giúp 3 phần nì gấp ạ, e cảm ơn ạ
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MD
1
ND
1
H
5 tháng 1 2022
bạn đăng tách ra tầm 10 câu mỗi lần đăng nha, chứ dài ntnay ngại làm lắm~
5 tháng 1 2022
e có tách 3 bài ra rồi ạ, phiền anh/chị/bạn giúp e với ạ, e cảm ơn ạ
10 tháng 6 2023
a: \(A=\dfrac{x}{x+3}-1=\dfrac{x-x-3}{x+3}=\dfrac{-3}{x+3}\)
\(B=\dfrac{9-x^2+x^2-9-x^2+4x-4}{\left(x+3\right)\left(x-2\right)}=\dfrac{-\left(x-2\right)^2}{\left(x-2\right)\left(x+3\right)}=\dfrac{-x+2}{x+3}\)
b: B nguyên
=>-x-3+5 chia hết cho x+3
=>x+3 thuộc {1;-1;5;-5}
=>x thuộc {-2;-4;-8}
c: P=A:B
=(-3/x+3):(-x+2)/(x+3)
=3/(x-2)
NH
0
MA
4
14 tháng 11 2021
Bài 6
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)
14 tháng 11 2021
Bài 7:
\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)
MA
1
15 tháng 11 2021
a) ĐKXĐ: \(\left\{{}\begin{matrix}2x+3\ne0\\2x+1\ne0\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
b) \(\Rightarrow P=\dfrac{2\left(2x+1\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)
\(\Rightarrow P=\dfrac{2}{2x+1}\)
c) \(P=-1\Rightarrow\dfrac{2}{2x+1}=-1\\ \Rightarrow2=-2x-1\\ \Rightarrow2x=-3\\ \Rightarrow x=-\dfrac{3}{2}\)
9 tháng 3 2022
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
=> \(n_{H_2SO_4}=0,15\left(mol\right)\)
=> mmuối = mKL + mSO4 = 5 + 0,15.96 = 19,4 (g)
e/\(E=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{575}\)
\(=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{23.25}\)
\(=\dfrac{7}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{23.25}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{23}-\dfrac{1}{25}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{25}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{24}{25}\)
\(=\dfrac{84}{25}\)
g)\(G=\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{100.102}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{100.102}\right)\)
\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{102}\right)\)
\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{102}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{25}{51}\)
\(=\dfrac{125}{102}\)
h/\(H=\dfrac{2024}{2.5}+\dfrac{2024}{5.8}+...+\dfrac{2024}{62.65}\)
\(=\dfrac{2024}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{62.65}\right)\)
\(=\dfrac{2024}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{62}-\dfrac{1}{65}\right)\)
\(=\dfrac{2024}{3}\left(\dfrac{1}{2}-\dfrac{1}{65}\right)\)
\(=\dfrac{2024}{3}\cdot\dfrac{63}{130}\)
\(=\dfrac{21252}{65}\)
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