Rút gọn biểu thức sau (1,0đ): (4x - 5)2 + 2(4x - 5)(3x + 5) + (3x + 5)2.
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a: A=(4x+5)^2-2*(4x+5)(4x-5)+(4x-5)^2
=(4x+5-4x+5)^2
=10^2=100
b: B=(3x-2)^2*(3x+2)^2-2(2x+3)(2x-3)
=(9x^2-4)^2-2(4x^2-9)
=81x^4-72x^2+16-8x^2+18
=81x^4-80x^2+34
\(a,A=\left(4x-5\right)^2+\left(4x+5\right)^2+2\left(5+4x\right)\left(5-4x\right)\)
\(=\left(5-4x\right)^2 +2\left(5-4x\right)\left(4x+5\right)+\left(4x+5\right)^2\)
\(=\left(5-4x+4x+5\right)^2\)
\(=10^2\)
\(=100\)
\(b,B=\left(3x-2\right)^2\left(3x+2\right)^2-2\left(2x+3\right)\left(2x-3\right)\)
\(=\left(9x^2-4\right)^2-2\left(4x^2-9\right)\)
\(=81x^4-72x^2+16-8x^2+18\)
\(=81x^4-80x^2+34\)
#\(Urushi\)
Ta có: P=|3x-6|-3x+5
=3x-6-3x+5
=-1
Ta có: Q=|8-2x|+3x+8
=2x-8+3x+8
=5x
(3x+4)2+(4x-1)2 + (2x+5)(2x-5)
=9x2+24x+16+16x2-8x+1+4x2-25
=29x2+16x-8
\(\left(3x+4\right)^2+\left(4x-1\right)^2+\left(2x+5\right)\left(2x-5\right)\)
\(=9x^2+24x+16+16x^2-8x+1+4x^2+25\)
\(=9x^2+16x^2+4x^2+24x-8x+16+1+25\)
\(=29x^2+16x+4x\)
\(\left(3x-5\right)^2+\left(x+2\right)^2+x\left(3-4x\right)\)
\(=9x^2-30x+25+x^2+4x+4+3x-4x^2\)
\(=6x^2-23x+29\)
a) \(\left(x-3\right)\left(3x+2\right)-3x\left(x-5\right)+3\)
\(=x.\left(3x+2\right)-3.\left(3x+2\right)-3x\left(x-5\right)+3\)
\(=x.3x+x.2-3.3x-3.2-3x.x+3x.5+3\)
\(=3x^2+2x-9x-6-3x^2+15x+3\)
\(=8x-3\)
b )
\(2x\left(x-3\right)-\left(x-5\right)\left(2x-1\right)\)
\(2x.x-2x.3-x.\left(2x-1\right)-5.\left(2x-1\right)\)
\(2x.x-2x.3-x.2x+x.1-5.2x+5.x\)
\(2x^3-6x-2x^2+x-10x+5x\)
\(2x^3-15x-2x^2\)
\(=\left(4x-5+3x+5\right)^2=\left(7x\right)^2=49x^2\)