Bn xem lại đề xem sao chứ chưa có kết quả chưa tìm đc x đâu nha

Ta có: \(x\left(x-4\right)+1=3x-5\)

\(\Leftrightarrow x^2-4x+1-3x+5=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\)

a: =>x^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

a: ĐKXD: x<>0

\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)

=>\(7x^2+6x-7=3x^2-4x+6x-8\)

=>\(7x^2+6x-7=3x^2+2x-8\)

=>\(4x^2+4x+1=0\)

=>\(\left(2x+1\right)^2=0\)

=>2x+1=0

=>x=-1/2(nhận)

b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)

=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)

=>\(24x^2-26x-5-24x^2+23x+12=15\)

=>-3x+7=15

=>-3x=8

=>\(x=-\dfrac{8}{3}\)

x2x2 là sao bn

Câu a) -3 phần 1/2 

Câu a) 2 mũ 2

a: =>-7/2:(4/5-1/2x)=4

=>4/5-1/2x=-7/2:4=-7/8

=>1/2x=4/5+7/8=67/40

=>x=67/20

b: =>5x=5

=>x=1

c: =>x(-2/3-1/3)=-2

=>-x=-2

=>x=2

d: =>-2/3(x+1)=-1/3+1/2=1/6

=>x+1=-1/6:2/3=-1/6*3/2=-3/12=-1/4

=>x=-1/4-1=-5/4

a) \(x\left(5-2x\right)-2x\left(1-x\right)=15\\ \Leftrightarrow5x-2x^2-2x+2x^2=15\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=5\)

Vậy x = 5 là nghiệm của pt.

b) \(\left(3x+2\right)^2+\left(1+3x\right)\left(1-3x\right)=2\\ \Leftrightarrow\left(9x^2+12x+4\right)+1-9x^2=2\\ \Leftrightarrow12x+5=2\\ \Leftrightarrow12x=-3\\ \Leftrightarrow x=\dfrac{-1}{4}\)

Vậy \(x=-\dfrac{1}{4}\) là nghiệm của pt.

a)x.(5-2x)-2x.(1-x)=15
   x [ 5 - 2x -2.(1-x) ] = 15
   x ( 5 - 2x -2 + 2x ) =15
   x . 3 =15
   x = 5
b)(3x+2)²+(1+3x).(1-3x)=2
   9x²+12x+4+1-9x²=2
   12x + 5 = 2
    12x = -3
        x = -1/4

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

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

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)