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a) \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\) (quy đồng mẫu chung)
Vì b,d > 0 nên bd > 0. Do đó ad < bc (đpcm)
b) ad < bc \(\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\) (cùng chia cho bd)
Vì b,d > 0 nên bd > 0. Do đó \(\frac{a}{b}< \frac{c}{d}\) (rút gọn tử và mẫu)
a, Ta có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{ad}{bd}< \frac{cb}{db}\Rightarrow ad< cb\)
b, Ta có: \(ad< bc\Rightarrow\frac{ad}{bd}< \frac{bc}{bd}\Rightarrow\frac{a}{b}< \frac{c}{d}\)
1a) 8x + 15 - 3x = -400 b) -32x + 12x - 5x = 900
=> 5x = -400 - 15 => -25x = 900
=> 5x = -415 => x = 900 : (-25)
=> x = -415 : 5 => x = -36
=> x = -83
c) 3(x - 1) - (x - 5) = -18
=> 3x - 3 - x + 5 = -18
=> 2x + 2 = -18
=> 2x = -18 - 2
=> 2x = -20
=> x = -10
d,e tự làm
a) \(8x+15-3x=-400\)
\(\Leftrightarrow8x-3x=-400-15\)
\(\Leftrightarrow5x=-415\)
\(\Leftrightarrow x=-415\div5\)
\(\Leftrightarrow x=-83\)
b) \(-32x+12x-5x=900\)
\(\Leftrightarrow-25x=900\)
\(\Leftrightarrow x=900\div\left(-25\right)\)
\(\Leftrightarrow x=-36\)
\(a,\left(a-b+c\right)-\left(a+c\right)=-b\)
Ta có: \(VT=\left(a-b+c\right)-\left(a+c\right)=a-b+c-a-c=-b=VP\left(đpcm\right)\)
\(b,\left(a+b\right)-\left(b-a\right)+c=2a+c\)
Ta có: \(VT=\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c=VP\left(đpcm\right)\)
\(c,-\left(a+b-c\right)+\left(a-b-c\right)=-2b\)
Ta có: \(VT=-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b=VP\left(đpcm\right)\)
\(d,a\left(b+c\right)-a\left(b+d\right)=a\left(c-d\right)\)
Ta có: \(VT=a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab-ad=ac-ad=a\left(c-d\right)=VP\left(đpcm\right)\)
\(e,a\left(b-c\right)+a\left(d+c\right)=a\left(b+d\right)\)
Ta có: \(VT=a\left(b-c\right)+a\left(d+c\right)=ab-ac+ad+ac=ab+ad=a\left(b+d\right)=VP\left(đpcm\right)\)
Bài 1:
a, (a - b + c) - (a + c)
= a - b + c - a - c
= ( a - a ) + ( c- c ) - b
= -b
b, ( a + b ) - (b - a ) + c
= a + b - b + a + c
= ( a + a ) + ( b- b ) + c
= 2a + c
c, - ( a + b - c) + (a - b -c)
= -a -b + c + a - b - c
= ( -a + a ) - ( b + b ) + ( c - c)
= - 2b
d, a(b + c) - a ( b + d )
= ab + ac - ab - ad
= ac - ad
= a( c-d )
e, a( b - c ) + a( d + c )
= ab - ac + ad + ac
= ab + ad
= a( b+d )
Bài 2:
a, ab + ac = a( b + c)
b, ab - ac + ad= a( b -c + d )
c, ax - bx - cx + dx = x( a - b - c + d)
d, a(b + c ) - d( b + c) = ( b+c ).(a - d )
e, ac - ad + bc - bd = a ( c - d ) + b( c - d) = (a+b).( c - d )
g, ax + by + bx + ay = a( x + y) + b(y + x) = ( a+b).( x+y )