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

\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)

\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)

\(\Leftrightarrow15x^2-4x-4=0\)​

\(\Leftrightarrow15x^2-10x+6x-4=0\)

Lỗi :vvvv

\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy ...

(3x-2)(4x+3)=(2-3x)(x-1)

=(3x-2)(4x+3)-(2-3x)(x-1)

=(3x-2)(4x+3)+(3x-2)(x-1)

=(3x-2)(4x+3+x-1)

=(3x-2)(5x+2)

=15x²-10x-4+6x=15x²-4x-4

\(a,=64x^3-48x^2+12x-1-\left(64x^3+12x-48x^2-9\right)\)

       \(=\left(64x^3-64x^3\right)+\left(48x^2-48x^2\right)+\left(12x-12x\right)+\left(9-1\right)\)

        \(=8\)  => ko phụ thuộc vào biến x

\(b,=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x^2+y^2\right)\)

thay x+y=1 vào 

\(=2\left(x^2-xy+y^2\right)-3\left(x^2+y^2\right)\)

\(=2x^2-2xy+2y^2-3x^2-3y^2\)

\(=-\left(x^2+2xy+y^2\right)=-\left(x+y\right)^2=-1\)  =>ko phụ thuộc vào biến

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

      \(=6x^2+2-6x^2+6=8\)

\(d,\frac{\left(2x+5\right)^2+\left(5x-2\right)^2}{x^2+1}=\frac{4x^2+20x+25+25x^2-20x+4}{x^2+1}=\frac{29\left(x^2+1\right)}{x^2+1}=29\)

Lời giải :

1. \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{a^3}{8}+\frac{3a^2b}{4}+\frac{3ab^2}{2}+b^3+\frac{a^3}{8}-\frac{3a^2b}{4}+\frac{3ab^2}{2}-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

Lời giải :

2. \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy...

3. \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(=64x^3-48x^2+12x-1-64x^3+48x^2-12x+9\)

\(=8\)

1) \(\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\left(\frac{a}{2}+b\right)^2+\left(\frac{a}{2}-b\right)^2\)

\(=\left(\frac{a}{2}+b\right)\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{b}b+b^2\right]+\left(\frac{a}{2}-b\right)\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a}{2}\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+b\left[\left(\frac{a}{2}\right)^2+2.\frac{a}{2}b+b^2\right]+\frac{a}{2}\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)\(-b\left[\left(\frac{a}{2}\right)^2-2.\frac{a}{2}b+b^2\right]\)

\(=\frac{a^3}{8}+\frac{a^2b}{2}+\frac{ab^2}{2}+\frac{ba^2}{4}+b^2a+b^3+\frac{a^3}{8}-\frac{a^2b}{2}+\frac{ab^2}{2}-\frac{ba^2}{4}+b^2a-b^3\)

\(=\frac{a^3}{4}+3ab^2\)

2) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3x^2.1+3.x.1^2-1^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

3) \(A=\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)

\(A=64x^3-32x^2+4x-16x^2+8x-1-64x^3-12x+48x^2+9\)

\(A=8\)

Vậy: biểu thức không phụ thuộc vào biến

1) \(\left(x+5\right)^3-x^3-125\)

\(=\left(x+5\right)\left(x^2+2x.5+5^2\right)-x^3-125\)

\(=x\left(x^2+2x.5+5^2\right)+5\left(x^2+2x.5+5^2\right)-x^3-125\)

\(=x^3+10x^2+25x+5x^2+50x+125-x^3-125\)

\(=15x^2+75x\)

2) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)

\(\Leftrightarrow x^3-4x^2+4x-2x^2+8x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow24x+10=0\)

\(\Leftrightarrow24x=0-10\)

\(\Leftrightarrow24x=-10\)

\(\Leftrightarrow x=-\frac{10}{24}=-\frac{5}{12}\)

\(\Rightarrow x=-\frac{5}{12}\)

3) \(\left(x-1\right)^3-x^3+3x^2-3x+1\)

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

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

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

\(=0\)

Vậy: biểu thức không phụ thuộc vào biến

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

d) đề là gì bn

$2x+3)\left(4{}^{}-6x+9\right)-2\left(4{}^{}-1\right)$

$=8{}^{}+27-8{}^{}+2=\mathrm{29\backslash }$

(4x−1)3−(4x−3)(16x2+3)(4x−1)3−(4x−3)(16x2+3)

=64x3−48x2+12x−1−(64x3+12x−48x2−9)=64x3−48x2+12x−1−(64x3+12x−48x2−9)

=64x3−48x2+12x−1−64x3−12x+

đề không rõ nên mình làm như này:

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

\(=2x^2+x-x^3-2x^2+x^3-x+3\)

\(=3\)

d) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2\)

\(=29\)