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a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x^2+3x-x-3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+12x-4x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
\(5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+4\left(x-1\right)\left(x+3\right)-2\left(25-3x+9x^2\right)\)
\(=20x^2-20x+5+4\left(x-1\right)\left(x+3\right)-2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+8x-12-50+60x-18x^2\)
\(=6x^2+48x-57\)
Đặt \(3x-1=y,x+2=z\)
\(\Rightarrow y^2-2yz+z^2=\left(y-z\right)^2\)
\(=\left(3x-1-x-2\right)^2=\left(2x-3\right)^2\)
b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)
c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)
\(\left(x^2-1\right)\left(x+2\right)-\left(x-4\right)\left(x^2+4x+16\right)\)
\(=x^3+2x^2-x-2-\left(x^3-4^3\right)\)
\(=x^3+2x^2-x-2-x^3+64\)
\(=2x^2-x+62\)
\(2x\left(3x-2\right)^2\)
\(=2x\left(9x^2-12x+4\right)\)
\(=18x^3-24x^2+8x\)
\(\left(x-3\right)\left(x^2-3x+9\right)\)
\(=x^3-3x^2+9x-3x^2+9x-27\)
\(=x^3-3x^2+18x-27\)
\(\left(x^2-1\right)\left(x+2\right)-\left(x-4\right)\left(x^2+4x+16\right)\)
\(=\left(x^2-1^2\right)\left(x+2\right)-x^3-4^3\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2\right)-x^3-64\)
\(P=4\left(\frac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\)
\(=4.\frac{3}{4}x-4.1+12x^2:\left(-3x\right)+\left(-3x\right):\left(-3x\right)-2x-1\)
\(=3x-4-4x+1-2x-1=-3x-4\)
Thay \(x=\frac{-4}{3}\)vào P ta được \(P=-3.\frac{-4}{3}-4=4-4=0\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
\(\left(3x-4\right)^2-2\left(3x-4\right)\left(x-4\right)+\left(x-4\right)^2\)
\(=\left(3x-4-x+4\right)^2\)
\(=4x^2\)
\(\left(3x-4\right)^2+\left(4-x\right)^2-2\left(3x-4\right)\left(x-4\right)=\left(3x-4\right)^2-2\left(3x-4\right)\left(x-4\right)+\left(x-4\right)^2=\left(3x-4-x+4\right)^2=\left(2x\right)^2=4x^2\)