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Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
a) Ta có:
\(\frac{5a+3b}{5a-3b}=\frac{5bk+3b}{5bk-3b}=\frac{b(5k+3)}{b(5k-3)}=\frac{5k+3}{5k-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d(5k+3)}{d(5k-3)}=\frac{5k+3}{5k-3}\)
\(\Rightarrow \frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
b)
\(\frac{2a-b}{2a+b}=\frac{2bk-b}{2bk+b}=\frac{b(2k-1)}{bb(2k+1)}=\frac{2k-1}{2k+1}\)
\(\frac{2c-d}{2c+d}=\frac{2dk-d}{2dk+d}=\frac{d(2k-1)}{d(2k+1)}=\frac{2k-1}{2k+1}\)
\(\Rightarrow \frac{2a-b}{2a+b}=\frac{2c-d}{2c+d}\) (đpcm)
a) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
Từ (1) và (2) suy ra \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=q\Rightarrow\left\{{}\begin{matrix}a=bq\\c=dq\end{matrix}\right.\)
Ta có:
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bq+b}{dq+d}\right)^2=\left[\dfrac{b\left(q+1\right)}{d\left(q+1\right)}\right]^2=\dfrac{b}{d}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bq\right)^2+b^2}{\left(dq\right)^2+d^2}=\dfrac{b^2.q^2+b^2}{d^2.q^2+d^2}=\dfrac{b^2\left(q^2+1\right)}{d^2\left(q^2+1\right)}=\dfrac{b}{d}\) (2)
Từ (1) và (2) suy ra \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)
áp dụng tính chất dãy tỉ số = nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)
= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
\(a,\)
Xét \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ad=bc\)
mà \(ad=bc\left(gt\right)\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)
\(b,\)
\(\dfrac{a}{b}=\dfrac{c}{d}\) (Chứng minh câu a)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a}{b}\)
\(c,\)
Xét \(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow ad=bc\)
mà \(ad=bc\left(gt\right)\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(d,\)
\(\dfrac{a}{c}=\dfrac{b}{d}\) (Chứng minh câu c)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
\(e,\)
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{2a+b}{2c+d}\)
\(\Rightarrow\dfrac{2a+b}{2c+d}=\dfrac{a}{c}\)
*a/b=c/d=k=>a=bk;c=dk
Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3
Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3
=>2a+3b/2a-3b=2c+3d/2c-3d
*a/b=c/d=>a/c=b/d=k
=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)
k^2=a/c.b/d=ab/cd (2)
Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2
*a/b=c/d=>a/c=b/d=k=a+b/c+d
=>k^2=(a+b/c+d)^2
k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2
=>(a+b/c+d)^2=a^2+b^2/c^2+d^2
Gọi \(\dfrac{a}{b}=\dfrac{c}{d}=k\).\(\Rightarrow a=bk,c=dk\)
a)Ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\)(1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}\dfrac{2k+3}{2k-3}\)(2)
Từ (1),(2)ta có:\(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)Ta có:\(\dfrac{ab}{cd}=\dfrac{bk\times b}{dk\times d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)(2)
Từ (1),(2) ta có:\(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
c)Ta có:\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{b^2}{d^2}\)(1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2}{d^2}\)(2)
Từ (1), (2) ta có \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{2a-3b}{2a+3b}=\dfrac{2bk-3b}{2bk+3b}=\dfrac{2k-3}{2k+3}\)
\(\dfrac{2c-3d}{2c+3d}=\dfrac{2dk-3d}{2dk+3d}=\dfrac{2k-3}{2k+3}\)
Do đó: \(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)
ta có :\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2\left(a+b+c+d\right)}=\dfrac{1}{2}\)
suy ra:\(a=b;b=c;c=d;d=a\)
\(A=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
\(A=\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}\)
\(A=\dfrac{c+c+c+c}{c+c}=2\)
vậy giá trị của A là 2
a)đặt \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=k\(\Rightarrow\)a=bk, c=dk
\(\dfrac{2a+3b}{2a-3b}=\dfrac{2bk+3b}{2bk-3b}=\dfrac{b\left(2k+3\right)}{b\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (1)
\(\dfrac{2c+3d}{2c-3d}=\dfrac{2dk+3d}{2dk-3d}=\dfrac{d\left(2k+3\right)}{d\left(2k-3\right)}=\dfrac{2k+3}{2k-3}\) (2)
từ (1),(2)\(\Rightarrow\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\)
b)ta có:
\(\dfrac{ab}{cd}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}\)
câu c bn tự giải nhé dễ mak ahihihi
chúc bn hc tốt
dễ mà :\
Ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\\ \Rightarrow\dfrac{a}{c}=\dfrac{b}{d} \)
\(\Rightarrow\dfrac{2a}{2c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{2a}{2c}=\dfrac{b}{d}=\dfrac{2a-b}{2c-d}=\dfrac{2a+b}{2c+d}\\ \Rightarrow\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
Vậy...