tìm x
a. x + 3/4 = 11/3
mọi người giúp mình nha
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x-3/-9 = -4/x-3
(x-3)^2=-9.-4=36
=>TH1 x-3= 6
TH2 x-3 = -6
=> TH1: X= 9
TH2: X=-3
`a)5/8x+2/5=1/5`
`=>5/8x=1/5-2/5`
`=>5/8x=-1/5`
`=>x=-1/5:5/8=-8/25`
`b)5/7:x+11/7=18/7`
`=>5/7:x=18/7-11/7`
`=>5/7:x=1`
`=>x=5/7`
`c)(-1,2).(-3/24)+(0,4-1 4/15):1 2/3`
`=(-6/5).(-1/8)+(2/5-19/15):5/3`
`=3/20+(-13/15)*3/5`
`=3/20-13/25=-37/100`
a)5/8.x+2/5=1/5
5/8.x=1/5 - 2/5
5/8.x=-1/5
x=(-1/5):5/8
x=(-1/5).8/5
x=-8/25. Vậy x=-8/25
b)5/7:x +11/7=18/7
5/7:x=1
x=5/7:1
x=5/7. Vậy x=5/7
x,y∈Z⇒x+1,xy-1∈Z và x+1,xy-1 thuộc Ư(3)
Ta có bảng:
x+1 | 1 | 3 | -1 | -3 |
xy-1 | 3 | 1 | -3 | -1 |
x | 0 | 2 | -2 | -4 |
y | loại | 1 | 1 | 0 |
Vậy \(\left(x,y\right)\in\left\{\left(2;1\right)\left(-2;1\right);\left(-4;0\right)\right\}\)
Vì x,y là chữ số nên \(x,y\in\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
Ta có \(\left(x+1\right)\left(xy-1\right)=3=3.1=1.3\)
Với \(\left\{{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\-1=3\end{matrix}\right.\left(loại\right)\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
a:ta có: \(2x^2\ge0\)
\(\Leftrightarrow2x^2+1>0\forall x\)
vậy: H(x) vô nghiệm
\(a.\dfrac{12}{3}=\dfrac{20}{5}=4\\ b.\dfrac{9}{-3}=\dfrac{-15}{5}=-3\)
a, Xét \(\dfrac{x}{3}=4\Rightarrow x=12;\dfrac{20}{y}=4\Rightarrow y=\dfrac{20}{4}=5\)
b, \(\dfrac{9}{-x}=-3\Rightarrow-x=-3\Leftrightarrow x=3\)
\(\dfrac{y}{5}=-3\Rightarrow y=-15\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)
\(a,\left|x-\dfrac{3}{4}\right|=\dfrac{1}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=\dfrac{-1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)
\(b,\dfrac{-1}{3}+\left|x\right|=0,5\)
\(\Leftrightarrow\left|x\right|=\dfrac{5}{6}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{-5}{6}\end{matrix}\right.\)
\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)
\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)
\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)
\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)
\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)
\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)
\(x=\dfrac{11}{3}-\dfrac{3}{4}=\dfrac{35}{12}\)