Bài 1:

a, \(2y.\left(y-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)

Vậy \(y\in\left\{0;\dfrac{1}{7}\right\}\)

b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)

\(\Rightarrow\dfrac{5}{6}y=\dfrac{-4}{15}+\dfrac{2}{5}\)

\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)

\(\Rightarrow y=\dfrac{4}{25}\)

Vậy \(y=\dfrac{4}{25}\)

Chúc bạn học tốt!!!

Bài 1:

a, \(2y\left(y-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2y=0\\y-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=0\\y=\dfrac{1}{7}\end{matrix}\right.\)

Vậy...

b, \(\dfrac{-2}{5}+\dfrac{2}{3}y+\dfrac{1}{6}y=\dfrac{-4}{15}\)

\(\Rightarrow\dfrac{5}{6}y=\dfrac{2}{15}\)

\(\Rightarrow y=\dfrac{4}{25}\)

Vậy...

Bài 2:

a, \(x\left(x-\dfrac{4}{7}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)

\(\Rightarrow x>\dfrac{4}{7}\left(x\ne0\right)\) hoặc \(x< \dfrac{4}{7}\left(x\ne0\right)\)

Vậy...

Các phần còn lại tương tự nhé

a, \(x.\left(x-\dfrac{4}{7}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\)

+, Xét \(\left\{{}\begin{matrix}x>0\\x-\dfrac{4}{7}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x>\dfrac{4}{7}\end{matrix}\right.\Rightarrow x>\dfrac{4}{7}\)(1)

+, Xét \(\left\{{}\begin{matrix}x< 0\\x-\dfrac{4}{7}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 0\\x< \dfrac{4}{7}\end{matrix}\right.\Rightarrow x< 0\)(2)

Vậy \(x>\dfrac{4}{7}\) và \(x< 0\)

b, \(\left(x-\dfrac{2}{5}x^2\right)< 0\)

\(\Rightarrow x.\left(1-\dfrac{2}{5}x\right)< 0\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\1-\dfrac{2}{5}x< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\\dfrac{2}{5}x>1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x>2,5\end{matrix}\right.\)

\(\Rightarrow x>2,5\)

Vậy \(x>2,5\)

c, \(\left(x-\dfrac{2}{5}\right).\left(x+\dfrac{3}{7}\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{2}{5}>0\\x+\dfrac{3}{7}>0\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-\dfrac{2}{5}< 0\\x+\dfrac{3}{7}< 0\end{matrix}\right.\)

+, Xét \(\left\{{}\begin{matrix}x-\dfrac{2}{5}>0\\x+\dfrac{3}{7}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>\dfrac{2}{5}\\x>\dfrac{-3}{7}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{2}{5}\)

+, Xét \(\left\{{}\begin{matrix}x-\dfrac{2}{5}< 0\\x+\dfrac{3}{7}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< \dfrac{2}{5}\\x< \dfrac{-3}{7}\end{matrix}\right.\) \(\Rightarrow x< \dfrac{-3}{7}\)

Vậy \(x>\dfrac{2}{5};x< \dfrac{-3}{7}\)

Chúc bạn học tốt!!!

Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)

\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)
 

cộng 2016 nhé

Dễ có: \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\)

\(gt\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)

Áp dụng BĐT Cauchy-Schwarz ta có: 

\(\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}=\frac{1}{\sqrt{\left(2+1\right)\left(2x^2+y^2\right)}}\le\frac{1}{2x+y}\)

\(\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{9}\left(\frac{2}{x}+\frac{1}{y}\right)\)

\(\Rightarrow P\le\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{1}{3}\sqrt{3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{Z^2}\right)}\le\sqrt{\frac{2016}{3}}\)

Câu 1: D

Câu 3: 53/144>9/170>9/230

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8