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Toán lớp 6? -_-
\(P=\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\)
*Áp dụng bất đẳng thức Cauchy, ta có:
\(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\ge\dfrac{9}{xy+yz+zx}\)
\(P\ge\dfrac{1}{x^2+y^2+z^2}+\dfrac{9}{xy+yz+xz}=\dfrac{1}{x^2+y^2+z^2}+\dfrac{4}{2\left(xy+yz+zx\right)}+\dfrac{7}{xy+yz+zx}\)
*Áp dụng bất đẳng thức Cauchy-Schwarz, ta có:
\(\dfrac{1}{x^2+y^2+z^2}+\dfrac{4}{2\left(xy+yz+zx\right)}\ge\dfrac{\left(1+2\right)^2}{\left(x+y+z\right)^2}\)
và \(\dfrac{7}{xy+yz+xz}\ge\dfrac{7}{\dfrac{1}{3}\left(x+y+z\right)}=21\)
\(\Rightarrow P\ge9+21=30\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)
b, \(\dfrac{x-3}{4}=\dfrac{15}{20}\)
<=> \(\dfrac{x-3}{4}=\dfrac{3}{4}\)
=> x-3=3
<=> x=6
Vậy x=6
\(a,\dfrac{x}{15}=\dfrac{4}{y}=\dfrac{-2}{5}\)
* \(\dfrac{x}{15}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{-6}{15}\)
\(\Rightarrow x=-6\)
*\(\dfrac{4}{y}=\dfrac{-2}{5}\)
\(\Rightarrow\dfrac{4}{y}=\dfrac{4}{-10}\)
\(\Rightarrow y=-10\)
Vậy x = - 6 ; y = - 10
\(b,\dfrac{x-3}{4}=\dfrac{15}{20}\)
=> ( x - 3 ) . 20 = 4. 15
=> 20x - 60 = 60
=> 20x = 60 + 60
=> 20x = 120
=> x = 120 : 20
=> x = 6
Vậy x = 6
\(c,\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{22}{-9}+\dfrac{-7}{15}< x\le\dfrac{-1}{3}+\dfrac{-1}{4}+\dfrac{-5}{12}\)
\(\Rightarrow\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{-22}{9}+\dfrac{-7}{15}< x\le\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-5}{12}\)
\(\Rightarrow\left(\dfrac{-5}{9}+\dfrac{-22}{9}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)< x\le-1\)
\(\Rightarrow-3+\left(-1\right)< x\le-1\)
\(\Rightarrow-4< x\le-1\)
\(\Rightarrow x=-3;-2;-1\)
Ta có : ( -1/4 ) : ( -3/4 ) + 1/2 = -1/4 * 4/-3 + 1/2 .
= -1/-3 + 1/2 .
= 2/6 + 3/6 .
= 5/6 .
Và 7/8 - 1/2 : ( -5/6 ) = 7/8 - 1/2 * 6/-5 .
= 7/8 + 3/5 .
= 35/40 + 24/40 .
= 59/40 .
Do đó : 5/6 < x < 59/40 .
⇒ 100/120 < x < 177/120 .
⇒ 100 < x < 177 .
⇒ x = 101 ; 102 ; .... ; 175 ; 176 ( vì x ∈ Z ) .
Vậy x = 101 ; 102 ; .... ; 175
\(\dfrac{-2}{3}\cdot\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\\ -\dfrac{2}{3}x+\dfrac{1}{6}-\dfrac{2}{3}x+\dfrac{1}{3}=0\)
\(-\dfrac{4}{3}x+\dfrac{1}{2}=0\\ -\dfrac{4}{3}x=-\dfrac{1}{2}\\ x=\dfrac{3}{8}\)
\(\dfrac{1}{5}2^x+\dfrac{1}{3}2^{x+1}=\dfrac{1}{5}2^7+\dfrac{1}{3}2^8\)
\(\dfrac{1}{5}2^x+\dfrac{1}{3}2^x\cdot2=\dfrac{1}{5}2^7+\dfrac{1}{3}2^7\cdot2\)
\(2^x\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)=2^7\left(\dfrac{1}{5}+\dfrac{1}{3}\cdot2\right)\)
\(2^x=2^7\\ x=7\)
\(y+30\%y=-1,3\\ 130\%y=-1,3\\ \Rightarrow y=\dfrac{-1,3}{130\%}=-1\)
\(x:\dfrac{4}{28}=\dfrac{13}{-19}+\dfrac{8}{25}\\ 7x=-\dfrac{173}{475}\\ x=-\dfrac{\dfrac{173}{475}}{7}=-\dfrac{173}{3325}\)