 1)

5a . \(\dfrac{1}{5}\) > 5b . \(\dfrac{1}{5}\) (nhân cả 2 vế cho \(\dfrac{1}{5}\) )

a > b

3.

a. 2x(x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\) 

\(S=\left\{0,-5\right\}\)

b. x² - 4 = 0 

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(S=\left\{0,4\right\}\)

d. (x - 5)(2x + 1) + (x - 5)(x + 6) = 0

\(\Leftrightarrow\left(x-5\right)\left(2x+1+x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)

\(S=\left\{5,\dfrac{-7}{3}\right\}\)

a)x-3/4=5/12 : 10/3

x-3/4= 1/8

x = 1/8 + 3/4

x = 7/8

b)7/6 - x=3/5 x 15/18

7/6 - x = 1/2

x = 7/6 - 1/2

x = 2/3

a)x-3/4=5/12 : 10/3

x-3/4= 1/8

x = 1/8 + 3/4

x = 7/8

b)7/6 - x=3/5 x 15/18

7/6 - x = 1/2

x = 7/6 - 1/2

x = 2/3

Làm ơn làm nhanh giúp mình đang cần gấp lắm

a)Thay \(x=\dfrac{-2}{3}\) vào\(x^3-6x^2-9x-3\):

\(\left(\dfrac{-2}{3}\right)^3-6\left(\dfrac{-2}{3}\right)^2+9.\dfrac{2}{3}-3\)

\(=\dfrac{-8}{27}-\dfrac{8}{3}+6-3\)

\(=\dfrac{-8-72}{27}+3=\dfrac{-80}{27}+3=\dfrac{1}{27}\)

b) Ta có: \(\dfrac{a}{b}=\dfrac{3}{4}\Rightarrow a=3k;b=4k\)

\(\Rightarrow\dfrac{2a-5b}{a-3b}=\dfrac{6k-20k}{3k-12k}=\dfrac{-14k}{-9k}=\dfrac{14}{9}\)

c) Có: a-b=7\(\Rightarrow a=b+7\)

Thay vào \(\dfrac{3a-b}{2a+7}+\dfrac{3b-a}{2b-7}=\dfrac{2b+21}{2b+21}+\dfrac{2b-7}{2b-7}\)

\(=1+1=2\)