cho A=a+b-5
B=-b-c+1
C=b-c-4
D=b-a
chứng minh A+B=C+D
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Ta có :\(\text{VT = A + B}\)
\(\text{= ( a + b + 5 ) + ( b – c – 9 )}\)
\(\text{= a + b + 5 + b – c – 9}\)
\(\text{= a + ( b + b ) – c + ( 5 – 9 )}\)
\(\text{= a + 2b – c – 4 (1)}\)
\(\text{VP = C – D}\)
\(\text{= ( b – c – 4 ) – ( -b – a )}\)
\(\text{= b – c – 4 + b + a}\)
\(\text{= ( b + b ) – c + a – 4}\)
\(\text{= 2b – c + a – 4}\)
\(\text{= a + 2b – c – 4 (2)}\)
\(\text{từ (1) và (2) suy ra}\)\(\text{ A + B = C – D ( đpcm ) }\)
a/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{a-b}{a+b}=\dfrac{bk-b}{bk+b}=\dfrac{b\left(k-1\right)}{b\left(k+1\right)}=\dfrac{k-1}{k+1}\left(1\right)\)
\(VP=\dfrac{c-d}{c+d}=\dfrac{dk-d}{dk+d}=\dfrac{d\left(k-1\right)}{d\left(k+1\right)}=\dfrac{k-1}{k+1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
b/ Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có :
\(VT=\dfrac{2a-5b}{3a+4b}=\dfrac{2bk-5b}{3bk+4b}=\dfrac{b\left(2k-5\right)}{b\left(3k+4\right)}=\dfrac{2k-5}{3k+4}\left(1\right)\)
\(VP=\dfrac{2c-5d}{3c+4d}=\dfrac{2dk-5d}{3dk+4d}=\dfrac{d\left(2k-5\right)}{d\left(3k+4\right)}=\dfrac{2k-5}{3k+4}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
a)a/b=c/d=a+b/c+d=a-b/c-d(tc day ti so bang nhau)
=>a+b/a-b=c+d/c-d
b)a/b=c/d=>5a/5b=2c/2d=5a+2c/5c+2d(*) va a/b=4c/4d=a-4c/c-4d(**)
c)a/b=c/d=a+b/c+d=>(a/b)^2=ab/cd=(a+b/c+d)^2
a. Ta có : ( a + b )( c - d ) = ac-ad+bc-bd (1)
( a - b )( c + d ) = ac+ad-bc+bd (2)
Từ giả thuyết : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow ad=bc\) (3)
Từ (1) , ( 2) và ( 3) \(\Rightarrow\)( a + b )( c - d) = ( a - b)( c + d )
\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\left(đpcm\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk;c=dk\)
Khi đó : \(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d\left(2k+5\right)}{d\left(3k-4\right)}=\frac{2k+5}{3k-4}\)
\(=>\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\left(=\frac{2k+5}{3k-4}\right)\)
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Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{3a}{3c}=\frac{4b}{4d}=\frac{2a+5b}{2c+5d}=\frac{3a-4b}{3c-4d}\)
Suy ra : \(\frac{2a+5b}{3c-4d}=\frac{2c+5d}{3a-4b}\) (đpcm)
đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=> a=bk, c=dk =>\(\frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b\left(2k+5\right)}{b\left(3k-4\right)}=\frac{2k+5}{3k-4}\)(1)
=> \(\frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{2k+5}{3k-4}\) ( 2)
từ (1)( 2)=> \(\frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
câu b c/m tg tự
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
a) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\frac{3a}{3c}=\frac{4b}{4d}=\frac{3a+4b}{3c+4d}=\frac{3a-4b}{3c-4d}.\)
\(\Rightarrow\frac{3a+4b}{3a-4b}=\frac{3c+4d}{3c-4d}\)
b) ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{5a}{5b}=\frac{2c}{2d}=\frac{4a}{4b}\)
Lại có: \(\frac{5a}{5b}=\frac{2c}{2d}=\frac{5a+2c}{5b+2d}\)
\(\Rightarrow\frac{4a}{4b}=\frac{5a+2c}{5b+2d}\Rightarrow\frac{5a+2c}{4a}=\frac{5b+2d}{4b}\)
c) ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Lại có: \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\frac{\left(a+b^2\right)}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\)
Thật ra là A+B=C-D