a,134

b,91

c,59

d,30

e,107

g,50

h,9

y,110

a, = 134

b, 91

c, 59

d, 30

e, 107

g, 50

h 9

p, 110

các bạn ấy trả lời đúng đó ha

a)P= 15+45x+33x^2-55x^3

a: \(P=\left(5x-1-5x-4\right)^2=\left(-3\right)^2=9\)

b: \(Q=\left(x+y\right)^3-3xy\left(x+y\right)=x^3+y^3\)

c: \(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)

\(=\dfrac{5^{32}-1}{2}\)

a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42

sửa đề đến đây thôi bạn nhé, do nếu thêm vào thì mình cũng ko biết có quy luật gì nữa :<

\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)

\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)

1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)

Suy ra: \(5x^2+3x-9=5x^2-5x\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(tm\right)\)

2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)

\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(6x=3x-15\)

\(\Leftrightarrow3x=-15\)

hay \(x=-5\left(loại\right)\)

2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)

Vậy pt vô nghiệm.

3. ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{-3(x+4)}{(x-4)(x+4)}-\frac{3-5x}{(x-4)(x+4)}=\frac{x-4}{(x-4)(x+4)}\)

\(\Rightarrow -3(x+4)-(3-5x)=x-4\)

\(\Leftrightarrow 2x-15=x-4\Leftrightarrow x=11\) (thỏa mãn)

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

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)