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=\(\frac{9^{2016}}{16^{2016}}.\frac{16^{2015}}{9^{2015}}.\frac{4}{3}\)
=\(\frac{9}{16}.\frac{4}{3}\)
=\(\frac{3}{4}\)
k cho mk nhoa
\(\left(\frac{9}{16}\right)^{2016}.\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)
\(=\left[\frac{9}{16}\left(\frac{9}{16}\right)^{2015}\right].\left(\frac{16}{9}\right)^{2015}.\frac{4}{3}\)
\(=\frac{9}{16}\left[\left(\frac{9}{16}\right)^{2015}.\left(\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)
\(=\frac{9}{16}\left[\left(\frac{9}{16}.\frac{16}{9}\right)^{2015}\right].\frac{4}{3}\)
\(=\frac{9}{16}.1^{2015}.\frac{4}{3}\)
\(=\frac{9}{16}.\frac{4}{3}\)
\(=\frac{3}{4}\)
giá trị x nguyên thỏa mãn:
\(\frac{6}{5}< x-\frac{3}{2}< \frac{12}{5}\)
các bn giải đầy đủ giúp mk nha
Ta có:
\(\frac{6}{5}< x-\frac{3}{2}< \frac{12}{5}\)
\(=>\frac{12}{10}< x-\frac{15}{10}< \frac{24}{10}\)
\(=>\frac{12+15}{10}< x< \frac{24+15}{10}\)
\(=>\frac{27}{10}< x< \frac{39}{10}\) (mà x là số nguyên)
\(=>x=3\)
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) nha bạn!
ko hỉu thì ib
\(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)\ge9\) với x,y,z dương hay jj đó chứ? (cái này t k bt -.-) VD: x=2, y=-2,z=4
=> \(\left(x+y+z\right).\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{x}\right)=\left(2-2+4\right).\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{4}\right)=1\)
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\(\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\frac{x+y+z}{x+y+z}=0\)
\(\Leftrightarrow\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
vì x+y+z khác 0 => \(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{xy+yz+xz}{xyz}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{\left(xy+yz+xz\right).\left(x+y+z\right)-xyz}{xzy.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\frac{x^2y+xy^2+xyz+zyx+y^2z+yz^2+x^2z+xyz+xz^2-xzy}{xyz.\left(x+y+z\right)}=0\)
\(\Leftrightarrow\left(x^2y+xyz\right)+\left(xy^2+y^2z\right)+\left(yz^2+xzy\right)+\left(x^2z+xz^2\right)=0\)
\(\Leftrightarrow xy.\left(x+z\right)+y^2.\left(x+z\right)+yz.\left(z+x\right)+xz.\left(x+z\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left(xy+y^2+yz+xz\right)=0\)
\(\Leftrightarrow\left(x+z\right).\left[x.\left(y+z\right)+y.\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x+y\right).\left(y+z\right).\left(x+z\right)=0\Leftrightarrow\orbr{\begin{cases}x=-y\\y=-z\end{cases}\text{hoặc }x=-z}\)
\(\Rightarrow P=\left(\frac{1}{x}-\frac{1}{y}\right).\left(\frac{1}{y}+\frac{1}{z}\right).\left(\frac{1}{z}+\frac{1}{x}\right)=0\)
ps: bài này t làm cách l8, ai có cách ez hơn giải vs ak :') morongtammat
Ta có : \(\frac{1+3y}{12}=\frac{1+6y}{16}\)
<=> (1 + 3y).16 = (1 + 6y).12
<=> 16 + 48y = 12 + 72y
<=> 16 - 12 = 72y - 48y
<=> 24y = 4
=> y = 1/6
Thay y = 1/6 vào ta có : \(\frac{1+6.\frac{1}{6}}{16}=\frac{1+9.\frac{1}{6}}{4x}\Rightarrow\frac{1}{8}=\frac{\frac{5}{2}}{4x}\)
=> x = \(\frac{5}{2}:\frac{1}{8}=20\)
+, Nếu x+y+z=0 => B = x+y/y. y+z/z . z+x/x = (-z/y).(-x/z).(-y/x) = -xyz/xyz = -1
+, Nếu x+y+z khác o thì :
Áp dụng tính chất dãy tỉ số bằng nhau ta có : y+z-x/x = z+x-y/y = x+y-z/z = y+z-x+z+x-y+x+y-z/x+y+z = 1
=> y+z-x=x ; z+X-y=y ; x+y-z=z
=> x=y=z
=> B = (1+1).(1+1).(1+!) = 8
Vậy .............
Tk mk nha
ADTCDTSBN
\(\frac{y+z-x}{x}\)=\(\frac{z+x-y}{y}\)=\(\frac{x+y-z}{z}\)=\(\frac{y+z-x+z+x-y+x+y-z}{x+y+z}\)=1
\(\Rightarrow\)\(\hept{\begin{cases}y=-z\\z=-x\\x=-y\end{cases}}\)
Khi đó B=\(\left(1+\frac{-y}{y}\right)\)\(\left(1+\frac{-z}{z}\right)\)\(\left(1+\frac{-x}{x}\right)\)=0
Vậy B=0 ........... hjhjh
\(\Leftrightarrow\dfrac{1}{6}< \left|\dfrac{2}{7}-x\right|< \dfrac{3}{4}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-\dfrac{2}{7}\right|>\dfrac{1}{6}\\\left|x-\dfrac{2}{7}\right|< \dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left(-\infty;\dfrac{10}{84}\right)\cup\left(\dfrac{38}{84};+\infty\right)\\x\in\left(-\dfrac{39}{84};\dfrac{87}{84}\right)\end{matrix}\right.\)
\(\Leftrightarrow x\in\left(\dfrac{38}{84};\dfrac{87}{84}\right)\)
\(-\frac{17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)
\(\Leftrightarrow-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}\)
\(\Leftrightarrow-\frac{17}{21}.\frac{20}{17}< x+\frac{4}{7}< \frac{7}{12}\)
\(\Leftrightarrow-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)
\(\Leftrightarrow-\frac{20}{21}< x< \frac{1}{84}\)
\(\Leftrightarrow-\frac{80}{84}< x< \frac{1}{84}\)
\(\Leftrightarrow-80< x< 1\Leftrightarrow x\in\left\{-79;-78;...;0\right\}\)
mà để Giá trị nguyên lớn nhất của x
\(\Rightarrow x=-1\)
Ta có : \(\frac{18}{39}< \frac{x}{13}< \frac{16}{26}\) => \(\frac{6}{13}< \frac{x}{13}< \frac{8}{13}\)
Vì 3 phân số có cùng mẫu nên 6 < x < 8 nên x = 7