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cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)(b,d>0). Chứng minh rằng: \(\frac{\left(a-b\right)^{2012}}{\left(c-d\right)^{2012}}=\frac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}\)
giúp mình nhanh nha
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(=> a=k\)x\(b\)
\(c=k\)x\(d\)
Rồi thay vào sẽ làm ra
CHÚC BẠN HOC
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Trả lời...............
Đặt a/b=c/d=k
Suy ra a=k . b ; c=k . d
Đó từ đấy bạn chỉ cần thay số vào mà tính thôi
......................học tốt........................
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M=\(\left[\frac{a+b}{\left(b-c\right)\left(c-a\right)}+\frac{b+c}{\left(c-a\right)\left(a-b\right)}+\frac{c-a}{\left(b-c\right)\left(a-b\right)}\right]^{2012}\)
CMR: M đạt giá trị bằng\(2013^{2012}\)khi\(a\ne b\ne c\)
Cho tỉ lệ thức a/b=c/d. Chứng minh rằng:
\(\left(\dfrac{a+b}{c+d}\right)^{2012}=\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^{2012}=\dfrac{a^{2012}}{c^{2012}}=\dfrac{b^{2012}}{d^{2012}}=\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}\) (đpcm)
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c\(\frac{\left(1+\frac{2012}{1}\right)\left(1+\frac{2012}{2}\right)....\left(1+\frac{2012}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)....\left(1+\frac{1000}{2012}\right)}\)
Rút gọn :
a/ \(A=\frac{\frac{1}{19}+\frac{2}{18}+\frac{3}{17}+...+\frac{19}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}}\)
b/ \(B=\frac{\left(1+\frac{2012}{1}\right)\left(1+\frac{2012}{2}\right)...\left(1+\frac{2012}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{2012}\right)}\)
1. Tinh a left(6^9.2^{10}+12^{10}right)+left(2^{19}.27^3+15.4^9.9^4right)2. So sanh A va B.a) Afrac{-2012}{4025};Bfrac{-1999}{3997}b) A3^{21};B2^{31}c) Afrac{2011}{1.2}+frac{2011}{3.4}+frac{2011}{5.6}+....+frac{2011}{1999.2000};Bfrac{2012}{1001}+frac{2012}{1002}+frac{2012}{1003}+....+frac{2012}{2000}
Đọc tiếp
1. Tinh a \(\left(6^9.2^{10}+12^{10}\right)+\left(2^{19}.27^3+15.4^9.9^4\right)\)
2. So sanh A va B.
a) \(A=\frac{-2012}{4025};B=\frac{-1999}{3997}\)
b) \(A=3^{21};B=2^{31}\)
c) \(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{5.6}+....+\frac{2011}{1999.2000};\)\(B=\frac{2012}{1001}+\frac{2012}{1002}+\frac{2012}{1003}+....+\frac{2012}{2000}\)
1/ (69.210+1210)+(219.273+15.49.94) = 29.39.210+310.220+219.39+5.3.218.38 = 219.39+310.220+219.39+5.218.39
= 218.39(2+3.22+5)=19.218.39
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sao bạn lại nhắn vớ va vớ vậy PHẠM ĐỨC PHÚC
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1/ (69
.210+1210
)+(219
.273+15.49
.94
) = 29
.39
.210+310
.220+219
.39+5.3.218
.38
= 219
.39+310
.220+219
.39+5.218
.39
= 2
18
.39
(2+3.22+5)=19.218
.39
Cho \(f\left(x\right)=\frac{x^3}{1-3x+3x^2}\)hãy tính giá trị biểu thức
\(A=f\left(\frac{1}{2012}\right)+f\left(\frac{2}{2012}\right)+...+f\left(\frac{2010}{2012}\right)+f\left(\frac{2011}{2012}\right)\)
Ta xét : \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{3x^2-3x+1}=\frac{\left(x+1-x\right)\left(x^2+x^2-2x+1+x^2-x\right)}{3x^2-3x+1}=\frac{3x^2-3x+1}{3x^2-3x+1}=1\)
Áp dụng ta có :
\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+\left[f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)\right]+...+\left[f\left(\frac{1006}{2012}\right)+f\left(\frac{1006}{2012}\right)\right]\)
\(=1+1+...+1\)(Có tất cả 1006 số 1)
\(=1006\)
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cho x>0 , y>0 , x+y =2012
a) Tìm Max \(B=\frac{2x^2+8xy+2y^2}{x^2+2xy+y^2}\)
b) Tìm Min \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\)
\(a)\) Có \(2012=x+y\ge2\sqrt{xy}\)\(\Leftrightarrow\)\(xy\le1006^2\)
\(B=\frac{2x^2+8xy+2y^2}{x^2+2xy+y^2}=\frac{2\left(x^2+2xy+y^2\right)}{x^2+2xy+y^2}+\frac{4xy}{x^2+2xy+y^2}=2+\frac{4xy}{\left(x+y\right)^2}\)
\(\le2+\frac{4.1006^2}{2012^2}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)
\(b)\) \(C=\left(1+\frac{2012}{x}\right)^2+\left(1+\frac{2012}{y}\right)^2\ge\left[2+2012\left(\frac{1}{x}+\frac{1}{y}\right)\right]^2\ge\left(2+\frac{2012.4}{x+y}\right)^2\)
\(=\left(2+\frac{2012.4}{2012}\right)^2=36\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=1006\)
...
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bạn ơi, mik học \(A^2+B^2\ge\left(A+B\right)^2d\text{ấu}"="\) xảy ra <=> \(A.B\ge0\) mà bạn?
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Xem thêm câu trả lời
\(\frac{\left(1+\frac{2012}{1}\right)\left(1+\frac{2012}{2}\right).......\left(1+\frac{2012}{100}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right).....\left(1+\frac{1000}{2012}\right)}\)