Question 1
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a x = a {\displaystyle ax=a\,}
a ⋅ ( x ⋅ a − 1 ) = a ⋅ a − 1 = 1 {\displaystyle a\cdot (x\cdot a^{-1})=a\cdot a^{-1}=1}
a ⋅ ( a − 1 ⋅ x ) = 1 {\displaystyle a\cdot (a^{-1}\cdot x)=1}
( a ⋅ a − 1 ) ⋅ x ) = 1 {\displaystyle (a\cdot a^{-1})\cdot x)=1}
1 ⋅ x = 1 {\displaystyle 1\cdot x=1}
x = 1 {\displaystyle x=1\,}
( x − y ) ( x + y ) = ( x ⋅ ( x − y ) + y ⋅ ( x − y ) ) {\displaystyle (x-y)(x+y)=(x\cdot (x-y)+y\cdot (x-y))}
= ( x 2 − x y ) + ( x y − y 2 ) {\displaystyle =(x^{2}-xy)+(xy-y^{2})\,}
= x 2 − x y + x y − y 2 {\displaystyle =x^{2}-xy+xy-y^{2}\,}
= x 2 − y 2 {\displaystyle =x^{2}-y^{2}\,}
x 2 = y 2 {\displaystyle x^{2}=y^{2}\,}
x 2 − y 2 = ( x + y ) ( x − y ) = 0 {\displaystyle x^{2}-y^{2}=(x+y)(x-y)=0\,}
Either ( x + y ) = 0 {\displaystyle (x+y)=0\,} x = − y {\displaystyle x=-y\,} ( x − y ) = 0 {\displaystyle (x-y)=0\,} x = y {\displaystyle x=y\,}
x 3 − y 3 = ( x − y ) ( x 2 + x y + y 2 ) {\displaystyle x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})\,}
= ( x − y ) x 2 + ( x − y ) x y + ( x − y ) y 2 {\displaystyle =(x-y)x^{2}+(x-y)xy+(x-y)y^{2}\,}
= ( x 3 − x 2 y + x 2 y − x y 2 + x y 2 − y 3 {\displaystyle =(x^{3}-x^{2}y+x^{2}y-xy^{2}+xy^{2}-y^{3}\,}
= x 3 − y 3 {\displaystyle =x^{3}-y^{3}\,}
x n − y n = ( x − y ) ( x n − 1 + x n − 2 y + . . . + x y n − 2 + y n − 1 ) {\displaystyle x^{n}-y^{n}=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})\,}
= x ( x n − 1 + x n − 2 y + . . . + x y n − 2 + y n − 1 ) − y ( x n − 1 + x n − 2 y + . . . + x y n − 2 + y n − 1 ) {\displaystyle =x(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})-y(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})\,}
= ( x n + x n − 1 y + . . . + x 2 y n − 2 + x y n − 1 ) − ( x n − 1 y + x n − 2 y 2 + . . . + x y n − 1 + y n ) {\displaystyle =(x^{n}+x^{n-1}y+...+x^{2}y^{n-2}+xy^{n-1})-(x^{n-1}y+x^{n-2}y^{2}+...+xy^{n-1}+y^{n})\,}
= x n − y n {\displaystyle =x^{n}-y^{n}\,}
Note that the two middle terms do no appear to match, but if you follow the logical development in each term, you will see that there is a matching additive inverse for each term on both sides.
Use the same method as iv , expand the expression and cancel.
Question 2
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Question 4
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All values of x satisfy the inequality, since it can be rewritten as − 3 < x 2 {\displaystyle -3<x^{2}\,} x 2 >= 0 {\displaystyle x^{2}>=0\,}
If 5 − x 2 < − 2 {\displaystyle 5-x^{2}<-2\,} x 2 > 7 {\displaystyle x^{2}>7\,}
x > 7 {\displaystyle x>{\sqrt {7}}\,} x < − 7 {\displaystyle x<-{\sqrt {7}}\,}
x 2 − 2 x + 2 {\displaystyle x^{2}-2x+2}
( x 2 − 2 x + 1 ) + 1 > 0 {\displaystyle (x^{2}-2x+1)+1>0\,} x 2 − 2 x + 1 = ( x − 1 ) 2 {\displaystyle x^{2}-2x+1=(x-1)^{2}\,}
( x − 1 ) 2 + 1 > 0 {\displaystyle (x-1)^{2}+1>0\,}
If x 2 + x + 1 > 2 {\displaystyle x^{2}+x+1>2\,} x 2 + x − 1 > 0 {\displaystyle x^{2}+x-1>0\,}
x 2 + x − 1 = ( x + 1 + 5 2 ) ( x − 5 − 1 2 ) {\displaystyle x^{2}+x-1=\left(x+{\frac {1+{\sqrt {5}}}{2}}\right)\left(x-{\frac {{\sqrt {5}}-1}{2}}\right)}
x < − 1 + 5 2 {\displaystyle x<-{\frac {1+{\sqrt {5}}}{2}}} x > 5 − 1 2 {\displaystyle x>{\frac {{\sqrt {5}}-1}{2}}}
Complete the square:
x 2 + x + 1 = x 2 + x + 1 4 + 3 4 = ( x + 1 2 ) 2 + 3 4 {\displaystyle x^{2}+x+1=x^{2}+x+{\frac {1}{4}}+{\frac {3}{4}}=\left(x+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}}
( x + 1 2 ) 2 + 3 4 > 0 {\displaystyle \left(x+{\frac {1}{2}}\right)^{2}+{\frac {3}{4}}>0}
Solve for ( x − π ) ( x + 5 ) > 0 {\displaystyle (x-\pi )(x+5)>0\,} ( x − 3 ) {\displaystyle (x-3)\,} − 5 < x < 3 {\displaystyle -5<x<3\,} x > π {\displaystyle x>\pi \,}
x > 2 {\displaystyle x>{\sqrt {2}}} x < 2 3 {\displaystyle x<{\sqrt[{3}]{2}}}
If 2 x < 8 {\displaystyle 2^{x}<8\,}
l o g 2 2 x < l o g 2 8 = x < 3 {\displaystyle log_{2}2^{x}<log_{2}8=x<3\,}
x < 1 {\displaystyle x<1\,}
NOTE: The answer in the 3rd Edition provides 0 < x < 1 {\displaystyle 0<x<1} x > 1 {\displaystyle x>1} x > 1 {\displaystyle x>1} 1 / 10 − 1 / 9 < 0 {\displaystyle 1/10-1/9<0}
1 x + 1 1 − x > 0 {\displaystyle {\frac {1}{x}}+{\frac {1}{1-x}}>0} 1 x − x 2 > 0 {\displaystyle {\frac {1}{x-x^{2}}}>0} 0 < x < 1 {\displaystyle 0<x<1} x 2 > x {\displaystyle x^{2}>x}
Question 5
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a + c < b + d {\displaystyle a+c<b+d\,}
= ( b + d ) − ( a + c ) > 0 {\displaystyle =(b+d)-(a+c)>0\,}
= ( b − a ) + ( d − c ) > 0 {\displaystyle =(b-a)+(d-c)>0\,} ( b − a ) , ( d − c ) > 0 {\displaystyle (b-a),(d-c)>0\,}
b − a > 0 {\displaystyle b-a>0\,}
= − a + b > 0 {\displaystyle =-a+b>0\,}
= − a − ( − b ) > 0 {\displaystyle =-a-(-b)>0\,}
= − b < − a {\displaystyle =-b<-a\,}
( b − a ) , c > 0 {\displaystyle (b-a),c>0\,} c ( b − a ) > 0 {\displaystyle c(b-a)>0\,}
c ( b − a ) = b c − a c {\displaystyle c(b-a)=bc-ac\,} a c < b c {\displaystyle ac<bc\,}
a > 1 {\displaystyle a>1\,} 1 − a > 0 {\displaystyle 1-a>0\,}
a ( 1 − a ) > 0 {\displaystyle a(1-a)>0\,}
= a 2 − a > 0 {\displaystyle =a^{2}-a>0\,}
= a 2 > a {\displaystyle =a^{2}>a\,}
If a {\displaystyle a} c {\displaystyle c} a c < b d {\displaystyle ac<bd}
Otherwise, we have already proved that a c < b c {\displaystyle ac<bc} a , c > 0 {\displaystyle a,c>0} c < d {\displaystyle c<d} a c < b d {\displaystyle ac<bd}
If a = 0 {\displaystyle a=0} a 2 < b 2 {\displaystyle a^{2}<b^{2}} b 2 > 0 {\displaystyle b^{2}>0}
If 0 < a < b {\displaystyle 0<a<b} a 2 < a b {\displaystyle a^{2}<ab} a b < b 2 {\displaystyle ab<b^{2}} a 2 < b 2 {\displaystyle a^{2}<b^{2}}
Question 6
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![{\displaystyle x<{\sqrt[{3}]{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61605d902da0e92dfbb747a2a8f8e9a04e2833e9)
