cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh: \(\frac{\left(a-b\right)^2}{\left(c+d\right)^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\), chứng minh rằng:
\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
cho \(\frac{a}{b}=\frac{c}{d}\) chung minh: \(\frac{\left(a+b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
1. Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng :
\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
2. Tìm x, y biết : x + y + 2xy = 83
1. Đặt \(\frac{a}{b}=\frac{c}{d}=k=>a=bk,c=dk\)
Thay vào 2 vế là sẽ CM được
1. Đặt \(\frac{a}{b}=\frac{c}{d}=k>a=bk.c=dk\)
Thay vào 2 vế để chứng minh
1 )
Ta có :
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)( Áp dụng t/c DTSBN )
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\left(1\right)\)
Lại có : \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{3a^2}{3c^2}=\frac{2b^2}{2d^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\) ( Áp dụng t/c DTSBN ) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\)
\(\Rightarrow\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{3a^2+2b^2}{3c^2+2d^2}\left(đpcm\right)\)
2 )
Ta có :
\(x+y+2xy=83\)
\(\Rightarrow2\left(x+y+2xy\right)=166\)
\(\Rightarrow2x+2y+4xy+1=167\)
\(\Rightarrow2x\left(2y+1\right)+\left(2y+1\right)=167\)
\(\Rightarrow\left(2x+1\right)\left(2y+1\right)=167\)
Do \(x;y\in Z\)
\(\Leftrightarrow2x+1;2y+1\in Z\)
\(\Leftrightarrow2x+1;2y+1\in\left\{\pm1;\pm167\right\}\)
Ta có bảng sau :
\(2x+1\) | \(1\) | \(167\) | \(-1\) | \(-167\) |
\(2y+1\) | \(167\) | \(1\) | \(-167\) | \(-1\) |
\(x\) | \(0\) | \(83\) | \(-1\) | \(-84\) |
\(y\) | \(83\) | \(0\) | \(-84\) | \(-1\) |
Vậy \(\left(x;y\right)\in\left\{\left(0;83\right),\left(83;0\right),\left(-1;-84\right),\left(-84;-1\right)\right\}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\), Chứng Minh rằng:
\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
M.N giải giúp mk, ai đúng mk chọn
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\). Ta có:
\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{\left(bk-b\right)^3}{\left(dk-d\right)^3}=\frac{b^3\left(k-1\right)^3}{d^3\left(k-1\right)^3}=\frac{b^3}{d^3}\)
\(\frac{3a^2+2b^2}{3c^2+2d^2}=\frac{3\left(bk\right)^2+2b^2}{3\left(dk\right)^2+2d^2}=\frac{3b^2k^2+2b^2}{3d^2k^2+2d^2}=\frac{b^2\left(3k^2+2\right)}{d^2\left(3k^2+2\right)}=\frac{b^2}{d^2}\)
Đến đây nhìn có vẻ đề sai
\(\frac{a}{b}=\frac{c}{d}=k\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)ta có:
\(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{\left(bk-b\right)^3}{\left(dk-d\right)^3}=\frac{\left[b\left(k-1\right)\right]^3}{\left[d\left(k-1\right)\right]^3}=\frac{b^3}{d^3}\)
\(\frac{2b^2+3a^2}{2d^2+3c^2}=\frac{4.b^2+9.k^2.b^2}{4.d^2+9.d^2.k^2}=\frac{b^2\left(4+k^2.9\right)}{d^2\left(4+9.k^2\right)}=\frac{b^2}{d^2}\)
\(Taco:\frac{b^3}{d^3}=\frac{b^2}{d^2}\Leftrightarrow b=d\)
Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh:
a) \(\frac{\left(a-b\right)^3}{\left(c-d\right)^3}=\frac{3a^2+2b^2}{3c^2+2d^2}\)
b)\(\frac{4a^4+5b^4}{4c^4+5d^4}=\frac{a^2b^2}{c^2d^2}\)
c)\(\left(\frac{a-b}{c-d}\right)^{2005}=\frac{2a^{2005}-b^{2005}}{2c^{2005}-d^{2005}}\)
d)\(\frac{2a^{2005}+5b^{2005}}{2c^{2005}+5d^{2005}}=\frac{\left(a+b\right)^{2005}}{\left(c+d\right)^{2005}}\)
e)\(\frac{\left(20a^{2006}+11b^{2006}\right)^{2007}}{\left(20a^{2007}-11b^{2007}\right)^{2006}}=\frac{\left(20c^{2006}+11d^{2006}\right)^{2007}}{\left(20c^{2007}-11d^{2007}\right)^{2006}}\)
f)\(\frac{\left(20a^{2007}-11c^{2007}\right)^{2006}}{\left(20a^{2006}+11c^{2006}\right)^{2007}}=\frac{\left(20b^{2007}-11d^{2007}\right)^{2006}}{\left(20b^{2006}+11d^{2006}\right)^{2007}}\)
ừ, bạn bik làm thì giúp mình nha ^^
Cho \(\frac{a}{b}=\frac{c}{d}\)Chứng minh rằng:
a) \(\frac{\left(2a+3b\right)^2}{\left(3a-4b\right)^2}=\frac{\left(2c+3d\right)^2}{\left(3c-4d\right)^2}\)
b) \(\frac{2a^2-3ab+4b^2}{2b^2+5ab}=\frac{2c^2-3cd+4d^2}{2d^2+5cd}\)
mk làm câu a thôi, b dài nhưng tương tự
Gọi a/b=c/d=k =>a=bk ; c=dk
=>\(\frac{\left(2a+3b\right)^2}{\left(3a-4b\right)^2}=\frac{\left(2bk+3b\right)^2}{\left(3bk-4b\right)^2}=\frac{\left[b\left(2k+3\right)\right]^2}{\left[b\left(3k-4\right)\right]^2}=\frac{b^2\left(2k+3\right)^2}{b^2\left(3k-4\right)^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(1)
=>\(\frac{\left(2c+3d\right)^2}{\left(3c-4d\right)^2}=\frac{\left(2dk+3d\right)^2}{\left(3dk-4d\right)^2}=\frac{\left[d\left(2k+3\right)\right]^2}{\left[d\left(3k-4\right)\right]^2}=\frac{\left(2k+3\right)^2}{\left(3k-4\right)^2}\)(2)
Từ (1);(2)=> đpcm
Chứng minh rằng nếu a/b=c/d\(a,\frac{a^2+2c^2}{b^2+2d^2}=\left(\frac{a+3c}{b+3d}\right)^2\) \(\frac{a^{2016}+3b^{2016}}{c^{2016}+3d^{2016}}=\left(\frac{a^2+2b^2}{c^2+2d^2}\right)^2\)
a) \(\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\) b)\(\frac{ac}{bd}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
cho \(\frac{a}{b}=\frac{c}{d}\) (a,b,c,d khác 0)
chứng minh ta có tỉ lệ thức trên.
b)\(\frac{ac}{bd}=\frac{bkdk}{bd}=k.k=k^2\)
\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{\left[k\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\frac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)
=> \(\frac{ac}{bd}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
Đặt k ( với k khác 0 , thuộc Z ) sao cho \(\frac{a}{b}=\frac{c}{d}=k\) => \(a=kb\) / \(c=dk\) .
a) Thế vào \(\frac{5a-b}{3a+2b}\) , ta có \(\frac{5kb-3b}{3kb+2b}\)\(=\frac{b\left(5k-3\right)}{b\left(3k+2\right)}\)\(=\frac{5k-3}{3k+2}\) / \(\frac{5c-3d}{3c+2d}=\frac{5dk-3d}{3dk-2d}=\frac{d\left(5k-3\right)}{d\left(3k+2\right)}=\frac{\left(5k+3\right)}{\left(3k+2\right)}\)
=> VT = VP
Cho \(\frac{a}{b}=\frac{c}{d}.\)Chứng minh.
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b)\(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a/ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{5b}{5d}=\frac{3a+5b}{3c+5d}=\frac{3a-5b}{3c-5d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b/ \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\left(\frac{a+b}{c+d}\right)^2\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{b^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)
\(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)