39+3=?
58-7=?
5+3x4=?
9x3=?
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a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
a: \(=\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21+10}{35}=\dfrac{31}{35}\)
b: \(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{1}{3}=\dfrac{6}{5}+\dfrac{1}{2}=\dfrac{12+5}{10}=\dfrac{17}{10}\)
c: \(=\dfrac{5\cdot3}{9}=\dfrac{5}{3}\)
d: =5-2/3*6/5
=5-12/15
=5-4/5
=21/5
\(\frac{7}{9}\times\frac{3}{5}+\frac{3}{5}\times\frac{5}{6}+\frac{3}{5}\times1\)
\(=\frac{3}{5}\times\left(\frac{7}{9}+\frac{5}{6}+1\right)\)
\(=\frac{3}{5}\times\frac{47}{18}\)
\(=\frac{47}{30}\)
#hoktot<3#
\(\frac{5}{9}x\frac{4}{7}+\frac{5}{9}x\frac{3}{7}\)
\(=\frac{5}{9}x\left(\frac{4}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}x1\)
\(=\frac{5}{9}\)
\(a,\dfrac{7}{9}\times\dfrac{3}{14}:\dfrac{5}{8}=\dfrac{7}{9}\times\dfrac{3}{14}\times\dfrac{8}{5}=\dfrac{4}{15}\)
\(b,\dfrac{3}{5}\times\dfrac{4}{21}\times\dfrac{25}{3}=\dfrac{20}{21}\)
\(c,\dfrac{15}{16}:\dfrac{5}{8}\times\dfrac{3}{4}=\dfrac{15}{16}\times\dfrac{8}{5}\times\dfrac{3}{4}=\dfrac{9}{8}\)
\(d,\dfrac{21}{4}\times\dfrac{16}{14}\times\dfrac{1}{2}\times\dfrac{8}{3}=8\)
\(a,=y\left(y-2\right)\\ b,=3x\left(x^2-2x+1\right)=3x\left(x-1\right)^2\\ c,=\left(y-1\right)\left(27x^2+9x^3\right)=9x^2\left(x+3\right)\left(y-1\right)\\ d,=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\\ e,=x\left(x^2+6x+9\right)=x\left(x+3\right)^2\\ f,=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\\ g,=\left(2-x\right)\left(x+1\right)\\ h,=\left(x-1\right)\left(3x-6\right)=3\left(x-1\right)\left(x-2\right)\)
a: =y(y-2)
b: \(=3x^2\left(x^2-2x+1\right)=3x^2\left(x-1\right)^2\)
d: \(=y\left(y^2-2y+1\right)=y\left(y-1\right)^2\)
`-1,5 \times (7/3-5/3 \times 4)`
`= -1,5 \times (7/3-20/3)`
`= -1,5 \times (-13/3)`
`= 13/2`
39 + 3 = 42
58 - 7 = 51
5 + 3 x 4 = 17
9 x 3 = 27
tk cho mk nha
39 + 3 = 42
58 - 7 = 51
5 + 3 X 4 = 5 + 12 = 17
9 X 3 = 27