Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\Rightarrow\frac{a+b+c}{b+c+d}=k\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=k^3\)
\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b+c}{b+c+d}\right)^3\left(=k^3\right)\)
Gợi ý :
Bước 1 : Cộng 6 vào các hạng tử đã cho ở đề bài
Bước 2 : xét 2 TH :
TH1 : a + b + c = 0
TH2 : a + b + c khác 0
Chúc học tốt !!!!
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
Bài 1
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Ta có:
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(1\right)\)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\left(đpcm\right)\)
Vậy .....
Bài 2
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Leftrightarrow\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
\(\Leftrightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(đpcm\right)\)
Vậy .....
Chúc bạn học tốt!
Ta có: \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{b+a+d}=\frac{d}{c+b+a}\)
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{b+a+d}+1=\frac{d}{c+b+a}+1\)
\(\Rightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{b+a+d}=\frac{a+b+c+d}{c+b+a}\)
Mà a+b+c+d khác 0
=> b+c+d = a+c+d = b+a+d = c+b+a
=> b = a = c = d
Ta có:
\(P=\frac{2a+5b}{3c+4d}-\frac{2b+5c}{3d+4a}-\frac{2c+5d}{3a+4b}-\frac{2d+5a}{3c+4b}\)
\(P=\frac{2a+5a}{3a+4a}-\frac{2b+5b}{3b+4b}-\frac{2c+5d}{3c+4c}-\frac{2d+5d}{3d+4d}\)
\(P=\frac{7a}{7a}-\frac{7b}{7b}-\frac{7c}{7c}-\frac{7d}{7d}\)
\(P=1-1-1-1=-2\)
- Bất đẳng thức tam giác: |x|+|y|≥|x+y|the absolute value of x end-absolute-value plus the absolute value of y end-absolute-value is greater than or equal to the absolute value of x plus y end-absolute-value|𝑥|+|𝑦|≥|𝑥+𝑦|.
- Bất đẳng thức tam giác mở rộng: |x|+|y|+|z|≥|x+y+z|the absolute value of x end-absolute-value plus the absolute value of y end-absolute-value plus the absolute value of z end-absolute-value is greater than or equal to the absolute value of x plus y plus z end-absolute-value|𝑥|+|𝑦|+|𝑧|≥|𝑥+𝑦+𝑧|.
Cách giải Sử dụng bất đẳng thức tam giác để chứng minh từng phần của bất đẳng thức.- Bước 1 . Chứng minh |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13|5a+4b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction the absolute value of 5 a plus 4 b end-absolute-value|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13|5𝑎+4𝑏|.
- Ta có 3(|a|+|b|+|a+b|)≥|3a+3b|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value close paren is greater than or equal to the absolute value of 3 a plus 3 b end-absolute-value3(|𝑎|+|𝑏|+|𝑎+𝑏|)≥|3𝑎+3𝑏|.
- Ta có 3(|a|+|b|+|c|+|a+b|+|b+c|+|c+a|)≥|3a+3b+3c+3a+3b+3c|=|6a+6b+6c|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value close paren is greater than or equal to the absolute value of 3 a plus 3 b plus 3 c plus 3 a plus 3 b plus 3 c end-absolute-value equals the absolute value of 6 a plus 6 b plus 6 c end-absolute-value3(|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|)≥|3𝑎+3𝑏+3𝑐+3𝑎+3𝑏+3𝑐|=|6𝑎+6𝑏+6𝑐|.
- Áp dụng bất đẳng thức tam giác:
- |a|+|a|+|a|+|b|+|b|≥|3a+2b|the absolute value of a end-absolute-value plus the absolute value of a end-absolute-value plus the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of b end-absolute-value is greater than or equal to the absolute value of 3 a plus 2 b end-absolute-value|𝑎|+|𝑎|+|𝑎|+|𝑏|+|𝑏|≥|3𝑎+2𝑏|.
- |a|+|b|+|a+b|≥|2a+2b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value is greater than or equal to the absolute value of 2 a plus 2 b end-absolute-value|𝑎|+|𝑏|+|𝑎+𝑏|≥|2𝑎+2𝑏|.
- Xét |5a+4b|=|(a+b)+(a+b)+(a+b)+a+a+b|the absolute value of 5 a plus 4 b end-absolute-value equals the absolute value of open paren a plus b close paren plus open paren a plus b close paren plus open paren a plus b close paren plus a plus a plus b end-absolute-value|5𝑎+4𝑏|=|(𝑎+𝑏)+(𝑎+𝑏)+(𝑎+𝑏)+𝑎+𝑎+𝑏|.
- Ta có |5a+4b|=|(a+b)+(a+b)+(a+b)+a+a+b|the absolute value of 5 a plus 4 b end-absolute-value equals the absolute value of open paren a plus b close paren plus open paren a plus b close paren plus open paren a plus b close paren plus a plus a plus b end-absolute-value|5𝑎+4𝑏|=|(𝑎+𝑏)+(𝑎+𝑏)+(𝑎+𝑏)+𝑎+𝑎+𝑏|.
- Áp dụng bất đẳng thức tam giác:
- |a|+|b|+|a+b|≥|2a+2b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value is greater than or equal to the absolute value of 2 a plus 2 b end-absolute-value|𝑎|+|𝑏|+|𝑎+𝑏|≥|2𝑎+2𝑏|.
- |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13|5a+4b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction the absolute value of 5 a plus 4 b end-absolute-value|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13|5𝑎+4𝑏|.
- Ta có 3(|a|+|b|+|a+b|)≥|3a+3b|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value close paren is greater than or equal to the absolute value of 3 a plus 3 b end-absolute-value3(|𝑎|+|𝑏|+|𝑎+𝑏|)≥|3𝑎+3𝑏|.
- Ta có 3(|a|+|b|+|c|+|a+b|+|b+c|+|c+a|)≥|5a+4b|+|5b+4c|+|5c+4a|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value close paren is greater than or equal to the absolute value of 5 a plus 4 b end-absolute-value plus the absolute value of 5 b plus 4 c end-absolute-value plus the absolute value of 5 c plus 4 a end-absolute-value3(|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|)≥|5𝑎+4𝑏|+|5𝑏+4𝑐|+|5𝑐+4𝑎|.
- Ta có |a|+|b|+|a+b|≥|2a+2b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value is greater than or equal to the absolute value of 2 a plus 2 b end-absolute-value|𝑎|+|𝑏|+|𝑎+𝑏|≥|2𝑎+2𝑏|.
- Ta có |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13(|5a+4b|+|5b+4c|+|5c+4a|)the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction open paren the absolute value of 5 a plus 4 b end-absolute-value plus the absolute value of 5 b plus 4 c end-absolute-value plus the absolute value of 5 c plus 4 a end-absolute-value close paren|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13(|5𝑎+4𝑏|+|5𝑏+4𝑐|+|5𝑐+4𝑎|).
Lời giải Bất đẳng thức cần chứng minh là |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13(|5a+4b|+|5b+4c|+|5c+4a|)the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction open paren the absolute value of 5 a plus 4 b end-absolute-value plus the absolute value of 5 b plus 4 c end-absolute-value plus the absolute value of 5 c plus 4 a end-absolute-value close paren|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13(|5𝑎+4𝑏|+|5𝑏+4𝑐|+|5𝑐+4𝑎|).