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a, Có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\)
Có: \(\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\Leftrightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b, Co: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{ab}{cd}\)
Lại có:\(\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\left(2\right)\)
Tu (1)&(2),có: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
a)\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)
\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(2)
Từ (1) và (2) \(\Rightarrow\)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b)\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2.k^2-b^2}{d^2.k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)
c)\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(ck+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2}{d^2}\)(1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2.k^2+b^2}{d^2.k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)(2)
Từ (1) và(2)\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
k cho mình nhé
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Xem ở lick này nhé (mình gửi cho)
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.......
Tạm thời giải phần a đã nhé -_-
a, Từ a/b = c/d => a/c=b/d
Đặt a/c=b/d=k thì a=ck, b=dk
Xét : 4a-3b/4a+3b=4ck-3dk/4ck+3dk=k.(4c-3d)/k.(4c+3d)=4c-3d/4c+3d
=> 4a-3b/4a+3b=4c-3d/4c+3d => 4a-3b/4c-3d=4a+3b/4c+3d
Nhìn trên máy khó lắm viết lại theo lời giải ra nháp trc' cho dễ nhìn nhé @@
\(a,\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{4a}{4c}=\frac{3b}{3d}=\frac{4a-3b}{4c-3d}\)\(\left(1\right)\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{4a}{4c}=\frac{3b}{3d}=\frac{4a+3b}{4a+3d}\)\(\left(2\right)\)
Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\left(đpcm\right)\)
\(b\)Đặt \(\frac{a}{c}=\frac{b}{d}=k\)\(\Rightarrow a=ck;b=dk\)
\(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{\left(ck\right)^2-\left(dk\right)^2}{c^2-d^2}=\frac{c^2k^2-d^2k^2}{c^2-d^2}=\frac{k^2\left(c^2-d^2\right)}{c^2-d^2}=k^2\)\(\left(3\right)\)
Mà \(\frac{ab}{cd}=\frac{ck.dk}{cd}=k^2\)\(\left(4\right)\)
Từ ( 3 ) và ( 4 ) \(\Rightarrow\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\left(đpcm\right)\)
\(c,\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\left(5\right)\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{5b}{5d}=\frac{2a+5b}{2c+5d}\left(6\right)\)
TỪ ( 5 ) và ( 6 ) \(\Rightarrow\frac{a-b}{c-d}=\frac{2a+5b}{2c+5d}\left(đpcm\right)\)
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{2a}{2c}=\dfrac{3b}{3d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{2a}{2c}=\dfrac{3b}{3d}=\dfrac{2a-3b}{2c-3d}=\dfrac{2a+3b}{2c+3d}\) ( đpcm )
b) Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\) ( đpcm ).
Theo đề bài ta có:
a/b=c/d=a/c=b/d
Áp dụng tính chất dãy tỉ số bằng nhau:
a/c=b/d=2a/2c=3b/3d=2a+3b/2c+3d
=2a-3b/2c-3d
=>2a+3b/2c+3d=2a-3b/2c-3d=2a+3b/2a-3b=2c+3d/2c-3d (đpcm)
b) Theo đề bài ta có:
a/b=c/d=ab/b^2=cd/d^2=ab/cd=b^2/d^2 (*)
Áp dụng tính chất dãy tỉ số bằng nhau :
a/b=c/d=a/c=b/d=a^2/c^2/b^2/d^2=a^2-b^2/c^2-d^2(**)
Từ (*) và(**) suy ra ab/cd=a^2-b^2/c^2-d^2 (đpcm)
(có thể trình bày theo cách khác)
a, a/b=c/d
<=>a/c=b/d
<=>2a/2c=3b/3d=2a+3b/2c+3d=2a-3b/2c-3d
<=>2a+3b/2a-3b=2c+3d/2c-3d(đpcm)
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 \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)