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NM
xy.(-8xy4) c) x3y.(-5y2 z) d) –x.(-3x3y)
2
6 tháng 9 2021
a) \(3xy^3\left(-2x^2yz^3\right)=-6x^3y^4z^3\)
b) \(xy\left(-8xy^4\right)=-8x^2y^5\)
c) \(x^3y\left(-5y^2z\right)=-5x^3y^3z\)
d) \(-x\left(-3x^3y\right)=3x^4y\)
VN
20 tháng 12 2022
a)
\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-12\times\dfrac{144}{1}\)
\(=-1728\)
b)
\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)
\(=3-17\)
\(=-14\)
S
20 tháng 12 2022
a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)
LV
2
22 tháng 6 2023
a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y
=5x^3-7x^2y+2xy^2+5x-2y
b: =(x^2-1)(x+2)
=x^3+2x^2-x-2
c: =1/2x^2y^2(4x^2-y^2)
=2x^4y^2-1/2x^2y^4
d: =(x^2-1/4)(4x-1)
=4x^3-x^2-x+1/4
e: =x^2-2x-35+(2x+1)(x-3)
=x^2-2x-35+2x^2-6x+x-3
=3x^2-7x-38
28 tháng 12 2021
\(\text{Câu 1:Thực hiện phép tính}\)
\(\text{a)}2,5+3-5,5\) \(\text{b)}2,6.2,7+2,6.7,3\)
\(=5,5-5,5\) \(=2,6.\left(2,7+7,3\right)\)
\(=0\) \(=2,6.10\)
\(=26\)
\(\text{Câu 2:}\)
\(\text{Xét }\Delta ABC\text{ có:}\)
\(\widehat{Â}+\widehat{B}+\widehat{C}=180^0\text{(tính chất tổng 3 góc 1 tam giác)}\)
\(\Rightarrow\widehat{C}=180^0-\left(\widehat{A}+\widehat{B}\right)\)
\(\widehat{C}=180^0-\left(40^0+70^0\right)=70^0\)
\(\text{Gọi x;y;z lần lượt là số vở lớp 7/1;7/2;7/3}\)
(đk:x;y;z\(\in\)N*,đơn vị:vở)
\(\text{Ta có:}\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}\text{ và }x+y+z=162\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x+y+z}{5+6+7}=\dfrac{162}{18}=9\)
\(\Rightarrow x=9.5=45\text{(vở)}\)
\(y=9.6=54\text{(vở)}\)
\(z=9.7=63\text{(vở)}\)
\(\text{Vậy số vở lớp 7/1 là:45 vở}\)
\(\text{lớp 7/2 là:54 vở}\)
\(\text{lớp 7/3 là:63 vở}\)
17 tháng 6 2023
1:
a: =7/5(40+1/4-25-1/4)-1/2021
=21-1/2021=42440/2021
b: =5/9*9-1*16/25=5-16/25=109/25
H9
HT.Phong (9A5)
CTVHS
30 tháng 8 2023
Bài 1:
a) \(\dfrac{5^{16}\cdot27^7}{125^5\cdot9^{11}}\)
\(=\dfrac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}\)
\(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}\)
\(=\dfrac{5}{3}\)
b) \(\left(0,2\right)^2\cdot5-\dfrac{2^3\cdot27}{4^6\cdot9^5}\)
\(=0,2\cdot5\cdot0,2-\dfrac{2^3\cdot3^3}{\left(2^2\right)^6\cdot\left(3^2\right)^5}\)
\(=\dfrac{1}{5}-\dfrac{2^3\cdot3^3}{2^{12}\cdot3^{10}}\)
\(=\dfrac{1}{5}-\dfrac{1}{2^9\cdot3^7}\)
\(=\dfrac{2^9\cdot3^7}{2^9\cdot3^7\cdot5}-\dfrac{5}{2^9\cdot3^7\cdot5}\)
\(=\dfrac{2^9\cdot3^7-5}{2^9\cdot3^7\cdot5}\)
c) \(\dfrac{5^6+2^2\cdot25^3+2^3\cdot125^2}{26\cdot5^6}\)
\(=\dfrac{5^6\cdot\left(1+2^2+2^3\right)}{26\cdot5^6}\)
\(=\dfrac{1+2^2+2^3}{26}\)
\(=\dfrac{1+4+8}{26}\)
\(=\dfrac{13}{26}\)
\(=\dfrac{1}{2}\)
Bài 2:
Theo đề ta có:
\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):-\dfrac{1}{4}=-\dfrac{15}{4}\)
\(\Rightarrow\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right)=-\dfrac{15}{4}\cdot-\dfrac{1}{4}\)
\(\Rightarrow a\cdot\dfrac{1}{2}+\dfrac{3}{4}=\dfrac{15}{16}\)
\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{15}{16}-\dfrac{3}{4}\)
\(\Rightarrow a\cdot\dfrac{1}{2}=\dfrac{3}{16}\)
\(\Rightarrow a=\dfrac{3}{16}:\dfrac{1}{2}\)
\(\Rightarrow a=\dfrac{3}{8}\)
30 tháng 8 2023
1:
a: \(=\dfrac{5^{16}\cdot3^{21}}{3^{22}\cdot5^{15}}=\dfrac{1}{3}\cdot5=\dfrac{5}{3}\)
b: \(=0.04\cdot5-\dfrac{2^3\cdot3^3}{3^6\cdot2^{12}}\)
\(=0.2-\dfrac{1}{3^3\cdot2^9}=\dfrac{1}{5}-\dfrac{1}{3^3\cdot2^9}=\dfrac{3^3\cdot2^9-5}{5\cdot3^3\cdot2^9}\)
c: \(=\dfrac{5^6+4\cdot5^6+2^3\cdot5^6}{26\cdot5^6}=\dfrac{1+4+8}{26}=\dfrac{13}{26}=\dfrac{1}{2}\)
2:
Theo đề, ta có:
\(\left(a\cdot\dfrac{1}{2}+\dfrac{3}{4}\right):\dfrac{-1}{4}=\dfrac{-15}{4}\)
=>\(\dfrac{1}{2}a+\dfrac{3}{4}=\dfrac{15}{16}\)
=>1/2a=15/16-12/16=3/16
=>a=3/8