Tính:
a) (-27)^5 : 9^8
b) 25 . (-\(\dfrac{1}{3}\))^3 +1/5 - 2 . (-\(\dfrac{1}{2}\))^2 - 1/2
Giải giúp mình với ạ
K
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Những câu hỏi liên quan
24 tháng 2 2022
a, \(=\dfrac{1+4}{5}+\dfrac{5+1+3}{9}=1+1=2\)
b, \(=\dfrac{1+4+2}{3}+\dfrac{1+2+5}{6}=\dfrac{6}{3}+\dfrac{8}{6}=2+\dfrac{4}{3}=\dfrac{6+4}{3}=\dfrac{10}{3}\)
S
1 tháng 4 2021
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
1 tháng 4 2021
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
Y
14 tháng 8 2023
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
NH
3 tháng 3 2023
a)
`2/3+5/2-3/4`
`=10/4-3/4+2/3`
`=7/4+2/3`
`=21/12+8/12`
`=29/12`
b)
`2/5xx1/2:1/3`
`=2/10xx3/1`
`=6/10=3/5`
c)
`2/9:2/9xx1/3`
`=2/9xx9/2xx1/3`
`=1xx1/3`
`=1/3`
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 3 2023
a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)
= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)
= \(\dfrac{38-9}{12}\)
= \(\dfrac{29}{12}\)
b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)
= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)
= \(\dfrac{3}{5}\)
c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)
= 1 x \(\dfrac{1}{3}\)
= \(\dfrac{1}{3}\)
NT
Nguyễn thành Đạt
CTVHS
27 tháng 8 2023
Ta có : \(B\text{=}4x^2-12x+9\)
\(B\text{=}\left(2x-3\right)^2\)
Với \(x\text{=}\dfrac{1}{2}\)
\(\Rightarrow B\text{=}\left(2.\dfrac{1}{2}-3\right)^2\)
\(B\text{=}\left(-2\right)^2\text{=}4\)
Ta có : \(A\text{=}5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)
\(A\text{=}5\left(x^2-9\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)
\(A\text{=}5x^2-45+4x^2+12x+9+x^2-12x+36\)
\(A\text{=}10x^2\)
Với \(x\text{=}-\dfrac{1}{5}\)
\(\Rightarrow A\text{=}10.\left(-\dfrac{1}{5}\right)^2\text{=}\dfrac{2}{5}\)
KV
27 tháng 8 2023
B = 4x² - 12x + 9
= (2x - 3)²
Tại x = 1/2 ta có:
B = (2.1/2 - 3)²
= (-2)²
= 4
-------------------
A = 5(x + 3)(x - 3) + (2x + 3)² + (x - 6)²
= 5x² - 45 + 4x² + 12x + 9 + x² - 12x + 36
= 10x²
Tại x = 1/5 ta có:
A = 10.(1/5)²
= 2/5
Y
25 tháng 4 2022
\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)
\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)
\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)
\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)
\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)
25 tháng 4 2022
mik xinloi ạ câu 50 mik viết sai đề nên mong bạn giải lại giúp ạ <3
29 tháng 12 2021
a: \(=\dfrac{-8}{9}-\dfrac{6}{5}+\dfrac{8}{9}=-\dfrac{6}{5}\)
c: \(=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
18 tháng 10 2021
\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)
\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)
\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\dfrac{7}{3}\)
12 tháng 9 2021
bài 1)
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3.3\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^7.3^7}\)
\(=\dfrac{\left(-3\right)^3.1.1}{5.1.3^2}\)
\(=\dfrac{-27.1.1}{5.1.9}\)
\(=\dfrac{-27}{45}\)
\(=\dfrac{-9}{15}\)
b)\(2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{2^2}.4+\left[\left(4:\dfrac{1}{2}\right)\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4.\dfrac{2}{1}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+8.8\)
\(=8+3-1+64\)
\(=11-1+64\)
\(=10+64\)
\(=74\)