Bài 7: Tỉ lệ thức
Các câu hỏi tương tự
Cho \(\dfrac{\text{a}}{b}=\dfrac{c}{d}.CM\)
\(\dfrac{3\text{a}+5b}{3\text{a}-5b}=\dfrac{3c+5d}{3c-5d}\)
\(\left(\dfrac{\text{a}+b}{c+d}\right)^2=\dfrac{\text{a}^2+b^2}{c^2+d^2}\)
\(\dfrac{\text{a}}{b}=\dfrac{b}{c}=\dfrac{c}{d}.CM\left(\dfrac{\text{a}+b+c}{b+c+d}\right)^3=\dfrac{\text{a}}{d}\)
cho tỉ lệ thức \(\dfrac{a}{b}\)chung minh \(\dfrac{a}{a-b}=\dfrac{a}{c-d}\)(giả thiet a khac b ,c khac d va a,b,c khac 0
Cho a^2+b^2tat ca/c^2+d^2 =ab/cd
va a,b,c,d khac 0
cm a/b=c/d hoac a/b=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
Cho tỉ lệ thức : \(\dfrac{a}{b}\) =\(\dfrac{c}{d}\). CMR ta có:
\(\dfrac{\text{2002a+2003b}}{\text{2002a - 2003b}}\) = \(\dfrac{\text{2002c+2003d}}{\text{2002c−2003d}}\)
1. Cho 2 số hữu tỉ dfrac{a}{b} và dfrac{c}{d} ( b 0, d 0 ). Chứng tỏ rằng:
a) Nếu dfrac{a}{b} dfrac{c}{d} thì ad bc
b) Nếu ad bc thì dfrac{a}{b} dfrac{c}{d}
2. Chứng tỏ rằng nếu dfrac{a}{b} dfrac{c}{d} ( b 0, d 0 ) thì dfrac{a}{b} dfrac{a+c}{b+d} dfrac{c}{d}
Đọc tiếp
1. Cho 2 số hữu tỉ \(\dfrac{a}{b}\) và \(\dfrac{c}{d}\) ( b > 0, d > 0 ). Chứng tỏ rằng:
a) Nếu \(\dfrac{a}{b}< \dfrac{c}{d}\) thì ad < bc
b) Nếu ad < bc thì \(\dfrac{a}{b}< \dfrac{c}{d}\)
2. Chứng tỏ rằng nếu \(\dfrac{a}{b}< \dfrac{c}{d}\) ( b > 0, d > 0 ) thì \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
cho a,b,c,x,y,z là các số thực khác 0 thỏa mãn: \(\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\). CMR: \(\dfrac{a^2-bc}{x}=\dfrac{b^2-ca}{y}=\dfrac{c^2-ab}{z}\)
Cho \(\dfrac{a}{c}=\dfrac{c}{b}\) với a, b, c ≠ 0. Chứng minh rằng: \(\dfrac{b-a}{a}=\dfrac{b^2-a^2}{a^2+c^2}\).