\(1-\left(\frac{73}{8}+X-\frac{173}{24}\right):\frac{50}{3}=0\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right):\frac{50}{3}=1-0\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right):\frac{50}{3}=1\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right)=1.\frac{50}{3}\)

\(\Leftrightarrow\left(\frac{73}{8}+X-\frac{173}{24}\right)=\frac{50}{3}\)

\(\Leftrightarrow\frac{73}{8}+X=\frac{50}{3}+\frac{173}{24}\)

\(\Leftrightarrow\frac{73}{8}+X=\frac{191}{8}\)

\(\Leftrightarrow X=\frac{191}{8}-\frac{73}{8}\)

\(\Leftrightarrow X=\frac{59}{4}\)

Dấu \(.\)là dấu X đấy ngen pn!!!!!Nhớ K cko mik nka

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_

(1-1/3).(1-1/5).(1-1/7).(1-1/9).(1-1/11).(1-1/13).(1-1/2).(1-1/4).(1-1/6).(1-1/8).(1-1/10)

=2/3.4/5.6/7.8/9.10/11.12/13.1/2.3/4.5/6.7/8.9/10

=8/15.48/63.120/143.3/8.35/48.9/10

=384/945.360/1144.315/480

=138240/1081080.315/480

=43545600/518918400=84/1001

khó quá

đơn giản 

nhưng trả lời câu hỏi của tớ đã

=\(\frac{9^{2016}}{16^{2016}}.\frac{16^{2015}}{9^{2015}}.\frac{4}{3}\)

=\(\frac{9}{16}.\frac{4}{3}\)

=\(\frac{3}{4}\)

k cho mk nhoa

\(\left(\frac{9}{16}\right)^{2016}.\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)

\(=\left[\frac{9}{16}\left(\frac{9}{16}\right)^{2015}\right].\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)

\(=\frac{9}{16}\left[\left(\frac{9}{16}\right)^{2015}.\left(\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)

\(=\frac{9}{16}\left[\left(\frac{9}{16}.\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)

\(=\frac{9}{16}.1^{2015}.\frac{4}{3}\)

\(=\frac{9}{16}.\frac{4}{3}\)

\(=\frac{3}{4}\)

2.

Áp dụng bất đẳng thức Cauchy - schwarz ( hay còn gọi là bất đẳng thức Cosi ):

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}=\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{9}{3+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi x = y = z = 1

1: 

Áp dụng bất đẳng thức Cô si:

\(x\left(y+\frac{x}{1+y}\right)+y\left(z+\frac{y}{1+z}\right)+z\left(x+\frac{z}{1+x}\right)\)

\(=\left(x+y+z\right)\left[\left(y+\frac{x}{1+y}\right)+\left(z+\frac{y}{1+z}\right)+\left(x+\frac{z}{1+x}\right)\right]\)

\(=1\left[\left(x+y+z\right)+\left(\frac{x}{1+y}+\frac{y}{1+z}+\frac{z}{1+x}\right)\right]\)

\(=1\left[1+\left(\frac{x+y+z}{1+y+1+z+1+x}\right)\right]\)

\(=1\left[1+\left(\frac{1}{3+\left(x+y+z\right)}\right)\right]\)

\(=1\left[1+\frac{1}{4}\right]\)

\(=1+\frac{5}{4}=\frac{9}{4}\)

Dấu "=" xảy ra khi x = y = z = \(\frac{1}{3}\)

2. áp dạng bất đẳng thức cauchy - schwarz dạng engel

\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{x+1}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3}=\frac{3^2}{3+3}=\frac{9}{6}=\frac{3}{2}\)

dấu bằng xay ra khi x=y=z=1

a) số số x là 4 nên ta có:

(x.4)+1/2+1/4+1/8+1/16=1 mà 1/2+1/4+1/8+1/16=15/16 nên x=1-15/16=1/16:4=1/64