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\(F=\left(\dfrac{-1}{2}-2\right)^3-\left(\dfrac{-1}{2}+3\right)^2+\left(-2+\dfrac{3}{2}\right)^3+\left(-\dfrac{1}{2}+1\right)^2\)
\(=\dfrac{-125}{8}-\dfrac{25}{4}+\dfrac{1}{8}+\dfrac{1}{4}\)
\(=\dfrac{-124}{8}-\dfrac{24}{4}\)
=-15,5-6=-21,5
1: \(F=\left(\dfrac{-1}{2}-2\right)^3-\left(-\dfrac{1}{2}+3\right)^3+\left(-2+\dfrac{3}{2}\right)^3+\left(-\dfrac{1}{2}+1\right)^2\)
\(=\dfrac{-125}{8}-\dfrac{125}{8}+\dfrac{-1}{8}+\dfrac{1}{4}\)
\(=\dfrac{-251}{8}+\dfrac{1}{4}=\dfrac{-249}{8}\)
2:\(N=\left(-1-1\right)^2-\left(-1+\dfrac{1}{8}\right)+\left(-1+1\right)^3\)
=4+1-1/8
=5-1/8=39/8
a) \(A=3x\left(x^2-2x+3\right)-x^2.\left(3x-2\right)+5\left(x^2-x\right)\)
\(=3x^3-6x^2+9x-3x^3+2x^2+5x^2-5x\)
\(=x^2+4x\)
Thay \(x=5\)vào biểu thức ta có: \(A=5^2+4.5=25+20=45\)
b) \(B=x\left(x^2+xy+y^2\right)-y\left(x^2+xy+y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
Thay \(x=10\); \(y=-1\)vào biểu thức ta có:
\(B=10^3-\left(-1\right)^3=1000+1=1001\)
Dễ mà bạn
\(P=\left(x-y\right)\left(x^2+xy+y^2\right)-2y^3=x^3-y^3-2y^3=x^3-3y^3=\left(\frac{1}{2}\right)^3-3.\left(\frac{2}{3}\right)^3=\frac{-55}{72}\)
A=[(x-y).(x2+xy+y2)] +2y3
= x3-y3+2y3=x3+y3
=(2/3)3+(1/3)3
=4/9 + 1/9 =5/9
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
\(U=x^3+9x^2y+27xy^2+27y^3-x^3-27y^3-2x\left(x^2-4x+4\right)\)
\(=9x^2y+27xy^2-2x^3+8x^2-8x\)
\(=9\cdot1\cdot2+27\cdot1\cdot2^2-2\cdot1^3+8\cdot1^2-8\cdot1\)
\(=18+108-2-8-8=108\)
1; \(x^2\) + 3\(x^2\) + 3\(x\) = 4\(x^2\) + 3\(x\) (1)
Thay \(x=99\) vào (1) ta có:
4.992 + 3.99 = 4.9801 + 297 = 39204 + 297 = 39501
Lời giải:
Những bài này sử dụng những hằng đẳng thức đáng nhớ.
Vì $x=-2$ nên $x+2=0$. Ta có:
\(A=(2x-3)^2-(x-3)^3+(4x+1)[(4x)^2-4x.1+1^2]\)
\(=(2x-3)^2-(x-3)^3+(4x)^3+1^3\)
\(=[2(x+2)-7]^2-(x+2-5)^3+8x^3+1\)
\(=(-7)^2-(-5)^3+8.(-2)^3+1=111\)
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\(B=(3x-y)^3-[x^3+(2y)^3]+(x+3)^2\)
\(=(3.1-2)^3-(1^3+8.2^3)+(1+3)^2=-48\)
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Vì $x=\frac{1}{2}; y=\frac{-1}{2}\Rightarrow x+y=0$
\(C=(x-5y)^2+(2x-3y)^3-(x-y)^3-[(2x)^3+(3y)^3]\)
\(=(x+y-6y)^2+[2(x+y)-5y]^3-(x+y-2y)^3-[8(x^3+y^3)+19y^3]\)
\(=(-6y)^2+(-5y)^3-(-2y)^3-19y^3\)
\(=36y^2-136y^3=36.(\frac{-1}{2})^2-136(\frac{-1}{2})^3=26\)
\(F=\left(-\dfrac{1}{2}-2\right)^3-\left(-\dfrac{1}{2}+3\right)^2+\left(-2+\dfrac{3}{2}\right)^3+\left(-\dfrac{1}{2}+1\right)^2\)
=-125/8+25/4-1/8+1/4
=-37/4