cho \(\frac{a}{b}\)=\(\frac{c}{d}\)chứng minh \(\frac{7a^2+3ad}{11a^2-7a^2}\)=\(\frac{7c^2+3cd}{11c^2-7d^2}\)
cho a/b= c/d thì
chứng minh\(\frac{7a^2-3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
cho \(\frac{a}{b}=\frac{c}{d}\)Chứng minh
Đọc lại lý thuyết Bài 8 sgk/28
chỉ cần có lý thuyết a=k.b và c=k.d thay vào biểu thức là xong
Cho a/b=c/d,Chứng minh: \(\frac{7a^2+ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
\(dat\frac{a}{b}=\frac{c}{d}=k\)->> a=b.k va c=d.k
về phai
\(\frac{7.\left(d.k\right)^2+d.k.d}{11.\left(d.k\right)^2-8.d^2}=\frac{d^2.k\left(7k+1\right)}{d^2.\left(11k^2-8\right)}=\frac{k.\left(7k+1\right)}{11k^2-8}\)
ve trai
\(\frac{7.\left(b.k\right)^2+b.k.b}{11.\left(b.k\right)^2-8b^2}=\frac{b^2k.\left(7k+1\right)}{b^2.\left(11k^2-8\right)}=\frac{k.\left(7k+1\right)}{11k^2-8}\)
vậy vế trái = vế phải --> dpcm ( để bạn ghi có sai cho 3.cd k??_)
Cho a/b=c/d . Hãy chứng minh \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8b^2}\)
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:
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
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 \(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng minh rằng :
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
cho\(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
\(\frac{a}{b}=\frac{c}{d}\)
\(=>\frac{a}{c}=\frac{b}{d}\)
\(=>\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)
\(=>\frac{7a^2}{7c^2}=\frac{11a^2}{11c^2}=\frac{8b^2}{8d^2}=\frac{3ab}{3cd}\)
\(=>\frac{7a^2+3ab}{7c^2+cd}=\frac{11a^2-8b^2}{11c^2-8d^2}\)
\(=>\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)(ĐPCM)