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c)(x2+x)2-2(x2+x)-15
đặt x2+x=a ta có
a2-2a-15
=a2+3a-5a-15
=(a2+3a)-(5a+15)
=a(a+3)-5(a+3)
=(a+3)(a-5)
thay a=x2+x
(x2+x+3)(x2+x-5)
\(a,\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\)
\(=x^2+2x+1-\left(x^2-2x+1\right)-3\left(x^2-1\right)\)
\(=x^2+2x+1-x^2+2x-1-3x^2+2=-3x^2+4x+2\)\(b,5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)
\(=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17\)
\(=5x^2-20-4x^2+12x-9-x^2+17=12x-12\)
a: \(A=2x-3-5x+2-3x+1=-6x=-6\cdot\dfrac{-2}{3}=4\)
b: \(B=x^{2n-2n+3}=x^3=\left(-3\right)^3=-27\)
\(S=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{3\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\cdot\dfrac{1}{2}}{2a^2+2b^2+2c^2-2ab-2bc-2ac}=\dfrac{3}{2}\)
c: \(\left(x^2+2x\right)^2+9x^2+18x+20\)
\(=\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20\)
\(=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)
d: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8+2x\right)\left(x^2+4x+8+x\right)\)
\(=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
\(=\left(x^2+5x+8\right)\left(x+4\right)\left(x+2\right)\)
a,\(20x^2+7x-6=20x^2-8x+15x-6\)
\(=\left(20x^2-8x\right)+\left(15x-6\right)=4x.\left(5x-2\right)+3.\left(5x-2\right)\)
\(=\left(5x-2\right).\left(4x+3\right)\)
b,\(12x^2-23xy+10y^2=12x^2-8xy-15xy+10y^2\)
\(=\left(12x^2-8xy\right)-\left(15xy-10y^2\right)\)
\(=4x.\left(3x-2y\right)-5y.\left(3x-2y\right)\)
\(=\left(3x-2y\right).\left(4x-5y\right)\)
Chúc bạn học tốt!!!
\(M=\dfrac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
\(=\dfrac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)}{2a^2+2b^2+2c^2+2ab-2bc+2ac}\)
\(=\dfrac{\left(a-b-c\right)\cdot\left(a^2+b^2+c^2-ab-bc+ac\right)}{2\cdot\left(a^2+b^2+c^2+ab-bc+ac\right)}=\dfrac{2}{2}=1\)
trình bày dài quá ; giờ chỉ cho cách làm thôi nha
dùng hằng đẳng thức : mũ 3
biền đổi
\(\left(a+b+c\right)^3=\left(a+\left(b+c\right)\right)^3\)
\(\left(b+c-a\right)^3=\left(b+\left(c-a\right)\right)^3\)
\(\left(a+c-b\right)^3=\left(a+\left(c-b\right)\right)^3\)
\(\left(a+b-c\right)^3=\left(a+\left(b-c\right)\right)^3\)
xong áp dụng hằng đẳng thức mũ 3
k có câu b ạ