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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
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:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{3a+2c}{3b+2d}\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{3a+2c}{3b+2d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a/ \(VT=\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1=\left(1\right)\)
\(VP=\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
b/ \(VT=\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
\(VP=\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
c/ \(VT=\dfrac{2a-5b}{2c-5d}=\dfrac{2bk-5b}{2dk-5d}=\dfrac{b\left(2k-5\right)}{d\left(2k-5\right)}=\dfrac{b}{d}\left(1\right)\)
\(VP=\dfrac{3a+4b}{3c+4d}=\dfrac{3bk+4b}{3dk+4d}=\dfrac{b\left(3k+4\right)}{d\left(3k+4\right)}=\dfrac{b}{d}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{2a-5b}{2c-5đ}=\dfrac{3a+4b}{3c+4d}\)
d/ \(VT=\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{\left(bk\right)^2-\left(dk\right)^2}{b^2-k^2}=\dfrac{k^2\left(b^2-d^2\right)}{b^2-d^2}=k^2\left(1\right)\)
\(VP=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
a: \(\dfrac{2}{3}:\left(6x+7\right)=0.2:1\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{2}{3}:\left(6x+7\right)=\dfrac{1}{5}:\dfrac{7}{6}=\dfrac{6}{35}\)
\(\Leftrightarrow6x+7=\dfrac{35}{9}\)
=>6x=-28/9
hay x=-28/54=-14/27
b: \(\dfrac{a}{a+2b}=\dfrac{c}{c+2d}\)
\(\Leftrightarrow a\left(c+2d\right)=c\left(a+2b\right)\)
\(\Leftrightarrow ac+2ad=ac+2bc\)
=>2ad=2bc
=>ad=bc
=>a/b=c/d
Đặt a/b=c/d=k
=>a=bk; c=dk
\(A=\dfrac{a^2\cdot d^2-4b^2\cdot c^2}{abcd}=\dfrac{b^2k^2\cdot d^2-4\cdot b^2\cdot d^2k^2}{bk\cdot b\cdot dk\cdot d}\)
\(=\dfrac{-3b^2k^2d^2}{b^2k^2d^2}=-3\)
bạn cứ đặt công thức gốc là k sau đó thay vào các câu là được thui
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 (1)
=> a=bk ,c=dk
a.Có \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a+c}{b+d}=\dfrac{a}{b}\left(=k\right)\)
b. Có \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\)
b,
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{d}=\dfrac{a}{c}=\dfrac{b+a}{d+c}\\ \Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có: \(a=bk;c=dk\)
\(\Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=\dfrac{k^2.\left(2b+3d\right)}{2b+3d}=k^2\\ \Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k^2.\left(2b-3d\right)}{2b-3d}=k^2\\ \Rightarrow\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e,
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
Ta có:\(a=bk;c=dk\)
\(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{k^2.\left(b-d\right)^2}{\left(b-d\right)^2}=k^2\\ \Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f,
(để hôm sau lm nha, mỏi tay quá)
a, \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)=> \(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)=\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)(1)
\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)=> \(\dfrac{a+b}{a-b}\)=\(\dfrac{c+d}{c-d}\)
Còn các phần còn lại làm giống thế
a: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}=\dfrac{a}{a-b}\)
b: \(\dfrac{a}{b}=\dfrac{bk}{b}=k\)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k=\dfrac{a}{b}\)
c \(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{k}{3k+1}\)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{k}{3k+1}=\dfrac{a}{3a+b}\)
d: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2=\dfrac{ac}{bd}\)
a,\(\dfrac{a-2b}{b}\)=\(\dfrac{c-2d}{d}\)
\(\Rightarrow\)\(\dfrac{a-b}{b}\)-1=\(\dfrac{c-d}{d}\)-1
\(\Rightarrow\)\(\dfrac{a-b}{b}\)=\(\dfrac{c-d}{d}\)
\(\Rightarrow\)\(\dfrac{a-2b}{b}\)=\(\dfrac{c-2d}{d}\)
Còn câu b dài quá mk sẽ ko chép lên.