Giúp em phần II ạ. Được bao nhiêu cx đc. Em xin cảm ơn
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KN
3
Những câu hỏi liên quan
LW
giúp em với ạ ai làm đc thì em xin cảm ơn ạ em đang cần gấp được ko ạ
4
2 tháng 7 2023
\(a,x+\dfrac{1}{2}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{2}\\ x=\dfrac{1}{2}\\ b,-\dfrac{2}{3}-x=1\\x=-\dfrac{2}{3}-1\\ x=-\dfrac{5}{3}\\ d,\dfrac{1}{4}+\dfrac{3}{4}:x=\dfrac{5}{2}\\ \dfrac{3}{4}:x=\dfrac{5}{2}-\dfrac{1}{4}\\ \dfrac{3}{4}:x=\dfrac{9}{4}\\ x=\dfrac{3}{4}:\dfrac{9}{4}\\ x=\dfrac{1}{3}\\ e,\left(x+\dfrac{1}{4}\right)\cdot\dfrac{3}{4}=-\dfrac{5}{8}\\ x+\dfrac{1}{4}=-\dfrac{5}{8}:\dfrac{3}{4}\\ x+\dfrac{1}{4}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{1}{4}\\ x=\dfrac{7}{12}\)
\(g,\dfrac{x-3}{15}=\dfrac{-2}{5}\\ 5\left(x-3\right)=-30\\ x-3=-6\\ x=-6+3\\ x=-3\\ h,\dfrac{x}{-2}=\dfrac{-8}{x}\\ x^2=16\\ x=\pm\sqrt{16}\\ x=\pm4\\ k,\dfrac{x+2}{3}=\dfrac{x-4}{5}\\ 5\left(x+2\right)=3\left(x-4\right)\\ 5x+10=3x-12\\ 5x-3x=-12-10\\ 2x=-22\\ x=-11\)
\(m,\left(2x-1\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
VD
29 tháng 5 2022
Hướng dẫn: A đạt GTLN khi \(\dfrac{1}{A}\) đạt GTNN
Ta có: \(x^2+2\ge0\forall x\)
\(\Rightarrow A=\dfrac{1}{x^2+2}\le\dfrac{1}{2}\forall x\)
Vậy GTLN của A là 1/2
=> A
DT
Đỗ Thanh Hải
CTVVIP
7 tháng 6 2021
B
1 is writing
2 is losing
3 is having
4 is staying
5 amnot telling
6 is always using
7 are having
8 Are you playing
C
1 are top musicians studying => Do top musician study
2 don't touch => aren't touching
3 does
4 is Christine listening
5 am usually buying => usually buy
6 is starting => starts
7 Does your team win => is your team winning
8 are enjoying => enjoy
HN
1
29 tháng 5 2022
Câu 1: A
Câu 2: B
Câu 3: D
Câu 4: A
Câu 5: C
Câu 6: B
Câu 7: A
Câu 9: B
26 tháng 1
p: \(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+...+\dfrac{5}{50\cdot51}\)
\(=5\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{50\cdot51}\right)\)
\(=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)
\(=5\cdot\left(1-\dfrac{1}{51}\right)=5\cdot\dfrac{50}{51}=\dfrac{250}{51}\)
q: \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{210}\)
\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{420}\)
\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\right)\)
\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{20\cdot21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{21}\right)=2\cdot\dfrac{19}{42}=\dfrac{19}{21}\)
29 tháng 5 2022
Câu 1: A
Câu 2: B
Câu 3: D
Câu 4: A
Câu 5: C
Câu 6: B
Câu 7: A
Câu 9: B
B2
0
NT
0
1) \(x^4-10x^2+1=0\)
\(\Leftrightarrow\left(x^2-5\right)^2=24\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5=2\sqrt{6}\\x^2-5=-2\sqrt{6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5+2\sqrt{6}}\\x=\sqrt{5-2\sqrt{6}}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\\x=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+\sqrt{2}\\x=\sqrt{3}-\sqrt{2}\end{matrix}\right.\)
Vậy \(x=\sqrt{2}+\sqrt{3}\) là một nghiệm của pt
2) Ta có: \(a-b=\sqrt{3}+1,b-c=\sqrt{3}-1\)
\(\Rightarrow a-c=a-b+b-c=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)
\(A=a^2+b^2+c^2-ab-bc-ac\)
\(\Rightarrow2A=2a^2+2b^2+2c^2-2ab-2ac-2bc\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)
\(\left(\sqrt{3}+1\right)^2+\left(\sqrt{3}-1\right)^2+\left(2\sqrt{3}\right)^2\)
\(=3+2\sqrt{3}+1+3-2\sqrt{3}+1+12=20\)
\(\Rightarrow A=10\in N\)