Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{4c}{4d}=\dfrac{a+4c}{b+4d}\left(đpcm\right)\)
b;c;d tương tự hết
b: a/b=c/d
nên 3a/3b=2c/2d
=>a/b=c/d=(3a+2c)/(3b+2d)
c: a/c=b/d nên a/c=2b/2d=(a-2b)/(c-2d)
d: a/c=b/d
nên 5a/5c=2b/2d
=>a/c=b/d=(5a-2b)/(5c-2d)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\\ \Rightarrow a=bk;c=dk\)
Ta có:
\(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2\left(bk\right)^2-2bkb+5b^2}{2b^2+3bkb}=\dfrac{2b^2k^2-2b^2k+5b^2}{2b^2+3b^2k}=\dfrac{b^2\left(2k^2-3k+5\right)}{b^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(1\right)\)
\(\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}=\dfrac{2\left(dk\right)^2-3dkd+5d^2}{2d^2+3dkd}=\dfrac{2d^2k^2-3d^2k+5d^2}{2d^2+3d^2k}=\dfrac{d^2\left(2k^2-3k+5\right)}{d^2\left(2+3k\right)}=\dfrac{2k^2-3k+5}{2+3k}\left(2\right)\)
Từ (1) và (2) suy ra:
\(\dfrac{2a^2-3ab+5b^2}{2b^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2d^2+3cd}\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{2c}{2d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\dfrac{a+c}{b+d}=\dfrac{a+2c}{b+2d}\)
\(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\left(đpcm\right)\)
Vậy...
Vì \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left[{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\) (!)
Thay (!) vào đề bài:
VT = \(c\left(k+2\right).d\left(k+1\right)\left(1\right)\)
\(VP=c\left(k+1\right).d\left(k+2\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow VT=VP\)
hay \(\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\).
\(\left(a-2c\right)\left(b+2d\right)=\left(b-2d\right)\left(a+2c\right)\)
\(\Leftrightarrow ab+2ad-2bc-4cd=ab+2bc-2ad-4cd\)
\(\Leftrightarrow2ad+2ad=2bc+2bc\Leftrightarrow4ab=4bc\)
\(\Leftrightarrow ad=bc\Rightarrow\dfrac{a}{b}=\dfrac{c}{d},\left(a,b,c,d\ne0\right)\)
*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 \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2(a+b+c+d)}=\frac{1}{2}\)
\(\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1\Leftrightarrow a=b=c=d\)
Do đó:
\(A=\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
Vậy \(A=2\)
Ta có: \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2\left(a+b+c+d\right)}=\dfrac{1}{2}\)
\(\Rightarrow a=b;b=c;c=d;d=a\)
\(A=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
\(A=\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}\)
\(A=\dfrac{c+c+c+c}{c+c}=2\)
Vậy ....................
Câu 1:
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{a^2}{c^2}=\dfrac{b^2k^2}{d^2k^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2a^2+3b^2}{2c^2+3d^2}=\dfrac{2b^2k^2+3b^2}{2d^2k^2+3d^2}=\dfrac{b^2}{d^2}\)
=>\(\dfrac{a^2}{c^2}=\dfrac{2a^2+3b^2}{2c^2+3d^2}\)
b: \(\dfrac{2a-3c}{c}=\dfrac{2bk-3dk}{dk}=\dfrac{2b-3d}{d}\)
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\dfrac{a}{b}=\dfrac{3a}{3b}\) ; \(\dfrac{c}{d}=\dfrac{2c}{2d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{3a+2c}{3b+2d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{3a+2c}{3b+2d}\)
Mình hướng dẫn thôi nhé:
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\) . Sau đó thế vào biểu thức tính rồi suy ra đpcm
Ví dụ bài đầu tiên: Thế a = kb; c=kd vào biểu thức,ta có:
\(\dfrac{a}{a+b}=\dfrac{kb}{kb+b}=\dfrac{kb}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (1)
\(\dfrac{c}{c+d}=\dfrac{kd}{kd+d}=\dfrac{kd}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
Từ (1) và (2) ,ta có đpcm: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Các bài sau làm tương tự:Thế a=kb ; c=kd vào biểu thức rồi tính từng vế . Sau đó so sánh hai vế. Thấy hai vế = nhau => đpcm