số dương ma \(\frac{x}{100}\) = \(\frac{25}{x}\) la so
Giải phương trình
a,\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
b, \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
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) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)
a) Tìm 3 số nguyên dương biết tổng của chúng bằng nửa tích của chúng
b) tìm các số tự nhiên x,y soa cho ƯCLN (x,y) = 1 và\(\frac{x+y}{x^2+y^2}=\frac{7}{25}\)
c) So sánh A =\(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\) và B =\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+......+\frac{1}{17}\)
mik fan Phong ca nè bạn
Bài3. Giải phương trình
a/ \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{102}{3}\)
b/ \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
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. \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Rightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+\frac{25-x}{25}+1+\frac{23-x}{27}+1+\frac{21-x}{29}+1=0\)
\(\Rightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Rightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Rightarrow50-x=0\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)
\(\Rightarrow x=50\)
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
Cho x,y,z la các số dương sao cho x+y+z\(\ge\)12
tìm Min M=\(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\)
\(P=4\left(\frac{x}{y+4}+\frac{y}{z+4}+\frac{z}{x+4}\right)=4\left(\frac{x^2}{xy+4x}+\frac{y^2}{yz+4y}+\frac{z^2}{zx+4z}\right)\)
\(\ge\frac{4\left(a+b+c\right)^2}{xy+4x+yz+4y+zx+4z}=\frac{4.12^2}{4.12+\left(xy+yz+zx\right)}\)
\(\ge\frac{4.12^2}{4.12+\frac{\left(x+y+z\right)^2}{3}}=\frac{4.12^2}{4.12+\frac{12^2}{3}}=6\)
Ta 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 \(M\ge6\)
Giải lại
Ta có
\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+2\left(\frac{xy}{\sqrt{yz}}+\frac{yz}{\sqrt{zx}}+\frac{zx}{\sqrt{xy}}\right)\)
Lại có
\(\hept{\begin{cases}\frac{xy}{\sqrt{yz}}+\sqrt{yz}\ge2\sqrt{xy}\\\frac{yz}{\sqrt{zx}}+\sqrt{zx}\ge2\sqrt{yz}\\\frac{zx}{\sqrt{xy}}+\sqrt{xy}\ge2\sqrt{zx}\end{cases}}\)
Cộng theo vế suy ra \(\frac{xy}{\sqrt{yz}}+\frac{yz}{\sqrt{zx}}+\frac{zx}{\sqrt{xy}}\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
Do đó
\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+2\left(\frac{xy}{\sqrt{yz}}+\frac{yz}{\sqrt{zx}}+\frac{zx}{\sqrt{xy}}\right)\)
\(\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(=\left(\frac{x^2}{y}+\sqrt{xy}+\sqrt{xy}\right)+\left(\frac{y^2}{z}+\sqrt{yz}+\sqrt{yz}\right)+\left(\frac{z^2}{x}+\sqrt{zx}+\sqrt{zx}\right)\)
\(\ge3\sqrt[3]{\frac{x^2}{y}.\sqrt{xy}.\sqrt{xy}}+3\sqrt[3]{\frac{y^2}{z}.\sqrt{yz}.\sqrt{yz}}+3\sqrt[3]{\frac{z^2}{x}.\sqrt{zx}.\sqrt{zx}}\)
\(=3\left(x+y+z\right)\ge36\)
Vậy \(M\ge6\)
ĐT xảy ra tại \(x=y=z=4\)
Cho x,y là các số thực dương thỏa x+y=1. CMR:
\(\left(x+\frac{1}{x}\right)^{2^{ }}+\left(y+\frac{1}{y}^{ }\right)^{2^{ }}\ge\frac{25}{2}\)
\(\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2\ge\frac{1}{2}\left(x+y+\frac{4}{x+y}\right)^2=\frac{25}{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
Cho x và y là các số dương có tổng bằng 1.
CMR: \(\left(y+\frac{1}{x}\right)^2+\left(x+\frac{1}{y}\right)^2\ge\frac{25}{2}\)
Tìm các cặp số nguyên dương x,y sao cho :
\(\frac{x+y}{x^2+y^2}=\frac{7}{25}\)
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
bạn chưa hiểu được chưa hiểu mk bày từ từ
1, Tìm x biết:
\(\frac{x+9}{x+5}=\frac{2}{7}\)
2,Số nguyên x để \(A=\left(x-\frac{23}{2}\right)\left(\frac{25}{2}-x\right)\)có giá trị dương
Tìm tổng của 3 số dương x,y,z biết:\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5};2x^2+2y^2-3z^2=-100\)
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:
$x=3k=3.2=6; y=4k=4.2=8; z=5k=5.2=10$