CMR nếu có (a+b+c+d)(a−b−c+d)=(a−b+c−d)(a+b−c−d) thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
a) Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) CMR: \(\dfrac{5a+3b}{5a-3b}\)=\(\dfrac{5c+3d}{5c-3d}\)
b) CMR: Nếu \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) thì : \(\dfrac{a}{b}\)=\(\dfrac{3a+2c}{3b+2d}\)
c) CMR: Nếu \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) thì \(\dfrac{7a^2+3ab}{11a^2-8b^2}\) = \(\dfrac{7c^2+3cd}{11c^{2^{ }}-8d^2}\)
\(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
\(\dfrac{a}{c}\) = \(\dfrac{b}{d}\)
\(\dfrac{a}{c}\) = \(\dfrac{5a}{5c}\) = \(\dfrac{3b}{3d}\) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{c}\) = \(\dfrac{5a+3b}{5c+3d}\) (1)
\(\dfrac{a}{c}\) = \(\dfrac{5a-3b}{5c-3d}\) (2)
Kết hợp (1) và (2) ta có:
\(\dfrac{5a+3b}{5c+3d}\) = \(\dfrac{5a-3b}{5c-3d}\)
⇒ \(\dfrac{5a+3b}{5a-3b}\) = \(\dfrac{5c+3d}{5c-3d}\) (đpcm)
b; \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\)
\(\dfrac{a}{b}\) = \(\dfrac{3a}{3b}\) = \(\dfrac{2c}{2d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}\) = \(\dfrac{3a+2c}{3b+2d}\) (đpcm)
Cho hai số hữu tỉ \(\dfrac{a}{b}\) và \(\dfrac{c}{d}\)(a,b,c,d ϵ Z, b,d ≠ 0) Chứng tỏ rằng:
a, Nếu \(\dfrac{a}{b}\) < \(\dfrac{c}{d}\) thì ad < bc
b, Nếu ad < bc thì \(\dfrac{a}{b}\) < \(\dfrac{c}{d}\)
Cho a, b, c, d > 0. CMR:
Nếu \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
Áp dụng, chứng minh BĐT sau:
a) \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
b) \(1< \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< 2\)
c) \(2< \dfrac{a+b}{a+b+c}+\dfrac{b+c}{b+c+d}+\dfrac{c+d}{c+d+a}+\dfrac{d+a}{d+a+b}< 3\)
\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow ab+ac< ba+bc\)
\(\Leftrightarrow ac< bc\)
\(\Leftrightarrow a< b\)(đúng)
a)Áp dụng
\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=2\left(1\right)\)
Lại có:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{b+c+a}+\dfrac{c}{c+a+b}=1\left(2\right)\)
Từ (1) và (2)=> đpcm
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow ac< bc\Rightarrow ac+ab< bc+ab\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow\dfrac{a\left(b+c\right)}{b\left(b+c\right)}< \dfrac{b\left(a+c\right)}{b\left(b+c\right)}\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)a) ta có
\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
b)
\(\dfrac{a}{a+b+c+d}+\dfrac{b}{b+c+d+a}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}< \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< \dfrac{a+d}{a+b+c+d}+\dfrac{a+b}{a+b+c+d}+\dfrac{b +c}{a+b+c+d}+\dfrac{d+c}{a+b+c+d}\)
\(\Leftrightarrow\dfrac{a+b+c+d}{a+b+c+d}< \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< \dfrac{2\left(a+b+c+d\right)}{a+b+c+d}\)\(\Leftrightarrow1< \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< 2\)
Chứng minh rằng nếu:
(a + b + c + d) (a - b - c + d) = (a - b + c - d) (a + b - c - d)
thì\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
(a, b, c, d khác 0)
Ta có: \(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)
\(\Leftrightarrow\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)
\(\Leftrightarrow\left(a+d-a+d\right)\left(a+d+a-d\right)=\left(b+c-b+c\right)\left(b+c+b-c\right)\)
\(\Leftrightarrow2d\cdot2a=2c\cdot2b\)
\(\Leftrightarrow ad=bc\)
hay \(\dfrac{a}{c}=\dfrac{b}{d}\)
Giúp mk với!Thank
Chứng minh nếu \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) thì
\(\dfrac{a}{b}\)=\(\dfrac{a+c}{b+d}\) và \(\dfrac{a}{c}\)=\(\dfrac{a-b}{c-d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
hay \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
hay \(\dfrac{a}{c}=\dfrac{a-b}{c-d}\)
Cho a,b,c,d>0. CMR :\(1< \dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}< 2\)
\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\)
Làm tương tự với 3 phân số còn lại và cộng vế với vế
\(\dfrac{a}{a+b+c}< \dfrac{a+d}{a+b+c+d}\)
Làm tương tự với 3 phân số còn lại và cộng vế với vế
Cho a,b,c,d>0. CMR: 1 <\(\dfrac{a}{a+b+c}\)+\(\dfrac{b}{b+c+d}\)+\(\dfrac{c}{c+d+a}\)+\(\dfrac{d}{d+a+b}\)< 2
Chứng minh rằng: Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
Chứng minh rằng : Nếu \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\) thì
a.\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\) b.\(\dfrac{a}{b}\)=\(\dfrac{a+c}{b+c}\) c.\(\dfrac{a}{c}\)=\(\dfrac{a-b}{c-d}\) d.\(\dfrac{a+b}{c+d}\)=\(\dfrac{a-b}{c-d}\)
a: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
d: Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)
hay \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
nên \(\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)
hay \(\dfrac{a}{c}=\dfrac{a-b}{c-d}\)
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 \(\dfrac{a}{b}\)<\(\dfrac{c}{d}\)
a) \(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow\dfrac{a}{b}-\dfrac{c}{d}< 0\Leftrightarrow\dfrac{ad-bc}{bd}< 0\)\(\Leftrightarrow ad-bc< 0\) ( do bc>0) \(\Leftrightarrow ad< bc\) (đpcm)
b) \(ad< bc\) \(\Leftrightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\) \(\Leftrightarrow\dfrac{a}{b}< \dfrac{c}{d}\)(đpcm)