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a: \(\Leftrightarrow6x^2-6x^2+4x-9x+6=7\)

=>-5x=1

hay x=-1/5

b: \(\Leftrightarrow5x\left(12x+7\right)-3x\left(80x-5\right)=-100\)

\(\Leftrightarrow60x^2+35x-240x^2+15x=-100\)

\(\Leftrightarrow-180x^2+50x+100=0\)

hay \(x\in\left\{\dfrac{5+\sqrt{745}}{36};\dfrac{5-\sqrt{745}}{36}\right\}\)

c: \(\Leftrightarrow21x-15x^2-35+25x-\left(10x-15x^2-4+6x\right)=4\)

\(\Leftrightarrow-15x^2+46x-35+15x^2-16x+4=4\)

=>30x-31=4

=>30x=35

hay x=7/6

31 tháng 12 2021

Answer:

\(6x^2-\left(2x+3\right)\left(3x-2\right)=7\)

\(\Rightarrow6x^2-\left(6x^2+9x-4x-6\right)=7\)

\(\Rightarrow6x^2-\left(6x^2+5x-6\right)=7\)

\(\Rightarrow6x^2-6x^2-5x+6=7\)

\(\Rightarrow-5x+6=7\)

\(\Rightarrow-5x=1\)

\(\Rightarrow x=\frac{-1}{5}\)

\(5x\left(12+7\right)-3x\left(80x-5\right)=-100\)

\(\Rightarrow5x.19-240x^2+15x=-100\)

\(\Rightarrow95x-240x^2+15x=-100\)

\(\Rightarrow-240x^2+110x+100=0\)

\(\Rightarrow-24x^2-11x-10=0\)

\(\Rightarrow24\left(x^2-\frac{11}{24}x+\frac{121}{2304}\right)-\frac{1081}{96}=0\)

\(\Rightarrow24\left(x-\frac{11}{48}\right)^2-\frac{1081}{96}=0\)

\(\Rightarrow24\left(x-\frac{11}{48}\right)^2=\frac{1081}{2304}\)

\(\Rightarrow\left(x-\frac{11}{48}\right)^2=\left(\frac{\pm\sqrt{1081}}{48}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{11}{48}=\frac{\sqrt{1081}}{48}\\x-\frac{11}{48}=\frac{-\sqrt{1081}}{48}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{1081}+11}{48}\\x=\frac{11-\sqrt{1081}}{48}\end{cases}}\)

\(\left(3x-5\right)\left(7-5x\right)-\left(5x-2\right)\left(2-3x\right)=4\)

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

\(\Rightarrow36x-15x^2-35-16x+15x^2+4-4=0\)

\(\Rightarrow\left(-15x^2+15x^2\right)+\left(36x-16x\right)+\left(-35+4-4\right)=0\)

\(\Rightarrow30x-35=0\)

\(\Rightarrow x=\frac{7}{6}\)

12 tháng 7 2016

a. \(x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)

\(x^3-25x-\left(x^3+8\right)=17\)

\(x^3-25x-x^3-8=17\)

\(-25x=25\)

\(x=-1\)

12 tháng 7 2016

c. \(6x^2-\left(6x^2-4x+15x-10\right)=7\)

\(6x^2-6x^2-11x+10=7\)

\(-11x=-3\)

\(x=\frac{3}{11}\)

10 tháng 9 2019

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. (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

13 tháng 6 2019

a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)-3=-3\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3-3=-3\)

\(\Leftrightarrow14x=0\)

\(\Leftrightarrow x=0\)

Vậy pt có nghiệm duy nhất x = 0.

b) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)

\(\Leftrightarrow6x^2+19x-7-6x^2-x+5=7\)

\(\Leftrightarrow18x-2=7\)

\(\Leftrightarrow18x=9\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy pt có nghiệm duy nhất \(x=\frac{1}{2}\)

c) \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)

\(\Leftrightarrow36x^2-24x+4+25x^2-20x+4-60x^2+33x-8=0\)

\(\Leftrightarrow x^2-11x=0\)

\(\Leftrightarrow x\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=11\end{matrix}\right.\)

Vậy pt có tập nghiệm \(S=\left\{0;11\right\}\)

d) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow x^2-6x+9-x^2-4x+32=1\)

\(\Leftrightarrow41-10x=1\)

\(\Leftrightarrow-10x=40\)

\(\Leftrightarrow x=-4\)

Vậy pt có nghiệm duy nhất x = -4.

e) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)

\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7x^2+36=36\)

\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2=0\)

\(\Leftrightarrow8x=-13\)

\(\Leftrightarrow x=-\frac{13}{8}\)

Vậy pt có nghiệm duy nhất \(x=-\frac{13}{8}\)