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Câu a)
\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
\(\Leftrightarrow\left(a+b\right).\left(c-2d\right)=\left(a-2b\right).\left(c+d\right)\)
\(\Leftrightarrow a.\left(c-2d\right)+b.\left(c-2d\right)=a.\left(c+d\right)-2b.\left(c+d\right)\)\(\)
\(\Leftrightarrow ac-2ad+bc-2bd=ac+ad-2bc-2bd\)
\(\Leftrightarrow bc-2ad=ad-2bc\)
\(\Leftrightarrow bc+2bc=ad+2ad\)
\(\Leftrightarrow3bc=3ad\)
\(\Leftrightarrow bc=ad\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
Câu b)
Ta có : \(a+d=b+c\Rightarrow\left(a+d\right)^2=\left(b+c\right)^2\)
\(\Leftrightarrow a^2+2ad+d^2=b^2+2bc+c^2\) (*)
Lại có : \(a^2+d^2=b^2+c^2\)
\(\Leftrightarrow2ad=2bc\) ( bớt cả hai vế của đẳng thức (*) đi \(a^2+d^2\) và \(b^2+c^2\))
\(\Leftrightarrow ad=bc\)
\(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
Vậy : 4 số a, b, c, d có thể lập được 1 tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\).
a
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)
b
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
c
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{5a^2}{5b^2}=\frac{3c^2}{3d^2}=\frac{5a^2+3c^2}{3d^2+5b^2}\)
1) Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Leftrightarrow\frac{a}{c}+1=\frac{b}{d}+1\)
\(\Leftrightarrow\frac{a+c}{c}=\frac{b+d}{d}\)(đpcm)
2) Để \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\) thì \(\frac{2a+3b}{2c+3d}=\frac{2a-3b}{2c-3d}\)
\(\Leftrightarrow\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a}{2c}=\frac{3b}{3d}\)
\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{a}{c}=\frac{b}{d}\)
\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
hay \(\frac{a}{b}=\frac{c}{d}\)(đpcm)
3) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\)(1)
Ta có: \(\frac{a^2-b^2}{c^2-d^2}\)
\(=\frac{k^2\cdot b^2-b^2}{k^2\cdot d^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
4) Ta có: \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
nên \(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2\cdot k^2+b^2}{d^2\cdot k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(3)
Ta có: \(\left(\frac{a+b}{c+d}\right)^2\)
\(=\left(\frac{bk+b}{dk+d}\right)^2\)
\(=\left(\frac{b\left(k+1\right)}{d\left(k+1\right)}\right)^2\)
\(=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\)(4)
Từ (3) và (4) suy ra \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
Bài 1:
a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)
\(\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\)
\(\Rightarrow\frac{a}{b}-\frac{b}{b}=\frac{c}{d}-\frac{d}{d}.\)
\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\left(đpcm\right).\)
Mình làm được thế thôi nhé.
Chúc bạn học tốt!
Ta có:
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a) \(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b) \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c) \(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2.\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2\right)+1}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) , (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
c) có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a^2}{^{c^2}}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
Lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(2\right)\)
Từ (1) và (2) có \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
các câu còn lại bạn tự làm đi! HI.......
Ta có:
\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{a.b}{c.d}\)=\(\frac{a^2+b^2+a.b}{c^2+d^2+c.d}\)=\(\frac{a^2+a.b+b^2+a.b}{c^2+c.d+d^2+c.d}\)
\(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{a.b}{c.d}\)=\(\frac{a\left(a+b\right)+b\left(a+b\right)}{c\left(c+d\right)+d\left(c+d\right)}\)\(\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\)
\(\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\)=\(\frac{a.b}{c.d}\)=) \(\frac{c\left(a+b\right)}{a\left(c+d\right)}\)=\(\frac{b\left(c+d\right)}{d\left(a+b\right)}\)
=) \(\frac{ca+cb}{ca+ad}\)=\(\frac{bc+bd}{ad+bd}\)=\(\frac{ca-bd}{ad-bd}\)=1
=) ca + cb = ca + ad
=) cb = ad
=) \(\frac{a}{b}\)= \(\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
1)\(VT=\frac{a}{b}=\frac{bk}{b}=k\left(1\right)\)
\(VP=\frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\left(2\right)\)
Từ (1) và (2) ->Đpcm
2)\(VT=\frac{a-b}{a}=\frac{bk-b}{bk}=\frac{b\left(k-1\right)}{bk}=\frac{k-1}{k}\left(1\right)\)
\(VP=\frac{c-d}{c}=\frac{dk-d}{dk}=\frac{d\left(k-1\right)}{dk}=\frac{k-1}{k}\left(2\right)\)
Từ (1) và (2) ->Đpcm
Hướng dẫn cách làm nè!
Đầu tiên làm ra nháp:
Xuất phát từ đầu bài: \(\frac{a}{b}\)=\(\frac{a+c}{b+d}\)
=> a.( b+d ) = b.( a+c ) {tích chéo}
=>ab+ad = ab+bc {phân phối}
=>ad = bc {rút gọn cùng chia cho ab}
=>\(\frac{a}{b}\)= \(\frac{c}{d}\) {tính chất của tlt}
_Đó là phần nháp, còn trình bày bạn chỉ cần chép từ dưới lên:
\(\frac{a}{b}\)=\(\frac{c}{d}\)
=> ad=bc
=> ab+ad=ab+bc
=> a.( b+d )= b. (a+c)
=> \(\frac{a}{b}\) = \(\frac{a+c}{b+d}\)
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Xem ở lick này nhé (mình gửi cho)
Học tốt!!!!!!!!!!!!!
Đặt:
\(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk\)
\(\Rightarrow c=dk\)
Thế vào vế phải:
\(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bk+b}{dk+d}\right)^2=\frac{bk^2+b^2}{dk^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}=\frac{b}{d}\)
Thế vào vế trái:
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}=\frac{b}{d}\)
=> Vế phải = vế trái
=> ĐPCM