a: =ab-ac+ad-ad=ab-ac

b:=(c+d)(a-b-a+b)=0

các bạn giúp mình làm bài toán này nhé !

a: =ab-ac+ad-ad=ab-ac

b:=(c+d)(a-b-a+b)=0

HT

a: \(\left(a-b-c\right)-\left(-c+b+a\right)-\left(a-b\right)\)

\(=a-b-c+c-b-a-a+b\)

\(=-a-b\)

b: \(a\left(b+c\right)-a\left(b+d\right)-\left(1+ac-ad\right)\)

\(=ab+ac-ab-ad-1-ac+ad\)

=-1

Bài 1:

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)

\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)

Bài 2:

a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)

b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)

a) = a - b + c - d - a - b - c - d

= -2b - 2d

b) = -a + b - c + a - b - a + b + c

= -a + b

c) = -a + b + c + b - c + d + a - b - d

= b 

a,= -2b - 2d

b,=-a + b

c,=b

a) = a - b + c - d - a - b - c - d

= -2b - 2d

b) = -a + b - c + a - b - a + b + c

= -a + b

c) = -a + b + c + b - c + d + a - b - d

= b 

a) - ( - a + c – d ) – ( c – a + d ) 

= a - c - d - c + a + d

= (a + a) + (-c - c) + (-d + d)

= 2a - 2c               

b) – ( a + b  - c + d ) + ( a – b – c –d )

= - a - b + c - d + a - b - c - d

= (-a + a) + (-b - b) + (c - c) + (-d - d)

= -2b - 2d

a) - ( - a + c - d) - ( c - a + d )

= a - c + d - c + a - d 

= 2a

b) - ( a+ b - c + d ) + ( a -b -c -d )

= - a-b+c-d+a-b-c-d

=-2d -2b

c) a(b-c-d) - a(b+c-d)

= a(b-c-d-b-c+d)

= ab-ac-ad-ab-ac+ad

= -2ab-2ac

d) (a+b)(c+d)-(a+d)(b+c)

= ac+ad+bc+bd - (ab+ac+bd+cd)

= ac+ad+bc+bd-ab-ac-bd-cd

=ad+bc-ab-cd

a) - ( - a + c - d) - ( c - a + d )

= a - c + d - c + a - d 

= 2a

b) - ( a+ b - c + d ) + ( a -b -c -d )

= - a-b+c-d+a-b-c-d

=-2d -2b

c) a(b-c-d) - a(b+c-d)

= a(b-c-d-b-c+d)

= ab-ac-ad-ab-ac+ad

= -2ab-2ac

d) (a+b)(c+d)-(a+d)(b+c)

= ac+ad+bc+bd - (ab+ac+bd+cd)

= ac+ad+bc+bd-ab-ac-bd-cd

=ad+bc-ab-cd

e)(a+b)(c-d)-(a-b)(c+d)

= ac-ad+bc-bd-ac-ad+bc+bd

= 2bc-2ad

f) ( a + b )² – ( a – b )²

= a²+2ab+b² - (a²+2ab-b²)

=a²+2ab+b² - a²-2ab+b²

=2b²

a, -( -a + c - d) - ( c - d + d) =  a - c + d - c + d - d =  a + d

b, - ( a+b-c+d) + (a-b-c-d) = -a -b+c-d + a-b-c-d = -2b + (-2c)= -2(b+c)

a, \(A=\frac{\left(x+2\right)^2}{x}\left(1-\frac{x^2}{x+2}\right)=\frac{\left(x+2\right)^2}{x}\left(\frac{x+2-x^2}{x+2}\right)\)

\(=\frac{-\left(x+2\right)^2\left(x-2\right)\left(x+1\right)}{x\left(x+2\right)}=\frac{-\left(x\pm2\right)\left(x+1\right)}{x}\)

c, Theo bài ra ta có : \(C=\frac{A}{B}\)hay \(\frac{\frac{-\left(x\pm2\right)\left(x+1\right)}{x}}{\frac{4}{\left(x-2\right)^2}}=\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}\)

d, Theo bài ra ta có : 

\(C>0\)hay \(\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}>0\)

\(\Leftrightarrow\frac{-\left(x+2\right)\left(x+1\right)}{x}.\frac{x-2}{4}>0\)

\(\Leftrightarrow-\left(x+2\right)\left(x+1\right)>0\Leftrightarrow\left(x+2\right)\left(x+1\right)>0\)

\(\Leftrightarrow x>-2;x>-1\Rightarrow x>-1\)

a) -(-a + c - d) - (c - a + d) = a - c + d - c + a - d = (a + a) - (c + c) + (d - d) = 2a - 2c

b) -(a + b - c + d) + (a - b - c - d) = -a - b + c - d + a  - b - c - d = (-a + a) - (b + b) + (c - c) - (d + d) = -2b - 2d

c) (a + b - c) - (b - c + d) = a + b - c - b + c - d = a + (b - b) - (c - c) - d = a - d

d) (b + a) + (c - d) - (c + a) - (b - d) = b + a + c - d - c - a - b + d = (b - b) + (a - a) + (c - c) - (d - d) = 0

e) (a - b) - (d + a) - (c - d) + (c + b) = a - b - d - a - c + d +  c + b = (a - a) - (b - b) - (d - d) - (c - c) = 0

f) -a + (c - b) - (c + a - b) = - a + c - b - c - a + b = (-a - a) + (c - c) - (b - b) = -2a

a ) a - c + d - c + a - d = 2a - 2c 

b )  -a - b + c - d + a - b - c - d = -2b - 2d 

 CÁC CÂU CÒN LẠI LÀM TƯƠNG TỰ NHÉ!!

a) - ( -a + c - d ) - ( c - a + d ) = a - c + d - c + a - d = ( a + a ) - ( c + c ) + ( d - d ) = 2a - 2c = 2 ( a - c )

b) - ( a + b - c + d ) + ( a - b - c - d ) = -a - b + c - d + a - b - c - d = ( -a + a ) - ( b + b ) + ( c - c ) - ( d - d ) = -2b

c) ( a + b - c ) - ( b - c + d ) = a + b - c - b + c - d = a + ( b + b ) - ( c - c ) - d = a + 2b -d

d) ( b + a ) + ( c - d ) - ( c + a ) - ( b - d ) = b + a + c - d - c - a - b + d = ( b - b ) + ( a - a ) + ( c - c ) - ( d - d ) = 0

e) ( a - b ) - ( d + a ) - ( c - d ) + ( c + b ) = a - b - d - a - c + d + c + b = ( a - a ) - ( b - b ) - ( d - d ) - ( c - c ) = 0

f) -a + ( c - b ) - ( c + a - b ) = -a + c - b - c - a + b = ( -a - a ) + ( c - c ) - ( b - b ) = -2a

=))

a)-(-a+c-d)-(c-a+d)=a-c+d-c+a-d=(a+a)-(c+c)+(d-d)=2a-2c=2(a-c)

b)-(a+b-c+d)+(a-b-c-d)=-a-b+c-d+a-b-c-d=(-a+a)-(b+b)+(c-c)-(d+d)=0-2b+0-2d=-2(b-d)

c)a(b-c-d)-a(b+c-d)=ab-ac-ad-ab-ac+ad=(ab-ac)-(ac+ac)-(ad-ad)=2ac

d)đề sai

e)(a+b)(c-d)-(a-b)(c+d)=ac+b-ad+b-(ac-b+ad-b)=ac+b-ad+b-ac+b-ad+b=(ac-ac)+(b+b+b+b)-(ad+ad)=4b-2ad=2(2b-ad)

f)(a+b)²-(a-b)²=a²+2ab+b²-(a²-2ab+b²)=a²+2ab+b²-a²+2ab-b²=(a²-a²)+(2ab+2ab)+(b²-b²)=4ab

mk k chắc đâu

-(a+b-c)+(b+c-d)-(c+d-a)