Tính :
a) \(\left(-4\right).\left(-5\right).\left(-6\right)\)
b) \(\left(-3+6\right).\left(-4\right)\)
c) \(\left(-3-5\right)\left(-3+5\right)\)
d) \(\left(-5-13\right):\left(-6\right)\)
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a, Ta thấy : \(\left\{{}\begin{matrix}\left(2a+1\right)^2\ge0\\\left(b+3\right)^2\ge0\\\left(5c-6\right)^2\ge0\end{matrix}\right.\)\(\forall a,b,c\in R\)
\(\Rightarrow\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\ge0\forall a,b,c\in R\)
Mà \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2\le0\)
Nên trường hợp chỉ xảy ra là : \(\left(2a+1\right)^2+\left(b+3\right)^2+\left(5c-6\right)^2=0\)
- Dấu " = " xảy ra \(\left\{{}\begin{matrix}2a+1=0\\b+3=0\\5c-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=-3\\c=\dfrac{6}{5}\end{matrix}\right.\)
Vậy ...
b,c,d tương tự câu a nha chỉ cần thay số vào là ra ;-;
a) ( − 3 ) . ( − 4 ) . ( − 5 )=-60
b) ( − 5 + 8 ) . ( − 7 )=-21
c) =27
d)
a) \(\left(-3\right).\left(-4\right)\left(-5\right)\) = \(12.\left(-5\right)\) = \(-60\)
b) \(\left(-5+8\right).\left(-7\right)\) = \(3.\left(-7\right)\) = \(-21\)
c) \(\left(-6-3\right).\left(-6+3\right)\) = \(-9.\left(-3\right)\) = 27
d) \(\left(-4-14\right):\left(-3\right)\) = \(-18:\left(-3\right)\) = 6
a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729
b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)
c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)
\(=\dfrac{2^3}{4^4}\cdot7\)
\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)
d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)
\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)
\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)
e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)
\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)
\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)
f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)
\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)
g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)
\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)
\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)
\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)
\(=54+8-32=30\)
\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)
\(=5-2\sqrt{2}\)
\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)
\(=2-2\sqrt{3}\)
\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)
\(=2\sqrt{6}\)
\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)
`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`
a) \(8 - \left( {x - 15} \right) = 2.\left( {3 - 2x} \right)\)
\(8 - x + 15 = 6 - 4x\)
\( - x + 4x = 6 - 8 - 15\)
\(3x = - 17\)
\(x = \left( { - 17} \right):3\)
\(x = \dfrac{{ - 17}}{3}\)
Vậy nghiệm của phương trình là \(x = \dfrac{{ - 17}}{3}\).
b) \( - 6\left( {1,5 - 2u} \right) = 3\left( { - 15 + 2u} \right)\)
\( - 9 + 12u = - 45 + 6u\)
\(12u - 6u = - 45 + 9\)
\(u = \left( { - 36} \right):6\)
\(6u = - 36\)
\(u = - 6\)
Vậy nghiệm của phương trình là \(u = - 6\).
c) \({\left( {x + 3} \right)^2} - x\left( {x + 4} \right) = 13\)
\(\left( {{x^2} + 6x + 9} \right) - \left( {{x^2} + 4x} \right) = 13\)
\({x^2} + 6x + 9 - {x^2} - 4x = 13\)
\(\left( {{x^2} - {x^2}} \right) + \left( {6x - 4x} \right) = 13 - 9\)
\(2x = 4\)
\(x = 4:2\)
\(x = 2\)
Vậy nghiệm của phương trình là \(x = 2\).
d) \(\left( {y + 5} \right)\left( {y - 5} \right) - {\left( {y - 2} \right)^2} = 5\)
\(\left( {{y^2} - 25} \right) - \left( {{y^2} - 4y + 4} \right) = 5\)
\({y^2} - 25 - {y^2} + 4y - 4 = 5\)
\(\left( {{y^2} - {y^2}} \right) + 4y = 5 + 4 + 25\)
\(4y = 34\)
\(y = 34:4\)
\(y = \dfrac{{17}}{2}\)
Vậy nghiệm của phương trình là \(y = \dfrac{{17}}{2}\).
a.-120
b. - 12
c. - 16
d. 3
a) (−4).(−5).(−6)= 20.(-6)= -120
b) (−3+6).(−4)= 3.(-4)= -12
c) (−3−5)(−3+5)= -8. 2 = -16
d) (−5−13):(−6)= -18:( -6)= 3