6x^3 + x + 4 = 11x^2

<=>6x³-11x²+x+4=0

<=>6x³+3x²-14x²-7x+8x+4=0

<=>3x²(2x+1)-7x(2x+1)+4(2x+1)=0

<=>(2x+1)(3x²-7x+4)=0

<=>(2x+1)(3x²-3x-4x+4)=0

<=>(2x+1)(3x-4)(x-1)=0

<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0

<=>x\(\in\){-1/2;1;4/3}

b)x^6 - 14x^4 + 49x^2 = 36

<=>x⁶-14x⁴+49x²-36=0

<=>x⁶-x⁴-13x⁴+13x²+36x²-36=0

<=>x⁴(x²-1)-13x²(x²-1)+36(x²-1)=0

<=>(x²-1)(x⁴-13x²+36)=0

<=>(x+1)(x-1)(x⁴-9x²-4x²+36)=0

<=>(x+1)(x-1)[x²(x²-9)-4(x²-9)]=0

<=>(x-1)(x+1)(x²-9)(x²-4)=0

<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0

<=>x\(\in\){-3;-2;-1;1;2;3}

phù.mệt

\(x^6-14x^4+49x^2>36\)
\(\Leftrightarrow x^6-x^5+x^5-x^4-13x^4+13x^3-13x^3+13x^2+36x^2-36x+36x-36>0\)

\(\Leftrightarrow x^5\left(x-1\right)+x^4\left(x-1\right)-13x^3\left(x-1\right)-13x^2\left(x-1\right)+36x\left(x-1\right)+36\left(x-1\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(x^5+x^4-13x^3-13x^2+36x+36\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left[x^4\left(x+1\right)-13x^2\left(x+1\right)+36\left(x+1\right)\right]>0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-13x^2+36\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-9x^2-4x^2+36\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left[x^2\left(x^2-9\right)-4\left(x^2-9\right)\right]>0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-9\right)\left(x^2-4\right) >0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0\)

Để \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\)

Vậy để \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0\) thì x>3 hoặc x<-3

a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)

\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)

\(< =>12-2+4x-2x^2=6x^2-13x+6\)

\(< =>10+4x-2x^2-6x^2+13x-6=0\)

\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)

b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)

\(< =>x-9=0< =>x=9\)

c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)

\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)

d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)

\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)

e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)

\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)

f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)

\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)

g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)

\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)

h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

\(< =>x^2-16-6x+4=x^2-8x+16\)

\(< =>x^2-6x-12-x^2+8x-16=0\)

\(< =>2x-28=0< =>x=\frac{28}{2}=14\)

q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)