gpt: 11a+8b=24+3ab
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Đề thiếu rồi bạn nhé. Bạn tham khảo ở đây.
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7^a+3ab/11a^2-8b^2=7c^2+3cd/11c^2-8d^2
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chứng minh
7a2+3ab/11a2-8b2=7c2+3cb/11a2-8d2
\(Cho\)\(\dfrac{a}{b}\)
\(Chứng\) \(Minh\)
\(\dfrac{7a^2+3ab}{11a^2-8b}\)\(=\)\(\dfrac{7c^2+3cd}{11c^2-8d}\)
Cho \(\dfrac{a}{b}\) như thế nào thì mới chứng minh được chứ em
cho a/b=c/d chứng minh 7a^2+3ab/11a^2-8b^2=7c^2+3cd/11c^2-8d^2
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3\cdot bk\cdot b}{11\cdot b^2k^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{7k^2+3k}{11k^2-8}\)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3\cdot dk\cdot d}{11d^2k^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Cho a/b=c/d . Hãy chứng minh \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8b^2}\)
chung minh rang 7a^2 + 3ab / 11a^2 -8b^2 = 7c^2 + 3cd / 11c^2 - 8d^2
Đề bài : Cho \(\frac{a}{c}=\frac{b}{d}.CMR:...\)
Ta có :
\(\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)
\(\Rightarrow\frac{7a^2}{7c^2}=\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}\)
ADTCDTSBN , ta có :
\(\frac{7a^2}{7c^2}=\frac{3ab}{3cd}=\frac{7a^2+3ab}{7c^2+3cd}\left(1\right)\)
\(\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{11a^2-8b^2}{11c^2-8d^2}\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\frac{7a^2+3ab}{7c^2+3cd}=\frac{11a^2-8b^2}{11c^2-8d^2}\)
\(\Rightarrow\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\left(đpcm\right)\)
Hẹp me
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) Chứng minh:
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT:\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\\ VP:\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\\ \Rightarrow VT=VP\\ \Rightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)
Ta có:
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7\left(kb\right)^2+3\left(kb\right).b}{11\left(kb\right)^2-8b^2}=\dfrac{7k^2+3k}{11k^2-8}\) (1)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(kd\right)^2+3\left(kd\right)d}{11\left(kd\right)^2-8d^2}=\dfrac{7k^2+3k}{11k^2-8}\) (2)
(1),(2) \(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
cho \(\dfrac{a}{b}=\dfrac{c}{d}\) Chứng minh rằng
\(\dfrac{7a^2+3ab}{11a^2+8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)