a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)

\(\Leftrightarrow6x+2-20+8x>8x-6-6\)

\(\Leftrightarrow14x-18-8x+12>0\)

\(\Leftrightarrow6x-6>0\)

\(\Leftrightarrow6x>6\)

hay x>1

Vậy: S={x|x>1}

b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)

\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)

\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)

\(\Leftrightarrow-1< 0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

a) \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x-4\right)\left(x+4\right)\le10\)

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)\le10\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240\le10\)

\(\Leftrightarrow\left(5x^3-5x^3\right)-\left(30x^2-15x^2-15x^2\right)-\left(45x-15x\right)+5-240\le10\)

\(\Leftrightarrow30x-235\le10\)

\(\Leftrightarrow30x\le10+235\)

\(\Leftrightarrow30x\le245\)

\(\Leftrightarrow30x:30\le245:30\)

\(\Leftrightarrow x\le\dfrac{49}{6}\)

Vậy nghiệm của bất phương trình là: \(x\le\dfrac{49}{6}\)

b) \(\left(3x-2\right)\left(9x^2+6x+4\right)+27x\left(\dfrac{1}{3}-x\right)\left(\dfrac{1}{2}+x\right)\ge1\)

\(\Leftrightarrow27x^3-8+27x\left(\dfrac{1}{9}-x^2\right)\ge1\)

\(\Leftrightarrow27x^3-8+3x-27x^3\ge1\)

\(\Leftrightarrow\left(27x^3-27x^3\right)-8+3x\ge1\)

\(\Leftrightarrow-8+3x\ge1\)

\(\Leftrightarrow3x\ge1+8\)

\(\Leftrightarrow3x\ge9\)

\(\Leftrightarrow3x:3\ge9:3\)

\(\Leftrightarrow x\ge3\)

Vậy nghiệm của bất phương trình là \(x\ge3\)

a: =>5x(x^2-6x+9)-5(x^3-3x^2+3x-1)+15(x^2-16)<=10

=>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240<=10

=>30x-235<=10

=>30x<=245

=>x<=49/6

b: =>27x^3-8+27x(1/9-x^2)>=1

=>27x^3-8+3x-27x^3>=1

=>3x>=9

=>x>=3

a: =>4(2x-1)-12x=3(x+3)+24

=>8x-4-12x=3x+9+24

=>-4x-4=3x+33

=>-7x=37

=>x=-37/7

b: =>(x-2)(x+2+x-9)=0

=>(2x-7)(x-2)=0

=>x=2 hoặc x=7/2

c: =>(x-1)(x+3)-x+3=3x+3

=>x^2+2x-3-x+3=3x+3

=>x^2+x-3x-3=0

=>x^2-2x-3=0

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

=>x=-1

1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(3x+9+4x-12=3x-7\)

\(\Leftrightarrow4x=-7+12-9=-4\)

hay \(x=-1\left(nhận\right)\)

2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)

\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(3x+12-4x+16=3x-4\)

\(\Leftrightarrow28-4x=-4\)

\(\Leftrightarrow4x=32\)

hay \(x=8\left(tm\right)\)

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

Suy ra: \(5x^2-12+3x+3=5x^2-5x\)

\(\Leftrightarrow3x-9+5x=0\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(nhận\right)\)

1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)

\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)

\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)

\(\Leftrightarrow-41x=-115\)

hay \(x=\dfrac{115}{41}\)

2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)

\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)

\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)

\(\Leftrightarrow x^3=64\)

hay x=4

3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)

\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)

\(\Leftrightarrow-5x-15=10x-20\)

\(\Leftrightarrow-5x-10x=-20+15\)

\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)

a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\) \(\Leftrightarrow\) \(6\left(\dfrac{2-x}{3}-x-2\right)\le6\left(\dfrac{x-17}{2}\right)\) \(\Leftrightarrow\) 4-2x-6x-12\(\le\)3x-51 \(\Leftrightarrow\) -2x-6x-3x\(\le\)-51-4+12 \(\Leftrightarrow\) -11x\(\le\)-43 \(\Rightarrow\) x\(\ge\)43/11.

b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\) \(\Leftrightarrow\) \(12\left(\dfrac{2x+1}{3}+\dfrac{4-x}{4}\right)\le12\left(\dfrac{3x+1}{6}+\dfrac{4-x}{12}\right)\) \(\Leftrightarrow\) 8x+4+12-3x\(\le\)6x+2+4-x \(\Leftrightarrow\) 8x-3x-6x+x\(\le\)2+4-4-12 \(\Leftrightarrow\) 0x\(\le\)-10 (vô lí).

a) \(\dfrac{2-x}{3}-x-2\le\dfrac{x-17}{2}\)

\(\Leftrightarrow2\left(2-x\right)-6\left(x+2\right)\le3\left(x-17\right)\)

\(\Leftrightarrow4-2x-6x-12\le3x-51\)

\(\Leftrightarrow-11x\le-43\)

\(\Leftrightarrow x\ge\dfrac{43}{11}\)

Vậy S = {\(x\) | \(x\ge\dfrac{43}{11}\) }

b) \(\dfrac{2x+1}{3}-\dfrac{x-4}{4}\le\dfrac{3x+1}{6}-\dfrac{x-4}{12}\)

\(\Leftrightarrow4\left(2x+1\right)-3\left(x-4\right)\le2\left(3x+1\right)-\left(x-4\right)\)

\(\Leftrightarrow8x+4-3x+12\le6x+2-x+4\)

\(\Leftrightarrow0x\le-10\) (vô lý)

Vậy \(S=\varnothing\)

giúp mình với ạ câu nào cũng được

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}=3\)

b) \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

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

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