số dương ma \(\frac{x}{100}\) = \(\frac{25}{x}\) la so
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DN
25 tháng 12 2017
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)
BL
19 tháng 3 2020
a. \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Rightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
\(\Rightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-105}{4}-\frac{x-105}{3}=0\)
\(\Rightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Rightarrow x-105=0\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)\)
\(\Rightarrow x=105\)
b. \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Rightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+\frac{25-x}{25}+1+\frac{23-x}{27}+1+\frac{21-x}{29}+1=0\)
\(\Rightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Rightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Rightarrow50-x=0\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)
\(\Rightarrow x=50\)
19 tháng 3 2020
a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\frac{x-5}{100}-1+\frac{x-4}{101}-1+\frac{x-3}{102}-1=\frac{x-100}{5}-1+\frac{x-101}{4}-1+\frac{x-102}{3}-1\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
Dễ dàng thấy nhân tử thứ hai luôn bé thua 0 nên \(x-105=0\)\(\Leftrightarrow x=105\)
b) Kĩ thuật làm tương tự câu a cộng mỗi phân số VT với 1 thì VP=0 và ta có nhân tử chung 50-x
TA
2 tháng 8 2017
\(P=4\left(\frac{x}{y+4}+\frac{y}{z+4}+\frac{z}{x+4}\right)=4\left(\frac{x^2}{xy+4x}+\frac{y^2}{yz+4y}+\frac{z^2}{zx+4z}\right)\)
\(\ge\frac{4\left(a+b+c\right)^2}{xy+4x+yz+4y+zx+4z}=\frac{4.12^2}{4.12+\left(xy+yz+zx\right)}\)
\(\ge\frac{4.12^2}{4.12+\frac{\left(x+y+z\right)^2}{3}}=\frac{4.12^2}{4.12+\frac{12^2}{3}}=6\)
TA
3 tháng 8 2017
Ta có
\(\frac{x}{\sqrt{y}}+\frac{x}{\sqrt{y}}+\frac{xy}{8}\ge3\sqrt[3]{\frac{x}{\sqrt{y}}.\frac{x}{\sqrt{y}}.\frac{xy}{8}}=\frac{3x}{2}\)
Tương tự cho 2 cái kia
Cộng lại theo vế:
\(2M\ge\frac{3}{2}\left(x+y+z\right)-\frac{xy+yz+zx}{8}\ge\frac{3}{2}\left(x+y+z\right)-\frac{\left(x+y+z\right)^2}{24}\ge12\)
Vậy \(M\ge6\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 11 2019
\(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{4}{x+y}\right)^2=\frac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
GH
2
LN
30 tháng 4 2015
vi x, y la nguyen duong nen ta thay lan luot x tư 1 den 6
tim ra cap x=3, y =4
va y=3, x=4
AH
Akai Haruma
Giáo viên
28 tháng 7
Lời giải:
Đặt $\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k(k>0)$
$\Rightarrow x=3k; y=4k; z=5k$.
Khi đó:
$2x^2+2y^2-3z^2=-100$
$\Rightarrow 2(3k)^2+2(4k)^2-3(5k)^2=-100$
$\Rightarrow -25k^2=-100$
$\Rightarrow k^2=4\Rightarrow k=2$ (do $k>0$)
Ta có:
$x=3k=3.2=6; y=4k=4.2=8; z=5k=5.2=10$
\(\frac{x}{100}=\frac{25}{x}\)
2x=25.100
2x=2500
x=2500:2
x=1250