Cho các số : a;b;c;d;e;f biết :
\(b^2=a.c\)
\(c^2=b.d\)
\(d^2=c.e\)
\(e=d.f\)
Chứng minh rằng : \(\dfrac{a}{f}=\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^2\)
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
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
Cho các số thực a,b,c,d,e thỏa mãn \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)chứng minh rằng: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)=\dfrac{a^2}{b.c}\)
Sửa: CMR: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\\ \Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3\left(1\right)\\ \dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\Rightarrow a=ck^2\\ \Rightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{ck\cdot c}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
1.Cho a,b,c,d,e,f \(\ne\) 0 thoả mãn : \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{e}{f}\)
Cmr:\(\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^5=\dfrac{a}{f}\) với (a+b+c+d+e+f \(\ne\)0)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{e}{f}=\dfrac{a+b+c+d+e}{b+c+d+e+f}=k\)
Ta có:
\(\dfrac{a}{f}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}.\dfrac{e}{f}=k^5=\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^5\)
cho 6 số a,b,c,d,e,g và
b^2=a.c
c^=b.d
d^2=c.e
e^2=d.g
cmr: a/e=(a+b+c+d+e)/(b+c+d+e+g
giúp nha,ai trả lời trước mình tích cho 2 bạn đầu tiên
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\) (1)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\) (2)
\(d^2=ce\Rightarrow\frac{c}{d}=\frac{d}{e}\) (3)
\(e^2=dg\Rightarrow\frac{d}{e}=\frac{e}{g}\) (4)
Từ (1),(2),(3),(4) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{g}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{g}=\frac{a+b+c+d+e}{b+c+d+e+g}\)
Ta có: \(\frac{a}{b}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (5)
\(\frac{b}{c}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (6)
\(\frac{c}{d}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (7)
\(\frac{d}{e}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (8)
\(\frac{e}{g}=\frac{a+b+c+d+e}{b+c+d+e+g}\) (9)
Nhân (5),(6),(7),(8),(9) vế với vế:
\(\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}\cdot\frac{d}{e}\cdot\frac{e}{g}=\frac{a}{g}=\left(\frac{a+b+c+d+e}{b+c+d+e+g}\right)^5\) (đpcm)
P/s: Mk nghĩ đề là c/m: a/g = (a+b+c+d+e/b+c+d+e+g)^5
Ta thấy A gồm có 99 số hạng nên ta nhóm mỗi nhóm 3 số hạng.
Ta có: A = 1 + 5 + 52 + 53 + 54 + 55 +...+ 597 + 598 + 599
= (1 + 5 + 52 )+ (53 + 54 + 55 )+...+( 597 + 598 + 599 )
=(1 + 5 + 52 )+ 53(1 + 5 + 52 ) +...+ 597(1 + 5 + 52 )
= ( 1 + 5 + 52)(1 + 53+....+597)
= 31(1 + 53+....+597)
Vì có một thừa số là 31 nên A chia hết cho 31.
P/s Đừng để ý câu trả lời của mình
tính giá trị biểu thức sau
a) \(A=\dfrac{25^6}{5^3}\)
b) \(B=32.\left(\dfrac{3}{2}\right)^5\)
c) \(C=\left(\dfrac{1}{3}\right)^4.3^{-3}\)
d) \(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4\)
e) \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4.25^2\)
f) \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)
g) \(G=\left(\dfrac{5}{3}\right)^3.\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)
a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)
c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)
d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)
\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)
\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)
e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)
\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)
\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)
f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)
\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)
\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)
g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)
\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)
\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)
\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)
\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)
\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)
\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)
\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)
\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)
\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)
Cho hình bình hành $A B C D$. Trên các tia $A D, A B$ lân lượt lây các điêm $F, E$ sao cho $A D=\dfrac{1}{2} A F, A B=\dfrac{1}{2} A E$. Chứng minh: a) Ba điểm $F, C, E$ thẳng hàng. b) Các tứ giác $B D C E, B D F C$ là hình bình hành.
Cho a,b,c,d,e là các số thực chứng minh rằng:
d) \(\dfrac{a^2+b^2}{2}>=\left(\dfrac{a+b}{2}\right)^2\)
e) \(\dfrac{a^2+b^2+c^2}{3}>=\left(\dfrac{a+b+c}{3}\right)^2\)
d) \(\dfrac{a^2+b^2}{2}\) \(\ge\) \(\left(\dfrac{a+b}{2}\right)^2\)
<=> \(\dfrac{a^2+b^2}{2}\) \(\ge\) \(\dfrac{a^2+2ab+b^2}{4}\)
<=> 4(a2 + b2 ) \(\ge\) 2 ( a2 + 2ab + b2 )
<=> 4a2 + 4b2 \(\ge\) 2a2 + 4ab +2b2
<=> 4a2 + 4b2 - 2a2 - 4ab - 2b2 \(\ge\) 0
<=> 2a2 - 4ab + 2b2 \(\ge\) 0
<=> a2 -2ab +b2 \(\ge\) 0
<=> (a-b)2 \(\ge\) 0 ( luôn đúng)
=> \(\dfrac{a^2+b^2}{2}\) \(\ge\) \(\left(\dfrac{a+b}{2}\right)^2\)
Và dấu bằng xảy ra <=> a = b
e) Làm tương tự nhé! Có gì ko hiểu thì hỏi lại mk! Ok??
cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{e}{f}\) chứng minh rằng \(\dfrac{a^4}{b^4}=\dfrac{a}{f}\)
Ta có: \(\dfrac{a^4}{b^4}=\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{a}{b}\)
\(=\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{d}\cdot\dfrac{e}{f}\)
\(=\dfrac{a}{f}\)
Cho a,b,c,d,e,f là các số dương. CMR:
\(\sqrt{\left(a+b+c\right)^2+\left(d+e+f\right)^2}\le\sqrt{a^2+d^2}+\sqrt{b^2+e^2}+\sqrt{c^2+f^2}\)
Mincopxki
\(\sqrt{a^2+d^2}+\sqrt{b^2+e^2}+\sqrt{c^2+f^2}\ge\sqrt{\left(a+b\right)^2+\left(d+e\right)^2}+\sqrt{c^2+f^2}\ge\sqrt{\left(a+b+c\right)^2+\left(d+e+f\right)^2}\)