Tìm giá trị của y biết
\(\left(\frac{3}{2}\right)^{10}=\left(\frac{3}{2}\right)^6\times\left(-\frac{3}{2}\right)^y\)
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`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`
`= 2xy`.
Thay `x = 2/3; y = -3/4` vào BT:
`2 . 2/3 . -3/4 = -1.`
`b, x(x-2y) - y(y^2-2x)`
`= x^2 - 2xy - y^3 + 2xy`
`= x^2 - y^3`
Thay `x = 5; y =3` vào BT:
`= 5^2 - 3^3 = 25 - 27 = -2`
a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)
\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)
\(=2xy\)
Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:
\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)
b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)
\(=x^2-2xy-y^3+2xy\)
\(=x^2-y^3\)
Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
\(P=\left(\dfrac{-1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{6}:2\)
\(=\left(\dfrac{1}{2}+\dfrac{3}{5}\right):3+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=\dfrac{11}{10}\cdot\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{22}{60}+\dfrac{15}{60}=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-\dfrac{126}{125}\right)\cdot\dfrac{7}{4}:\left[\dfrac{-119}{36}\cdot\dfrac{36}{17}\right]\)
\(=\dfrac{-116}{125}\cdot\dfrac{7}{4}:\left(-7\right)\)
\(=\dfrac{116}{125}\cdot\dfrac{7}{4}\cdot\dfrac{1}{7}=\dfrac{29}{125}\)
1. a) Ta có: M = |x + 15/19| \(\ge\)0 \(\forall\)x
Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19
Vậy MinM = 0 <=> x = -15/19
b) Ta có: N = |x - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x
Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7
Vậy MinN = -1/2 <=> x = 4/7
2a) Ta có: P = -|5/3 - x| \(\le\)0 \(\forall\)x
Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3
Vậy MaxP = 0 <=> x = 5/3
b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x
Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10
Vậy MaxQ = 9 <=> x = 1/10
\(\Rightarrow\left(-\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^{10}:\left(\frac{3}{2}\right)^6\Rightarrow\left(-\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^4=\left(-\frac{3}{2}\right)^4\)
\(\Rightarrow y=4\)
\(\left(\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^{10}:\left(\frac{3}{2}\right)^6\Rightarrow\left(\frac{3}{2}\right)^y=\left(\frac{3}{2}\right)^4\)
\(\Rightarrow\)y=4