Tính
3/5x2/7x5/6x1/2
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\(\dfrac{3}{5}-\dfrac{20}{9}:\dfrac{5}{18}+\dfrac{2}{4}\)
\(=\dfrac{3}{5}-\dfrac{20}{9}\times\dfrac{18}{5}+\dfrac{1}{2}\)
\(=\dfrac{3}{5}-\dfrac{360}{45}+\dfrac{1}{2}\)
\(=\dfrac{3}{5}-8+\dfrac{1}{2}\)
\(=-\dfrac{37}{5}+\dfrac{1}{2}\)
\(=-\dfrac{69}{10}\)
\(\dfrac{3}{5}-\dfrac{20}{9}:\dfrac{5}{18}+2,4\)
\(=\dfrac{3}{5}-8+\dfrac{7}{5}\)
\(=\left(\dfrac{3}{5}+\dfrac{7}{5}\right)-8\)
\(=2-8\)
\(=-6\)
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
\(1\frac{1}{5}x1\frac{1}{6}x...x1\frac{1}{2014}\)
=\(\frac{6}{5}x\frac{7}{6}x...x\frac{2015}{2014}\)
= \(\frac{2015}{5}\)= 403.
x= 2015 hoặc 403 vì hai người kia nói hai kết quả nên tui cho là 2 đáp án
thay tui sieu khong
nếu thấy tui giỏi thì tick cho tui nhé bạn hiền
3/4+2/3+5/12=11/6
7/8-7/16-11/32=3/32
3/5*2/7*5/6=1/7
15/16:3/8:3/4=10/3
bạn nhé
-Câu cuối thôi nha bạn :v
\(B=-5x^2-4x-\dfrac{19}{5}=-5\left(x^2+\dfrac{4}{5}x+\dfrac{19}{25}\right)=-5\left(x^2+2.\dfrac{2}{5}x+\dfrac{4}{25}+\dfrac{15}{25}\right)=-5\left(x+\dfrac{2}{5}\right)^2-\dfrac{15}{5}\le-3\)\(B_{max}=-3\Leftrightarrow x=\dfrac{-2}{5}\)
`(x-3/5)xx3/8-5/8=1/6`
`(x-3/5)xx3/8=1/6+5/8`
`(x-3/5)xx3/8=4/24+15/24=19/24`
`x-3/5=19/24:3/8=19/9`
`x=19/9+3/5=122/45`
_______________________________________________________
`1 1/6-5/6:(x+1/4)=1/2`
`7/6-5/6:(x+1/4)=1/2`
`5/6:(x+1/4)=7/6-1/2=2/3`
`x+1/4=5/6:2/3=5/4`
`x=5/4-1/4=4/4=1`
a: \(\left(x-\dfrac{3}{5}\right)\cdot\dfrac{3}{8}-\dfrac{5}{8}=\dfrac{1}{6}\)
\(\Leftrightarrow\left(x-\dfrac{3}{5}\right)\cdot\dfrac{3}{8}=\dfrac{1}{6}+\dfrac{5}{8}=\dfrac{4}{24}+\dfrac{15}{24}=\dfrac{19}{24}\)
\(\Leftrightarrow x-\dfrac{3}{5}=\dfrac{19}{24}:\dfrac{3}{8}=\dfrac{19}{24}\cdot\dfrac{8}{3}=\dfrac{19}{9}\)
hay x=122/45
b: \(\Leftrightarrow\dfrac{7}{6}-\dfrac{5}{6}:\left(x+\dfrac{1}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{6}:\left(x+\dfrac{1}{4}\right)=\dfrac{7}{6}-\dfrac{3}{6}=\dfrac{2}{3}\)
=>x+1/4=5/6:2/3=5/6x3/2=15/12=5/4
=>x=1
3/5x2/7x5/6x1/2=(3/5x5/6)x(2/7x1/2)
=1/2x7=7/2
Cho mik tick nha
\(\frac35\cdot\frac27\cdot\frac56\cdot\frac12\)
=\(\frac{3\cdot2\cdot5\cdot1}{5\cdot7\cdot6\cdot2}\)
=\(\frac{1}{14}\)