1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

Thế mày làm đi

cho ít thôi thì làm

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

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

câu còn lại đâu bạn 

a: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

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

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

Sao lm mỗi câu a, c thế

Câu c viết dấu hoặc đi

1: \(\left(2x-2\right)\left(3x+1\right)-\left(3x-2\right)\left(2x-3\right)=5\)

=>\(6x^2+2x-6x-2-\left(6x^2-9x-4x+6\right)=5\)

=>\(6x^2-4x-2-6x^2+13x-6=5\)

=>9x-8=5

=>9x=13

=>\(x=\frac{13}{9}\)

2: \(\left(1-3x\right)\left(3x-5\right)-\left(2x-4\right)\left(2-3x\right)=x-6\)

=>\(3x-5-9x^2+15x+\left(2x-4\right)\left(3x-2\right)=x-6\)

=>\(-9x^2+18x-5+6x^2-4x-12x+8=x-6\)

=>\(-3x^2+2x+3-x+6=0\)

=>\(-3x^2+x+9=0\)

=>\(3x^2-x-9=0\)

=>\(x^2-\frac13x-3=0\)

=>\(x^2-2\cdot x\cdot\frac16+\frac{1}{36}-\frac{109}{36}=0\)

=>\(\left(x-\frac16\right)^2=\frac{109}{36}\)

=>\(x-\frac16=\pm\frac{\sqrt{109}}{6}\)

=>\(x=\frac16\pm\frac{\sqrt{109}}{6}\)

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

=>\(8x^3-1-8x^3-1=5x+6\)

=>5x+6=-2

=>5x=-8

=>\(x=-\frac85\)

* Trả lời:

\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)

\(\Leftrightarrow-3+6x-4-12x=-5x+5\)

\(\Leftrightarrow6x-12x+5x=3+4+5\)

\(\Leftrightarrow x=12\)

\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)

\(\Leftrightarrow6x-15-6+24x=-3x+7\)

\(\Leftrightarrow6x+24x+3x=15+6+7\)

\(\Leftrightarrow33x=28\)

\(\Leftrightarrow x=\dfrac{28}{33}\)

\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)

\(\Leftrightarrow1-3x-6x+12=-4x-5\)

\(\Leftrightarrow-3x-6x+4x=-1-12-5\)

\(\Leftrightarrow-5x=-18\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)

\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)

\(\Leftrightarrow-x-5x=-7\)

\(\Leftrightarrow-6x=-7\)

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

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

\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)

\(\Leftrightarrow-15x+3x=4\)

\(\Leftrightarrow-12x=4\)

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

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

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

giải phần còn lại giúp mình được ko?