Cho \(\frac{a}{b}=\frac{c}{d}\)Chứng minh \(\frac{3a+7b}{3a-5b}=\frac{3c+7d}{3c-5d}\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\) chứng minh \(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
Ta có:
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)
cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh:\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
Cho tỉ lệ thức : \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh
\(\frac{3a+5b}{3a-5b}\)= \(\frac{3c+5d}{3c-5d}\)
ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{d}{b}=\frac{c}{a}\Rightarrow\frac{c+d}{a+b}\Rightarrow\frac{3c+3d}{3a+3b}=\frac{3c-3d}{3a-3b}\)
\(\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)\(\left(điềuphảichứngminh\right)\)
cho \(\frac{a}{b}=\frac{c}{d}cmr\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{3a+5b}{3a-5b}=\frac{3bk+5b}{3bk-5b}=\frac{b\left(3k+5\right)}{b\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
\(\frac{3c+5d}{3c-5d}=\frac{3dk+5d}{3dk-5d}=\frac{d\left(3k+5\right)}{d\left(3k-5\right)}=\frac{3k+5}{3k-5}\)
Vậy từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
cho tỷ lệ thức a/b=c/d. chứng minh:
a, 2a+5b/3a-4b=2c+5d/3c-4d
b. 3a+7b/5a-7b=3c+7d/5c-7d
d. 4a+9b/4a-7b=4c+9d/4c-7d
giúp mình với ạ
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh:
1) \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2) \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3) \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4) \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\). Chứng minh
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
1.Cho tỉ lệ thức : \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng:
a,\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b,\(\frac{a}{b}=\frac{4a+7c}{4b+7d}\)
2. 7A,7B,7C có 130 học sinh cùng tham gia trồng cây. Mỗi học sinh lớp 7A trồng được 2 cây,7B trồng được 3 cây,7C trồng được 4 cây. Hỏi mỗi lớp có bao nhiêu bạn học sinh tham gia biết rằng số cây trồng đc của ba lớp bằng nhau.
giúp minh với!
1/
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{3a}{3c}=\frac{5b}{5d}=\frac{3a+5b}{3c+5d}=\frac{3a-5b}{3c-5d}\Rightarrow\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b,\(\frac{a}{b}=\frac{c}{d}=\frac{4a}{4b}=\frac{7c}{7d}=\frac{4a+7c}{4b+7d}\)
2/
Gọi số học sinh tham gia của mỗi lớp lần lượt là a,b,c
Ta có: \(2a=3b=4c\)
\(\Rightarrow\frac{2a}{12}=\frac{3b}{12}=\frac{4c}{12}\Rightarrow\frac{a}{6}=\frac{b}{4}=\frac{c}{3}=\frac{a+b+c}{6+4+3}=\frac{130}{13}=10\)
=> a/6 = 10 => a = 60
b/4 = 10 => b = 40
c/3 = 10 => c = 30
Vậy số học sinh mỗi lớp lần lượt là 60 hs, 40 hs, 30hs
cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\).chứng minh
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b) \(\frac{2a+3b}{2a-3b}\)=\(\frac{2c+3d}{2c-3d}\)