Tính giá trị của biểu thức:
a) \(32-6.\left(8-2^3\right)+18;\)
b) \(\left(3.5-9\right)^3.\left(1+2.3\right)^2+4^2.\)
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QN
26 tháng 3 2021
a) 9 234 : [3 . 3. (1 + 83)] = 9 234 : [3 . 3 . (1 + 512)]
= 9 234 : [3 . 3 . 513] = 9 234 : 4617 = 2
b) 76 - {2 . [2 . 52 - (31 - 2 . 3)]} + 3 . 25
= 76 - {2 . [2 . 25 - (31 - 6)]} + 75
= 76 - {2 . [50 - 25]} + 75 = 76 - {2 . 25} + 75 = 76 - 50 + 75 = 101
1 tháng 2 2021
Ta có: \(A=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{28}\cdot3^{18}\cdot2}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2\cdot2^{28}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(5-7\cdot2\right)}\)
\(=\dfrac{2^{28}\cdot3^{18}\cdot\left(5\cdot2^2-2\right)}{2^{28}\cdot3^{18}\cdot\left(5-14\right)}\)
\(=\dfrac{20-2}{-9}=\dfrac{18}{-9}=-2\)
8 tháng 9 2021
a) 32 – 6 . (8 – 23) + 18
= 32 – 6 . ( 8 – 8) + 18
= 32 – 6 . 0 + 18
= 32 – 0 + 18
= 32 + 18 = 50.
b) (3 . 5 – 9)3 . (1 + 2 . 3)2 + 42
= (15 – 9)3 . (1 + 6)2 + 16
= 63 . 72 + 16
= 216 . 49 + 16
= 10 584 + 16
= 10 600.
22 tháng 7 2023
`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`
22 tháng 7 2023
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:
\(5^2-3^3=-3\)
29 tháng 6 2021
\(a,A=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=4\sqrt{2}\)
\(b,B=\left|1-\sqrt{5}\right|+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|1-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left|1-\sqrt{5}\right|+\left|\sqrt{5}+1\right|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(c,C=\dfrac{2+\sqrt{6}+2-\sqrt{6}}{\left(2+\sqrt{6}\right)\left(2-\sqrt{6}\right)}=\dfrac{4}{4-6}=-2\)
AH
Akai Haruma
Giáo viên
29 tháng 6 2021
Lời giải:
a.
\(A=2\sqrt{2}-3\sqrt{18}+4\sqrt{32}-\sqrt{50}=2\sqrt{2}-9\sqrt{2}+16\sqrt{2}-5\sqrt{2}\)
\(=(2-9+16-5)\sqrt{2}=4\sqrt{2}\)
b.
\(B=\sqrt{(1-\sqrt{5})^2}+\sqrt{(\sqrt{5}+1)^2}=|1-\sqrt{5}|+|\sqrt{5}+1|=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
c.
\(C=\frac{2+\sqrt{6}+2-\sqrt{6}}{(2-\sqrt{6})(2+\sqrt{6})}=\frac{4}{2^2-6}=-2\)
3 tháng 8 2021
a) Ta có: \(-\dfrac{3}{2}\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2\cdot\left(1+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{2}\left(\sqrt{5}-2\right)+4\cdot\left(\sqrt{5}+1\right)\)
\(=\dfrac{-3}{2}\sqrt{5}+3+4\sqrt{5}+4\)
\(=\dfrac{5}{2}\sqrt{5}+7\)
b) Ta có: \(\left(1+\dfrac{1}{\tan^225^0}\right)\cdot\sin^225^0-\tan55^0\cdot\tan35^0\)
\(=\dfrac{\tan^225^0+1}{\tan^225^0}\cdot\sin25^0-1\)
\(=\left(\dfrac{\sin^225^0}{\cos^225^0}+1\right)\cdot\dfrac{\cos^225^0}{\sin^225^0}\cdot\sin25^0-1\)
\(=\dfrac{\sin^225^0+\cos^225^0}{\cos^225^0}\cdot\dfrac{\cos^225^0}{\sin25^0}-1\)
\(=\dfrac{1}{\sin25^0}-1\)
\(=\dfrac{1-\sin25^0}{\sin25^0}\)
2 tháng 7 2021
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
2 tháng 7 2021
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600
a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600