rút gọn biểu thức
a) ( a + b ) - ( a - b ) + ( a - c ) - ( a + c )
b) (a + b - c ) + ( a - b + c ) - ( b + c - a ) - ( a - b - c )
c) - { - ( a + b ) - [ - (a - b ) - (a + b) ] -}
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a.\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ac-2\left(a^2+2ab+b^2\right)=2a^2+2b^2+2c^2+4ab-2a^2-2ab-2b^2=2c^2+2ab\)
b. \(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\left(a^2+b^2-c^2+a^2-b^2+c^2\right)=\left(2b^2-2c^2\right).2a^2=4a^2\left(b^2-c^2\right)=4a^2b^2-4a^2c^2\)
\(a,=x^3-16x-x^2-1-x^2+1=x^3-2x^2-16x\\ b,=y^4-81-y^4+4=-77\\ d,=a^2+b^2+c^2+2ab-2bc-2ac+a^2-2ac+c^2-2ab-2ac\\ =2a^2+b^2+2c^2-2bc-6ac\)
\(log_{a^3}b.log_ba=\dfrac{1}{3}.log_ab.log_ba=\dfrac{1}{3}\)
\(log_{a^{10}}b^5.log_{b^3}a^9=\dfrac{1}{10}.5.log_ab.\dfrac{1}{3}.9.log_ba=\dfrac{3}{2}\)
\(log_{a^{107}}b^{101}.log_{b^{303}}a^{428}=\dfrac{1}{107}.101.log_ab.\dfrac{1}{303}.428.log_ba=\dfrac{4}{3}.log_ab.log_ba=\dfrac{4}{3}\)
a: \(log_{a^3}b\cdot log_ba=\dfrac{1}{3}\cdot log_ab\cdot log_ba=\dfrac{1}{3}\)
b: \(log_{a^{10}}b^5\cdot log_{b^3}a^9\)
\(=\dfrac{1}{10}\cdot log_ab^5\cdot\dfrac{1}{3}\cdot log_ba^9\)
\(=\dfrac{1}{30}\cdot5\cdot log_ab\cdot9\cdot log_ba=\dfrac{45}{30}=\dfrac{3}{2}\)
c: \(log_{a^{107}}b^{101}\cdot log_{b^{303}}a^{428}\)
\(=\dfrac{1}{107}\cdot log_ab^{101}\cdot\dfrac{1}{303}\cdot log_ba^{428}\)
\(=\dfrac{1}{107}\cdot101\cdot log_ab\cdot\dfrac{1}{303}\cdot428\cdot log_ba\)
\(=4\cdot\dfrac{1}{3}=\dfrac{4}{3}\)
\(log_{a^4}b^4.log_ba^5=\dfrac{1}{4}.4.log_ab.5.log_ba=5.log_ab.log_ba=5\)
\(log_{a^3}b^2.log_ba^4=\dfrac{1}{3}.2.log_ab.4.log_ba=\dfrac{8}{3}.log_ab.log_ba=\dfrac{8}{3}\)
\(log_{a^{15}}b^7.log_{b^{49}}a^{30}=\dfrac{1}{15}.7.log_ab.\dfrac{1}{49}.30.log_ba=\dfrac{2}{7}log_ab.log_ba=\dfrac{2}{7}\)
\(log_{a^{2021}}b^{2020}.log_{b^{4040}}a^{6063}=\dfrac{1}{2021}.2020.log_ab.\dfrac{1}{4040}.6063.log_ba=\dfrac{3}{2}\)
a: =b-c-a+c+1-a-b+c
=-2a+1
b: =a-b-c-b+c+a+c-b-a
=c-3b+a
c: =2(a-b-b+c-c+a)
=2(2a-2b)
=4a-4b
a) \(\left(b-c\right)-\left(a-c-1\right)-\left(a+b-c\right)\)
\(=b-c-a+c+1-a-b+c\)
\(=c-2a+1\)
b) \(\left(a-b-c\right)-\left(b-c-a\right)+\left(c-b-a\right)\)
\(=a-b-c-b+c+a+c-b-a\)
\(=a-3b+c\)
c) \(2\cdot\left(a-b\right)-2\cdot\left(b-c\right)-2\cdot\left(c-a\right)\)
\(=2\cdot\left(a-b-b+c-c+a\right)\)
\(=2\cdot\left(2a-2b\right)\)
\(=4a-4b\)
A=(a-b)+(a+b+c)-(a-b-c)
=a-b+a+b+c-a+b+c
=(a+a-a)+(-b+b+b)+(c+c)
= a+b+c.2
= a+b+2c
B=(a-b)-(b-c)+(c-a)-(a-b-c)
=a-b-b+c+c-a-a+b+c
=(a-a-a)+(-b-b+b)+(c+c+c)
= (-a)+ (-b) +c.3
= (-a)+(-b)+3c
C=(-a+b+c)-(a-b+c)-(a+b-c)
= (-a)+b+c-a+b-c-a-b+c
=(-a-a-a)+(b+b-b)+(c-c+c)
= (-a.3) +b+c
B1:
a, a+b+(-a)+b+a+(-c)+(-a)+(-c)=[a+(-a)+a+(-a)]+(b+b)+[(-c)+(-c)]=0+2.b+(-2).c
b, a+b+(-c)+a+(-b)+c+(-b)+(-c)+a+(-a)+b+c=[a+a+a+(-a)]+[b+(-b)+(-b)+b]+[(-c)+c+(-c)+c]=2.a+0+0=2a
B2:
N=(a+b)-(a-b)+(a+b)=a+b+(-a)+b+a+b=[a+(-a)+a)+(b+b+b)=a+3.b
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rút gọn biểu thức
a,A=(a-b)-(a-b+c)
b,B=-(a+b+c)-(a+b-5)
a)\(\left(a+b\right)-\left(a-b\right)+\left(a-c\right)-\left(a+c\right)=a+b-a+b+a-c-a-c\)
\(=2a-2a+2b-2c=0+2\left(b-c\right)=2\left(b-c\right)\)
b)\(\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)\)
\(=a+b-c+a-b+c-b-c+a-a+b+c\)
\(=3a-a+2b-2b+2c-2c=2a+0+0=2a\)
c)\(-\left\{-\left(a+b\right)-\left[-\left(a-b\right)-\left(a+b\right)\right]\right\}=\left\{-\left(a+b\right)-\left[b-a-a-b\right]\right\}\)
\(=-\left\{-\left(a+b\right)-b+a+a+b\right\}\)
\(=-\left\{b-a-b+2a+b\right\}\)
\(=-\left\{b-b+b+2a-a\right\}\)
\(=-\left\{b+a\right\}=-a-b\)