Bài 7: Tỉ lệ thức
Các câu hỏi tương tự
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\)và b, d khác 0. CMR \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Cho dfrac{a}{b}dfrac{c}{d} chứng minh rằng:
a) dfrac{a}{a-b}dfrac{c}{c-d}
b) dfrac{a}{b}dfrac{a+c}{b+d}
c)dfrac{a}{3a+b}dfrac{c}{3c+b}
d) dfrac{a.c}{b.c}dfrac{a^2+c^2}{b^2+d^2}
e) dfrac{a.b}{c.d}dfrac{a^2-b^2}{c^2-d^2}
f) dfrac{a.b}{c.d}dfrac{left(a-bright)^2}{left(c-dright)^2}
Đọc tiếp
Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) chứng minh rằng:
a) \(\dfrac{a}{a-b}\)=\(\dfrac{c}{c-d}\)
b) \(\dfrac{a}{b}\)=\(\dfrac{a+c}{b+d}\)
c)\(\dfrac{a}{3a+b}\)=\(\dfrac{c}{3c+b}\)
d) \(\dfrac{a.c}{b.c}\)=\(\dfrac{a^2+c^2}{b^2+d^2}\)
e) \(\dfrac{a.b}{c.d}\)=\(\dfrac{a^2-b^2}{c^2-d^2}\)
f) \(\dfrac{a.b}{c.d}\)=\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
30 tháng 10 2021 lúc 21:49
Cho a, b, c, d là 4 số khác 0 thỏa mãn \(b^2\) = ac; \(c^2\) = bd và \(b^3+c^3+d^3\ne0\)
Chứng minh rằng: \(\dfrac{a}{d}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
Biết \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\) với a,b,c,d khác 0. Chứng minh rằng: \(\dfrac{a}{b}=\dfrac{c}{d}\) hoặc \(\dfrac{a}{b}=\dfrac{d}{c}\)
cho dfrac{a}{b} dfrac{c}{d} cm rằng a) dfrac{a}{a-b} dfrac{c}{c-d} b)dfrac{a}{b} dfrac{a+c}{b+d} c) dfrac{a}{3a+d} dfrac{c}{3c+d} d)dfrac{a.c}{b.d} dfrac{a^2+c^2}{b^2+c^2} e)dfrac{a.b}{c.d} dfrac{a^2-b^2}{c^2-d^2} f)dfrac{a.b}{c.d} dfrac{left(a-bright)^2}{left(c-dright)^2} mn giúp mk vs ạ! thanks
Đọc tiếp
cho \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\) cm rằng
a) \(\dfrac{a}{a-b}\) =\(\dfrac{c}{c-d}\) b)\(\dfrac{a}{b}\) =\(\dfrac{a+c}{b+d}\) c) \(\dfrac{a}{3a+d}\) =\(\dfrac{c}{3c+d}\) d)\(\dfrac{a.c}{b.d}\) =\(\dfrac{a^2+c^2}{b^2+c^2}\) e)\(\dfrac{a.b}{c.d}\) =\(\dfrac{a^2-b^2}{c^2-d^2}\) f)\(\dfrac{a.b}{c.d}\) =\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
mn giúp mk vs ạ! thanks
Bài 1 Cho \(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}\left(b\ne0\right)\) CMR \(c=0\)
Bài 2 Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}CMR\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
\(Cho\dfrac{a}{b}=\dfrac{c}{d}CMR\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2-b^2}{c^2-d^2}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
CMR \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\) và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Chứng minh rằng từ tỉ lệ thức dfrac{a}{b}dfrac{c}{d} (b, d ≠ 0) ta suy ra được các tỉ lệ thức:
a/ dfrac{a+b}{a-b}dfrac{c+d}{c-d}
b/ dfrac{a}{a+b}dfrac{c}{c+d}
c/ dfrac{2a+3c}{2b+3d}dfrac{2a-3c}{2b-3d}
d/ dfrac{a^2+c^2}{b^2+d^2}dfrac{ac}{bd}
e/ dfrac{a^2+c^2}{b^2+d^2}dfrac{a^2-c^2}{b^2-d^2}
f/ dfrac{left(a-bright)^2}{left(c-dright)^2}dfrac{a^2+b^2}{c^2+d^2}
Đọc tiếp
Chứng minh rằng từ tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) (b, d ≠ 0) ta suy ra được các tỉ lệ thức:
a/ \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b/ \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
c/ \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
d/ \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
e/ \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{a^2-c^2}{b^2-d^2}\)
f/ \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)