a) x = 2 

b) x = 2     

c) x = 2

d) x = 1.

1) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\Leftrightarrow\left(2x-5\right).-2=0\)

\(\Leftrightarrow-4x+10=0\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}.\)

Vậy \(S=\left\{\frac{5}{2}\right\}\)

2)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right).\left(x^2-3x+9+x-9\right)=0\)

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

\(\Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\)

\(\Leftrightarrow x+3=0\)hoặc \(x=0\)hoặc \(x-2=0\)

\(\Leftrightarrow x=-3\)hoặc \(x=0\)hoặc \(x=2\)

Vậy \(S=\left\{-3;0;2\right\}\)

Lời giải:

a. $121-3(x-5)=6$

$3(x-5)=121-6=115$

$x-5=115:3=\frac{115}{3}$

$x=\frac{115}{3}+5=\frac{130}{3}$

b.

$2x-138=2^3.3^2=72$

$2x=72+138=210$

$x=210:2=105$

c.

$x-3\vdots 7$

$\Rightarrow x-3\in\left\{0;7;14;21;28;35;42;49; 56;...\right\}$

Mà $10< x< 50$ nên $x\in\left\{14;21;28;35;42;49\right\}$

d.

$27\vdots x+1$

$\Rightarrow x+1\in\left\{\pm 1; \pm 3; \pm 9; \pm 27\right\}$

$\Rightarrow x\in\left\{0; -2; -4; 2; 8; -10; 26; -28\right\}$

a ) 121-3.(x - 5 ) = 6

3.(x-5) = 121 -6

3. (x-5)=115

x-5  = 115:3

x-5=35

x=35+5

x = 40

b) 2x - 138 = 2'3. 3'2

2x -138=8.9

2x-138=72

2x=72+138

2x=210

x=210:2

x=105

c) theo bài ra : x-3 ∈ B(7)

ta có B(7)=(0,7,14,21,28,35,49,56,...)

=) x-3 ∈ ( 0,7,14,21,28,35,49,56,...)

=) x ∈( 3 , 10,17,24,31,38,42,58,..)

mà 10 <x<50 nên x ∈ ( 17 , 24 ,31,38,42 )

vậy x ∈(17,24,31,38,42)

Lời giải:

a. $121-3(x-5)=6$

$3(x-5)=121-6=115$

x-5=115:3=\frac{115}{3}x−5=115:3=3115​

x=\frac{115}{3}+5=\frac{130}{3}x=3115​+5=3130​

b.

2x-138=2^3.3^2=722x−138=23.32=72

$2x=72+138=210$

$x=210:2=105$

c.

$x-3⋮7$

$\Rightarrow x-3\in \left\{0;7;14;21;28;35;42;49;56;...\right\}$

Mà $10<x<50$ nên $x\in \left\{14;21;28;35;42;49\right\}$

d.

$27⋮x+1$

$\Rightarrow x+1\in \left\{\pm 1;\pm 3;\pm 9;\pm 27\right\}$

$\Rightarrow x\in \left\{0;-2;-4;2;8;-10;26;-28\right\}$

a, x = 28

b, x = 1

c, x = 24

d, x = 3

a) x+3/22=-27/121.11/9

⇒ x+3/22=-3/11

⇒ x=-9/22

b) (3/2/20-2/1/20)x+2/7=3/5

⇒ (31/10-41/20)x=11/35

⇒ 21/20x=11/35

⇒x=44/147

c) |3/2x-5|-1/2=-1/4

⇒ |3/2x-5|=1/4

⇒ hoặc 3/2x-5=1/4⇒3/2x=21/4⇒x=7/2

hoặc 3/2x-5=-1/4⇒3/2x=19/4⇒x=19/6