Cho \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng : \(\frac{7a^2+3ab}{2a^2-ab}=\frac{7c^2+3cd}{2c^2-cd}\)
GIÚP MK VS MN !!!
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng
\(1.\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\) \(2.\frac{2a+b}{2a-b}=\frac{2c+d}{2c-d}\) \(3.\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\) \(4.\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho\(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng:
a)\(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
b)\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
c)\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
a)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)(T/C...)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
b)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{c}{d}\right)^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)(T/C...)
\(\Rightarrow\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
c)\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{7a^2}{7c^2}=\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}\)
\(\Rightarrow\frac{7a^2}{7c^2}=\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}=\frac{7a^2+3ab}{7c^2+3cd}=\frac{11a^2-8b^2}{11c^2-8d^2}\)
\(\Rightarrow\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)
Cho \(\frac{a}{b}=\frac{c}{d}\), chứng minh rằng
a) \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
b) \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
bạn ơi viết phân số như bạn thế nào rồi mk giải cho
Nguyễn Thị Thu Huyền ơi , viết phân số = dấu /
a) \(\frac{a}{b}=\frac{c}{d}=k\) => a=bk;c=dk
\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\) (1)
(2)\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2}{d^2}\)
từ (1) và (2) suy ra \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
cho \(\frac{a}{b}\) = \(\frac{c}{d}\) chứng minh rằng :\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
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Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\cdot b^2k^2+3\cdot bk\cdot b}{11\cdot b^2\cdot k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\cdot d^2k^2+3dk\cdot d}{11\cdot d^2k^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)
Do đó: VT=VP(đpcm)
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng : (a+2015c)(b+d)=(a+c)(b+2015d)
Cho \(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng: \(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)
đg cần gấp nka m.ng záng zúp mk
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
\(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2.\left(bk\right)^2-3.bk.b+5.b^2}{2b^2+3.bk.b}\)=\(\frac{2.b^2.k^2-3.k.b^2+5.b^2}{2.b^2+3.b^2.k}=\frac{b^2\left(2.k^2-3.k+5\right)}{b^2\left(2+3.k\right)}=\frac{2.k^2-3.k+5}{2+3.k}\)
\(\frac{2c^2-3cd+5d^2}{2d^2+3cd}=\frac{2.\left(dk\right)^2-3.dk.d+5.d^2}{2.d^2+3.dk.d}\)\(=\frac{2.d^2.k^2-3.d^2.k+5.d^2}{2.d^2+3.d.k.d}\)=\(\frac{d^2\left(2.k^2-3.k+5\right)}{d^2\left(2+3.k\right)}=\frac{2.k^2-3.k+5}{2+3.k}\)
=> bằng nhau
Bằng nhau pạn nhé. Mjk ko tjện giảj nha pạn tự làm nha.
Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\). Chứng minh: \(\frac{2a^2-3ab+5b^2}{2b^2+3ab}=\frac{2c^2-3cd+5d^2}{2d^2+3cd}\)
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1.chứng minh \(\frac{a}{b}=\frac{c}{d}thì\frac{7a^2+3ab}{7c^2+3cd}=\frac{7a^2+3cd}{7c^2+8d^2}\)
2.\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
chứng minh\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
1. xem lại đề bài nhé bạn.
2. nhân cả tử và mẫu với lần lượt a, b, c
sau đó sẽ nhận thấy chúng bằng nhau
Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng:
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)
\(=\frac{11a^2}{11c^2}=\frac{7a^2}{7c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}=\frac{7a^2+3ab}{7c^2+3cd}=\frac{11a^2-8b^2}{11c^2-8d^2}\)
\(\Rightarrow\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Ta có: a/b=c/d => a2/d2=>a2/c2=b2/d2=ab/cd
=11a2/11c2=7a2/7c2=8b2/8d2=3ab/3cd=7a2+3ab/7c2+3cd=11a2-8b2/11c2-8d2
=>7a2+3ab/11a2-8b2=7c2+3cd/11c2-8d2
Hàng ta có cái cuối sai r
tỉ lệ thức không có nhân
Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng :
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)