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A =(a+b-2c) -(-a+b+c) -(2a-b-c)
= a+b-2c+a-b-c-2a+b+c
= b-2c
B=-(2a-b+c) + (b-2c-3a) -(-5a-3c+b)
= -2a+b-c+b-2c-3a+5a+3c-b
= b-c
C=(3a-b-2c)-( 2b+3c-a) +(2a-3b)
= a-b-2c-2b-3c+a+2a-3b
= -6b-5c
D=(5a-3b+c) +( 2a-3b+5) -( b-c+a)
= 5a-3b+c+2a-3b+5-b+c-a
= 6a-7b+2c
\(A=\left(a+b-2c\right)-\left(-a+b+c\right)-\left(2a-b-c\right)\)
\(=a+b-2c+a-b-c-2a+b+c=b-2c\)
\(B=-\left(2a-b+c\right)+\left(b-2c-3a\right)-\left(-5a-3c+b\right)\)
\(=-2a+b-c+b-2c-3a+5a+3c-b=b\)
\(C=\left(3a-b-2c\right)-\left(2b+3c-a\right)+\left(2a-3b\right)\)
\(=3a-b-2c-2b-3c+a+2a-3b=6a-6b-5c\)
\(D=\left(5a-3b+c\right)+\left(2a-3b+5\right)-\left(b-c+a\right)\)
\(=5a-3b+c+2a-3b+5-b+c-a=6a-7b+2c\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\) (\(k\in N\)*)
\(\Rightarrow\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\frac{2bk-3b}{2bk+3b}=\frac{2dk-3d}{2dk+3d}\)
Xét vế trái \(\frac{2a-3b}{2a+3b}=\frac{2bk-3b}{2bk+3b}=\frac{b\left(2k-3\right)}{b\left(2k+3\right)}=\frac{2k-3}{2k+3}\left(1\right)\)
Xét vế phải \(\frac{2c-3d}{2c+3d}=\frac{2dk-3d}{2dk+3d}=\frac{d\left(2k-3\right)}{d\left(2k+3\right)}=\frac{2k-3}{2k+3}\left(2\right)\)
Từ (1) và (2) ta có Đpcm
Đặt ab=cd=kab=cd=k (k∈Nk∈N*)
⇒{a=bkc=dk⇒{a=bkc=dk⇒2bk−3b2bk+3b=2dk−3d2dk+3d⇒2bk−3b2bk+3b=2dk−3d2dk+3d
Xét vế trái 2a−3b2a+3b=2bk−3b2bk+3b=b(2k−3)b(2k+3)=2k−32k+3(1)2a−3b2a+3b=2bk−3b2bk+3b=b(2k−3)b(2k+3)=2k−32k+3(1)
Xét vế phải 2c−3d2c+3d=2dk−3d2dk+3d=d(2k−3)d(2k+3)=2k−32k+3(2)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b}{d}\)
\(\dfrac{2a-3b}{2c-3d}=\dfrac{2bk-3b}{2dk-3d}=\dfrac{b}{d}\)
Do đó: \(\dfrac{a-b}{c-d}=\dfrac{2a-3b}{2c-3d}\)
a) 3a + 4b - 5c - 2a - 3b + 5c
= ( 3a - 2a ) + ( 4b - 3b ) - ( 5c - 5c )
= a + b
b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c
= ( 7a - 3a - 4a ) + ( 3b + 2b + b ) - ( 4c + 2c + 2c )
= 6b - 8c
a) 3a + 4b - 5c - 2a - 3b + 5c
= (3a - 2a) + (4b - 3b) - (5c - 5c)
= a + b - 0 = a + b
b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c
= (7a - 3a - 4a) + (3b + 2b + b) - ( 4c + 2c + 2c)
= 0 + 6b - 8c = 6b - 8c
B=(-5c+3a-4b)-(3a-4b+7c)-(-12b-6a+15c)+(-3c+21a-10b)
=-5c+3a-4b-3a+4b-7c+12b+6a-15c
=6a +12b -27c
C=-(-32b-12c+5a)+(2c-4b-23a)-(17a-16c-31b)-(-6b+3c)
=32b+12c-5a+2c-4b-23a-17a+16c+31b+6b-3c
=-45a+65b+9c
1 , a - ( a - b - c ) - ( b - c -a ) - ( c - b -a )
= a - a + b + c - b + c + a - c + b + a
= (a-a+a) + (b-b+b) + (c-c+c)
= a+b+c
2 , - ( a + b + c ) - ( b - c -a ) + ( 1 - a - b ) - ( c - 3b )
= -a - b - c - b + c + a + 1 - a - b - c + 3b
= (a+a-a) - (b+b+b) + (c-c+c) + 3b
= a - 3b + c + 3b
= a+c + (3b - 3b)
= a+c + 0
= a+c
3 , ( b - c - 6 ) - ( 7 - a + b ) + c
= b - c - 6 - 7 + a - b + c
= (b-b) + (c-c) - (6+7) + a
= 0 + 0 - 13 + a
= -13 + a
4 , - ( 3b - 2a - c ) - ( a - b - c ) - ( a - 2b -+ 2c )
= -3b + 2a + c - a + b + c - a + 2b - 2c
= -3b + (2b + b) + (c + c) - (a+a) +2a - 2c
= -3b + 3b + 2c - 2a + 2a - 2c
= (3b - 3b) + (2c - 2c) + (2a + 2a)
= 0 + 0 + 0
= 0
chỉ bt lm đến đây thoy
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