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a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}=3\)
b) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
`<=>(x-1)/2023-1+(x-2)/2022-1=(x-3)/2021-1+(x-4)/2020-1`
`<=>(x-2024)/2023+(x-2024)/2022=(x-2024)/2021+(x-2024)/2020`
`<=>(x-2024)(1/2023+1/2022-1/2021-1/2020)=0`
`<=>x-2024=0(1/2023+1/2022-1/2021-1/2020>0)`
`<=>x=2024`
=>\(\left(\dfrac{x-1}{2023}-1\right)+\left(\dfrac{x-2}{2022}-1\right)=\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-4}{2020}-1\right)\)
=>x-2024=0
=>x=2024
a: =>-3x=-12
=>x=4
b: =>3(3x+2)-3x-1=12x+10
=>9x+6-3x-1=12x+10
=>12x+10=6x+5
=>6x=-5
=>x=-5/6
c: =>x(x+1)+x(x-3)=4x
=>x^2+x+x^2-3x-4x=0
=>2x^2-6x=0
=>2x(x-3)=0
=>x=3(loại) hoặc x=0(nhận)
=>5(4x-1)-2+x<=3(10x-3)
=>20x-5+x-2<=30x-9
=>21x-7<=30x-9
=>-9x<=-2
=>x>=2/9
\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{5\left(x+3\right)+4\left(x-3\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\\ \Leftrightarrow5x+15+4x-12=x-5\\ \Leftrightarrow5x+4x-x=-5-15+12\\ \Leftrightarrow8x=-8\\ \Leftrightarrow x=-1\left(TM\right)\\ Vậy:S=\left\{-1\right\}\)
Câu 1:
1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{7}{x^2+3}\)
Câu 2a đề sai, pt này ko giải được
2b.
\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)
Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)
\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)
3a.
\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)
\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)
3b
Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6
\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6
\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)
Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)
\(\Rightarrow2021^{2022}-1⋮6\)
\(\Rightarrow S-1⋮6\)
Ta có: VT = \(\dfrac{x+1}{2021}\)+1 - (\(\dfrac{x+2}{2020}\)+1) = \(\dfrac{x+3}{2019}\)+1=VP
=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}-\dfrac{x+2022}{2019}=0\)
=>\(\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}\right)=0\)
=>x +2022 = 0=> x =-2022
a) \(3\left(2x-x\right)=5x+1\)
\(\Leftrightarrow6x-3x=5x+1\)
\(\Leftrightarrow6x-3x-5x=1\)
\(\Leftrightarrow-2x=1\)
\(\Leftrightarrow x=\dfrac{1}{-2}=-\dfrac{1}{2}\)
b) \(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}+\dfrac{x+3}{2019}+\dfrac{x+4}{2018}=0\)
\(\Leftrightarrow\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)
\(\Leftrightarrow\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)
\(\Leftrightarrow\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}\right)\)
\(\Leftrightarrow x+2022=0\)
\(\Leftrightarrow x=-2022\)
