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) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

\(a,\Leftrightarrow x^2+6x+9-x^2+3x+10=1\\ \Leftrightarrow9x=-18\Leftrightarrow x=-2\\ b,\Leftrightarrow4x^2-4x+1-4x^2+17x+15=3\\ \Leftrightarrow13x=-13\Leftrightarrow x=-1\\ c,\Leftrightarrow3x\left(x-2\right)+4\left(x-2\right)=0\\ \Leftrightarrow\left(3x+4\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=2\end{matrix}\right.\\ d,\Leftrightarrow2x\left(3x+5\right)-6\left(3x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(D\left(x\right)=-4x^3-4x^3-x^2-x^2+2x+3x+5=0\)

\(-8x^3-2x^2+5x+5=0\)

\(\left(-8x^2-10x-5\right)\left(x-1\right)=0\)

TH1 : \(x=1\)

TH2 : cj phân tích như vậy nhé 

 \(\Delta=\left(-2\right)^2-4.\left(-8\right).\left(-5\right)=4-160=-156< 0\)

Nên phương trình vô nghiệm  (P/s chỗ này : đừng chép vào bài TH2 nhé, cj thử thôi !) 

Vậy x = 1 

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

\(< =>-8x^3-2x^2+5x+5=0\left(1\right)\)

Nháp : dùng pp nhẩm nghiệm ta thấy \(-8-2+5+5=0\)

Nên phương trình nhận 1 là nghiệm 

Dùng lược đồ hóc-ne 

\(\left(1\right)< =>\left(x-1\right)\left(-8x^2-10x-5\right)=0\)

\(< =>\orbr{\begin{cases}x-1=0\\-8x^2-10x-5=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=1\\\Delta=\left(-10\right)^2-4.\left(-5\right)\left(-8\right)=100-160=-60\end{cases}}\)

\(< =>\orbr{\begin{cases}x=1\\vo-nghiem\end{cases}}\)

Vậy nghiệm của đa thức trên là 1

a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\)

hay \(x=-\dfrac{1}{2}\)

c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)

hay x=0

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.

\(a,=3x-9-4x+12=-x+3=0\)

\(\Leftrightarrow x=3\)

Vậy ..

\(b,=\left(x+2\right)\left(x+2-x+2\right)=4\left(x+2\right)=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy ..

\(c,=x^3-3x^2+3x-1=\left(x-1\right)^3=0\)

\(\Leftrightarrow x=1\)

Vậy ..

\(d,\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)

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

Vậy ..

\(e,=\left(2x-3-5\right)\left(2x-3+5\right)=\left(2x-8\right)\left(2x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{2}=4\\x=-\dfrac{2}{2}=-1\end{matrix}\right.\)

Vậy ...

a) Ta có: 3(x-3)-4x+12=0

\(\Leftrightarrow3\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

hay x=3

Vậy: S={3}

b) Ta có: \(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\)

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

\(\Leftrightarrow4x=-8\)

hay x=-2

Vậy: S={-2}

c) Ta có: \(x^3+3x=3x^2+1\)

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

\(\Leftrightarrow x-1=0\)

hay x=1

Vậy: S={1}

d) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

Vậy: S={0;2;-2}