Biế rằng \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\).CMR \(\frac{a}{d}=\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}\)
cho b2= ac; c2=bd với b,c,d khác 0; b+c khác d, b3+c3 khác d3
chứng minh rằng \(\frac{a}{b}\)=\(\frac{b}{c}\)=\(\frac{c}{d}\) và \(\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}\)=\(\frac{a}{d}\)
ơ kìa đăng 2 hôm r ko ai giúp
Cho b2 = ac; c2 = bd. Chứng minh rằng:
a,\(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)
b,\(\frac{3a^2+5b^4-7c^6}{3b^2+5c^4-7d^6}=\frac{2a^3+4b^5-6c^7}{2b^3+4c^5-6d^7}\)
Giúp mk nha, thứ 3 mình nộp ùi
a/ Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=k^3\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Áp dụng tính chất của tỉ lệ thức ta có:\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=k^3\)
Mặt khác: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(=k^3\right)\)
giup minh nha: Tinh nhanh lop 4
42 x 43 - 12 x 9 - 42 x 3
Cho b2 = a.c; c2 = b.d
Chứng minh rằng \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)
\(\frac{3a^2+5b^4-7c^6}{3b^2+5c^4-7d^6}=\frac{2a^3+4b^5-6c^7}{2b^3+4c^5-6d^7}\)
cho \(\frac{a}{b}\)=\(\frac{c}{d}\) CMR
A)\(\frac{3a-2c}{3b+2d}\)= \(\frac{2a+5c}{2b+5d}\)
B)\(\frac{a^3}{b^3}\) =\(\frac{\left(a+c\right)^3}{\left(b+d\right)^5}\)
Đặt
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
a)
Sửa đề nhá :
\(\frac{3a+2c}{3b+2d}=\frac{3bk+2dk}{3b+2d}=\frac{k\left(3b+2d\right)}{\left(3b+2d\right)}=k\)(1)
\(\frac{2a+5c}{2b+5d}=\frac{2bk+5dk}{2b+5d}=\frac{k\left(2b+5d\right)}{2b+5d}=k\)(2)
Từ (1) và (2)
=> \(\frac{3a+2c}{3b+2d}=\frac{2a+5c}{2b+5d}\)
b)
\(\frac{a^3}{b^3}=\frac{b^3k^3}{b^3}=k^3\)(3)
\(\frac{\left(a+c\right)^3}{\left(b+d\right)^3}=\frac{\left(bk+dk\right)^3}{\left(b+d\right)^3}=\frac{k^3\left(b+d\right)^3}{\left(b+d\right)^3}=k^3\)(4)
Từ (3) và (4)
=> \(\frac{a^3}{b^3}=\frac{\left(a+c\right)^3}{\left(b+d\right)^3}\)
cho \(\frac{a}{b}=\frac{c}{d}\): CMR :
\(a,\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\)
\(b,\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\Leftrightarrow\frac{4a}{4c}=\frac{3b}{3d}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{4a}{4c}=\frac{3b}{3d}=\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\)
Vậy \(\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\left(ĐPCM\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\left(k\ne0\right)\)
a) \(\frac{4a-3b}{4c-3d}=\frac{4bk-3b}{4dk-3d}=\frac{b.\left(4k-3\right)}{d.\left(4k-3\right)}=\frac{b}{d}\)
\(\frac{4a+3b}{4c+3d}=\frac{4bk+3b}{4dk+3d}=\frac{b.\left(4k+3\right)}{d.\left(4k+3\right)}=\frac{b}{d}\)
\(\Rightarrow\frac{4a-3b}{4c-3d}=\frac{4a+3b}{4c+3d}\)
b) \(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3k^3+b^3}{d^3k^3+d^3}=\frac{b^3.\left(k^3+1\right)}{d^3.\left(k^3+1\right)}=\frac{b^3}{d^3}\)
\(\frac{a^3-b^3}{c^3-d^3}=\frac{\left(bk\right)^3-b^3}{\left(dk\right)^3-d^3}=\frac{b^3k^3-b^3}{d^3k^3-d^3}=\frac{b^3.\left(k^3-1\right)}{d^3.\left(k^3-1\right)}=\frac{b^3}{d^3}\)
\(\Rightarrow\frac{a^3+b^3}{c^3+d^3}=\frac{a^3-b^3}{c^3-d^3}\)
Cho a/b = c/d với a, b, c, d khác 0. Chứng minh rằng : \(\frac{3a-5c}{4a+7c}=\frac{3b-5d}{4b+7d}\)
Gọi \(\frac{a}{b}=\frac{c}{d}=x\Rightarrow a=bx;c=dx\)
Thay vào vế trái ta được
\(\frac{3a-5c}{4a+7c}=\frac{3.bx-5.dx}{4.bx+7.dx}=\frac{x\left(3b-5d\right)}{x\left(4b+7d\right)}=\frac{3b-5d}{4b+7d}\)
Vậy vế trái bằng vế phải
Ta có:\(\frac{a}{b}=\frac{c}{d}=\frac{3a-5c}{3b-5d}\left(1\right)\)
Ta lại có:\(\frac{a}{b}=\frac{c}{d}=>\frac{4a+7c}{4b+7d}\left(2\right)\)
Từ (1) và (2),suy ra : \(\frac{3a-5c}{4a+7c}=\frac{3b-5d}{4b+7d}\)
Cách của mình cũng đúng nhưng khác cách làm của thang Tam thôi
Cho b^2=ac; c^2=bd với b,c,d khác 0; b+c khác d, b^3+c^3 khác d^3Chứng mỉnh rằng a/b=b/c=c/d và 3a^3-4b^3+5c^3/3b^3-4c^3+5d^3=a/d
giúp ;-;
Tính \(C=\frac{2.a}{3.b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}\) biết \(\frac{2.a}{3.b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\)
Đặt: \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=k\)
=> 2a = k .3b; 3b = k. 4c; 4c = k. 5d; 5d = k.2a
Mà \(1=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=\frac{k.3b+k4c+k.5d+k.2a}{3b+4c+5d+2a}=\frac{k.\left(3b+4c+5d+2a\right)}{3b+4c+5d+2a}=k\)
=> C = 1+1+1+1 = 4
Cho các số thực dương a,b,c. Chứng minh rằng :
\(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\)< \(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\)
Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:
\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)
\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)
\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)
Cộng (1),(2) và (3) có:
\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)
\(\Rightarrow2VP\ge2VT\)
\(\RightarrowĐPCM\)