Giúp e với ạa. E đang cần gấpp🥲
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27 tháng 12 2023
Câu 2.
Nhiệt lượng bếp tỏa ra trong thời gian \(t=3min=180s\) là:
\(Q=UIt=RI^2t=60\cdot2,5^2\cdot180=675000J\)
Câu 3.
\(I_{Đ1}=\dfrac{U_{Đ1}}{R_{Đ1}}=\dfrac{6}{6}=1A\)
\(I_{Đ2}=\dfrac{U_{Đ2}}{R_{Đ2}}=\dfrac{1,5}{8}=\dfrac{3}{16}A\)
\(I_b=I_{Đ1}-I_{Đ2}=1-\dfrac{3}{16}=\dfrac{13}{16}A\)
\(R_b=\dfrac{U_b}{I_b}=\dfrac{1,5}{\dfrac{13}{16}}=\dfrac{24}{13}\Omega\)
19 tháng 12 2023
a: \(VP=a^3+b^3+c^3-3bac\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VT\)
b: \(VT=\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)\)
\(=3a^2+15a+2ab+10b-a-5-2ab+4b\)
\(=3a^2+14a+14b-5\)
\(VP=\left(3a+5\right)\left(a+3\right)+2\left(7b-10\right)\)
\(=3a^2+9a+5a+15+14b-20\)
\(=3a^2+14a+14b-5\)
=>VT=VP
c: \(VT=a\left(b-x\right)+x\left(a+b\right)\)
\(=ab-ax+ax+bx\)
\(=ab+bx=b\left(a+x\right)=VP\)
d: \(VT=a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)
\(=ab-ac-ab-bc+ca-cb\)
\(=-2bc\)
=VP
15 tháng 7 2021
1)
ĐKXĐ: x>4
Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)
\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)
\(\Leftrightarrow8x+6x=8-15\)
\(\Leftrightarrow14x=-7\)
hay \(x=-\dfrac{1}{2}\)(loại)
2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)
\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
7 tháng 11 2021
Bài 1:
\(a,A=\dfrac{22}{7}-\dfrac{22}{7}-0,25-0,75-4=-1-4=-5\\ b,B=\dfrac{1\cdot\left(\dfrac{2}{5}\right)^3\cdot\left(\dfrac{15}{4}\right)^2}{\left(\dfrac{15}{2^2}\right)^2\cdot\left(\dfrac{2}{5}\right)^3}=1\)
Bài 2:
\(a,\Rightarrow2^{x+3}=13,9+2,1=16=2^4\\ \Rightarrow x+3=4\Rightarrow x=1\\ b,\Rightarrow\left|x-\dfrac{3}{2}\right|=0,5=\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{3}{2}=2\\x=-\dfrac{1}{2}+\dfrac{3}{2}=1\end{matrix}\right.\\ c,\Rightarrow\dfrac{x}{2}=\dfrac{y}{7}=\dfrac{2x-5y}{2\cdot2-5\cdot7}=\dfrac{93}{-31}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-6\\y=-21\end{matrix}\right.\)
30 tháng 7 2023
14:
a: Xét ΔHNM vuông tại H và ΔMNP vuông tại M có
góc N chung
=>ΔHNM đồng dạng với ΔMNP
b: NP=căn 3^2+4^2=5cm
MH=3*4/5=2,4cm
NH=3^2/5=1,8cm
13:
a: 3x+5=x-5
=>2x=-10
=>x=-5
b: (x-2)(2x+5)=0
=>x-2=0 hoặc 2x+5=0
=>x=2 hoặc x=-5/2
c: =>2(5x-2)=3(3x+1)
=>10x-4=9x+3
=>x=7
d: =>(3x+6-x+1)/(x+2)(x-1)=17-3x/(x+2)(x-1)
=>2x+7=17-3x
=>5x=10
=>x=2
31 tháng 1
a: \(\dfrac{2}{x+5}=\dfrac{2\cdot4\cdot\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}=\dfrac{8\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{-3}{4x-20}=\dfrac{-3}{4\left(x-5\right)}=\dfrac{-3\left(x+5\right)}{4\left(x-5\right)\left(x+5\right)}=\dfrac{-3x-15}{4\left(x-5\right)\left(x+5\right)}\)
\(\dfrac{-x+2}{x^2-25}=\dfrac{-x+2}{\left(x-5\right)\left(x+5\right)}=\dfrac{4\left(-x+2\right)}{4\left(x-5\right)\left(x+5\right)}=\dfrac{-4x+8}{4\left(x-5\right)\left(x+5\right)}\)
b: \(\dfrac{1}{3x-6y}=\dfrac{1}{3\left(x-2y\right)}=\dfrac{\left(x-2y\right)\left(x+2y\right)}{3\left(x-2y\right)^2\cdot\left(x+2y\right)}\)
\(\dfrac{-x}{x^2-4y^2}=\dfrac{-x}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{-x\cdot3\cdot\left(x-2y\right)}{3\left(x-2y\right)^2\cdot\left(x+2y\right)}\)
\(\dfrac{-2y^2}{x^2-4xy+4y^2}=\dfrac{-2y^2}{\left(x-2y\right)^2}=\dfrac{-2y^2\cdot3\left(x+2y\right)}{3\left(x+2y\right)\left(x-2y\right)^2}\)
\(=\dfrac{-6y^2\left(x+2y\right)}{3\left(x+2y\right)\left(x-2y\right)^2}\)
