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a: Ta có: \(x^2-4-\left(x+2\right)^2\)
\(=x^2-4-x^2-4x-4\)
=-4x-8
b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-x^2+2x+3\)
=2x-1
c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)
\(=\left(x-2\right)\left(x+2-x-5\right)\)
\(=-3x+6\)
d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
=4
e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)
\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)
\(=29a^2-45a-3-36a^2+24a-4\)
\(=-7a^2-21a-7\)
g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)
\(=25y^2-9-25y^2+40y-16\)
=40y-25
h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)
\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)
\(=35x^3+15x^2+15x\)
i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=16x^2\)
\(=\dfrac{2x-x-5+x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{2}{x+5}\)
a)\(\left(6x+1\right)^2+\left(6x+1\right)^2-2\left(1-6x\right)\left(6x-1\right)\)
\(\Leftrightarrow36x^2+1+36x^2+1-2\left(36x^2-1\right)\)
\(\Leftrightarrow1+1+2=4\)
b) \(\Leftrightarrow4x^2-9-\left(x^2+10x+25\right)-\left(x^2+x-2\right)\)
\(\Leftrightarrow2x^2-11x-32\)
a: Ta có: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x^2=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2=5\)
\(\Leftrightarrow13x=13\)
hay x=1
(-6x + 1)(x + 5) + 6x2
= -6x(x + 5) + 1(x + 5) + 6x2
= -6x2 - 30x + x + 5 + 6x2
= (-6x2 + 6x2) + (-30x + x) + 5
= -29x + 5
\(\left(-6x+1\right)\left(x+5\right)+6x^2\)
\(=-6x.\left(x+5\right)+1\left(x+5\right)+6x^2\) ( nhân đa thức với đa thức)
\(=-6x^2-30+x+5+6x^2\)
\(=-29x+5\)