Giải giúp mình bài 2/ với ạ! MÌNH ĐANG CẦN GẤP Ạ!!!
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
PN
0
Những câu hỏi liên quan
MP
2 tháng 5 2023
2.
a. \(\dfrac{-4}{9}\) . \(\dfrac{7}{15}+\dfrac{4}{-9}.\dfrac{8}{15}\) = \(\dfrac{-4}{9}.\left(\dfrac{7}{15}+\dfrac{8}{15}\right)\) = \(\dfrac{-4}{9}\) . 1 = \(\dfrac{-4}{9}\)
b. \(\dfrac{5}{-4}.\dfrac{16}{25}+\dfrac{-5}{4}.\dfrac{9}{25}\) = \(\dfrac{-5}{4}.\left(\dfrac{16}{25}+\dfrac{6}{25}\right)\) = \(\dfrac{-5}{4}.1\) = \(\dfrac{-5}{4}\)
c. \(4\dfrac{11}{23}-\dfrac{9}{14}+2\dfrac{12}{23}-\dfrac{5}{4}\) = \(\left(4\dfrac{11}{23}+2\dfrac{12}{23}\right)\) \(-\dfrac{9}{14}-\dfrac{5}{4}\) = \(\dfrac{68}{23}-\dfrac{9}{14}-\dfrac{5}{4}\) = \(\dfrac{745}{322}\) - \(\dfrac{5}{4}=\dfrac{685}{644}\)
d. \(2\dfrac{13}{27}-\dfrac{7}{15}+3\dfrac{14}{27}-\dfrac{8}{15}\) = \(\left(2\dfrac{13}{27}+3\dfrac{14}{27}\right)\) - \(\left(\dfrac{7}{15}-\dfrac{8}{15}\right)\) = \(\dfrac{68}{27}\) - \(\dfrac{-1}{15}\) =
e. \(11\dfrac{1}{4}-\left(2\dfrac{7}{5}+5\dfrac{1}{4}\right)\) = \(11\dfrac{1}{4}\) - \(\dfrac{81}{20}\) = \(\dfrac{-13}{10}\)
g. \(\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\) = \(\dfrac{7}{9}.\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\) = \(\dfrac{7}{9}.1+\dfrac{12}{19}\) = \(\dfrac{7}{19}+\dfrac{12}{19}\) = \(1\)
JW
0
PQ
2
N
19 tháng 10 2021
Bài 3:
a. \(R=R1+R2=15+30=45\Omega\)
b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)
Bài 4:
\(I1=U1:R1=6:3=2A\)
\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)
\(U=R.I=\left(3+15\right).2=36V\)
\(U2=R2.I2=15.2=30V\)
BT
3
N
cảm ơn
1
29 tháng 10 2021
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt[]{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt{81-80}.x=18+3x\)\(\Rightarrow x^3-3x=18\left(1\right)\)
\(y=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
\(\Rightarrow y^3=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{9-8}.y=6+3y\)\(\Rightarrow y^3-3y=6\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow P=x^3+y^3-3\left(x+y\right)+1996=x^3-3x+y^3-3y+1996\)
\(=18+6+1996=2020\)
5 tháng 12 2021
\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{24\cdot12}{24+12}=8\Omega\)
\(I=\dfrac{U}{R}=\dfrac{12}{8}=1,5A\)
\(P=\dfrac{U^2}{R}=\dfrac{12^2}{8}=18W\)
\(Q_{tỏa1}=A_1=U_1\cdot I_1\cdot t=12\cdot\dfrac{12}{24}\cdot1\cdot3600=21600J\)
\(Q_{tỏa2}=A_2=U_2\cdot I_2\cdot t=12\cdot\dfrac{12}{12}\cdot1\cdot3600=43200J\)
MA
các bạn giải giúp m với ạ! Mình đang cần gấp bài 2 và bài 3II
0
KV
16 tháng 7 2023
Bài 2
a) 3x(x - 1) - 3(x - 1) = 0
(x - 1)(3x - 3) = 0
3(x - 1)(x - 1) = 0
3(x - 1)² = 0
x - 1 = 0
x = 1
b) x² - x = 0
x(x - 1) = 0
x = 0 hoặc x - 1 = 0
*) x - 1 = 0
x = 1
Vậy x = 0; x = 1
c) 25x² - 100x = 0
25x(x - 4) = 0
25x = 0 hoặc x - 4 = 0
*) 25x = 0
x = 0
*) x - 4 = 0
x = 4
Vậy x = 0; x = 4
d) (2x - 1)² - 64 = 0
(2x - 1 - 8)(2x - 1 + 8) = 0
(2x - 9)(2x + 7) = 0
*) 2x - 9 = 0
2x = 9
x = 9/2
*) 2x + 7 = 0
2x = -7
x = -7/2
Vậy x = -7/2; x = 9/2