25*(-1/5)^3+1/5-2*(-1/2)^2-1/2
giup mih nha
K
Khách
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B
1
TP
Tinh bang hai cach:
a. (1/3 + 1/5) x 1/2 b. (1/3 - 1/5) x 1/2
giup minh voi, tra loi minh tick cho nha
2
15 tháng 3 2022
a.\(=\dfrac{8}{15}\times\dfrac{1}{2}=\dfrac{4}{15}\)
b.\(=\dfrac{1}{3}\times\dfrac{1}{2}-\dfrac{1}{5}\times\dfrac{1}{2}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\)
15 tháng 9 2021
\(a,=4x^2+4x+1\\ b,=9-12y+4y^2\\ c,=\dfrac{x^2}{4}-xy+y^2\\ d,=\dfrac{25}{4}-5x+x^2\\ e,=4x^2+32xy+64y^2\\ f,=9x^2-30xy+25y^2\)
7 tháng 4 2016
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
suy ra S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
1 tháng 7 2019
a)12.(\(\frac{-5}{3}+\frac{7}{4}\) ) \(-16\left[\left(\frac{-1}{2}\right)^2-\left(\frac{-5}{2}\right)^3\right]\)
=-20+21-4-1000
=-1005
b)\(\frac{-3}{4}:\left(\frac{1}{2}+\frac{-2}{3}\right)-\frac{16}{25}\)
=\(\frac{-3}{4}:\frac{-1}{6}-\frac{4}{5}\)
=\(\frac{9}{2}-\frac{4}{5}\)
=\(\frac{37}{10}\)
27 tháng 2 2021
\(\left(-x^2y\right)^3\cdot\dfrac{1}{2}\cdot x^2y^3\cdot\left(-2xy^2z\right)^2\\ =-x^6y^3\cdot\dfrac{1}{2}x^2y^3\cdot4x^2y^4z^2\\ =\left(-1\cdot\dfrac{1}{2}\cdot4\right)\cdot\left(x^6\cdot x^2\cdot x^2\right)\cdot\left(y^3\cdot y^3\cdot y^4\right)\cdot z^2\\ =-2x^{10}y^{10}z^2\)
LT
0
AH
Akai Haruma
Giáo viên
23 tháng 10 2018
a)
\(2x+3=(2x+3)^2\)
\(\Leftrightarrow (2x+3)^2-(2x+3)=0\)
\(\Leftrightarrow (2x+3)(2x+3-1)=0\)
\(\Leftrightarrow (2x+3)(2x+2)=0\Rightarrow \left[\begin{matrix} 2x+3=0\\ 2x+2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-\frac{3}{2}\\ x=-1\end{matrix}\right.\)
b) \((x-5)^2=5-x\)
\(\Leftrightarrow (x-5)^2+(x-5)=0\)
\(\Leftrightarrow (x-5)(x-5+1)=0\)
\(\Leftrightarrow (x-5)(x-4)=0\)
\(\Rightarrow \left[\begin{matrix} x-5=0\\ x-4=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=5\\ x=4\end{matrix}\right.\)
c) \((x+2)^3=x+2\)
\(\Leftrightarrow (x+2)^3-(x+2)=0\)
\(\Leftrightarrow (x+2)[(x+2)^2-1]=0\)
\(\Leftrightarrow (x+2)(x+2-1)(x+2+1)=0\)
\(\Leftrightarrow (x+2)(x+1)(x+3)=0\)
\(\Rightarrow \left[\begin{matrix} x+2=0\\ x+1=0\\ x+3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=-1\\ x=-3\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
23 tháng 10 2018
d)
\(|3x-1|=(1-3x)^2\)
\(\Leftrightarrow |3x-1|=|3x-1|^2\)
\(\Leftrightarrow |3x-1|^2-|3x-1|=0\)
\(\Leftrightarrow |3x-1|(|3x-1|-1)=0\)
\(\Rightarrow \left[\begin{matrix} |3x-1|=0\\ |3x-1|-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} |3x-1|=0\\ |3x-1|=1\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} 3x-1=0\\ 3x-1=\pm 1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{3}\\ x=\frac{2}{3}\\ x=0\end{matrix}\right.\)
e)
\(2x+(x+3)(3-x)+(x+1)(x-1)=7\)
\(\Leftrightarrow 2x+(3^2-x^2)+(x^2-1^2)=7\)
\(\Leftrightarrow 2x=-1\Rightarrow x=-\frac{1}{2}\)
\(25.\left(-\frac{1}{5}\right)^3+\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2-\frac{1}{3}\)
\(\Leftrightarrow25.\left(-\frac{1}{5}\right)^3-2.\left(-\frac{1}{2}\right)^2+\frac{1}{5}-\frac{1}{2}\)
\(\Leftrightarrow-\frac{1}{5}-2.\left(-\frac{1}{2}\right)^2+\frac{1}{5}-\frac{1}{2}\)
\(\Leftrightarrow-\frac{1}{5}+\frac{1}{5}-2\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(\Leftrightarrow-2.\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(\Leftrightarrow-\frac{1}{2}-\frac{1}{2}\)
\(\Leftrightarrow\frac{-1-1}{2}\)
\(\Rightarrow-1\)