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\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = \(\dfrac{-3}{14}\) : \(\dfrac{5}{7}\)
\(\dfrac{3}{5}\)\(x\) - \(\dfrac{11}{5}\) = - \(\dfrac{3}{10}\)
\(\dfrac{3}{5}\)\(x\) = - \(\dfrac{3}{10}\) + \(\dfrac{11}{5}\)
\(\dfrac{3}{5}\)\(x\) = \(\dfrac{19}{10}\)
\(x\) = \(\dfrac{19}{10}\) : \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{19}{6}\)
\(\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}:\dfrac{5}{7}\)
\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{14}\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{3}{5}x-\dfrac{11}{5}=-\dfrac{3}{10}\)
\(\Rightarrow\dfrac{3}{5}x=-\dfrac{3}{10}+\dfrac{11}{5}\)
\(\Rightarrow\dfrac{3}{5}x=\dfrac{19}{10}\)
\(\Rightarrow x=\dfrac{19}{10}:\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{19}{6}\)
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a) Các góc kề bù nhau là:
1. \(\widehat{xOy}\) và \(\widehat{xOt}\)
2. \(\widehat{yOz}\) và \(\widehat{zOt}\)
b) Ta có: \(\widehat{yOt}\) là góc bẹt \(\Rightarrow\widehat{yOt}=180^o\)
Mà \(\widehat{xOy}\) và \(\widehat{xOt}\) kề bù \(\Rightarrow\widehat{xOy}+\widehat{xOt}=\widehat{yOt}\)
\(\Rightarrow\widehat{xOt}=\widehat{yOt}-\widehat{xOy}=180^o-45^o=135^o\)
Ta có: \(\widehat{xOz}=\widehat{xOy}+\widehat{yOz}=45^o+30^o=75^o\)
Mà \(\widehat{yOz}\) và \(\widehat{zOt}\) kề bù \(\Rightarrow\widehat{yOz}+\widehat{zOt}=\widehat{yOt}=180^o\)
\(\Rightarrow\widehat{zOt}=\widehat{yOt}-\widehat{yOz}=180^o-30^o=150^o\)
![](https://rs.olm.vn/images/avt/0.png?1311)
AB<AC
=>góc B>góc C
=>90 độ-góc B<90 độ-góc C
=>góc HAB<góc HAC
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a: x=3/4-1/3=9/12-4/12=5/12
b: 1/3x=2/3-5/9=6/9-5/9=1/9
=>x=1/3
c: =>3x=6/5-3/4-7/4=6/5-5/2=12/10-25/10=-13/10
=>x=-13/30
d: =>x+2/5-2/3=5/3
=>x=5/3+2/3-3/5=7/3-3/5=35/15-9/15=26/15
e: =>1/4:x=2/5-3/4=8/20-15/20=-7/20
=>x=-1/4:7/20=-1/4*20/7=-20/28=-5/7
f: =>1/4x-3/4=1/2-25/4=2/4-25/4=-23/4
=>1/4x=-20/4
=>x=-20
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`a)` Cho `x-1/2x^2=0`
`=>x(1-1/2x)=0`
`@TH1: x = 0`
`@TH2: 1-1/2x=0=>1/2x=1=>x=2`
Nghiệm của đa thức là `x=0` hoặc `x=2`
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`b)\overline{X}=[6.3+7.6+8x+9.4]/[3+6+x+4]`
Mà `\overline{X}=7,6`
`=>[96+8x]/[13+x]=7,6`
`=>96+8x=7,6(13+x)`
`=>96+8x=98,8+7,6x`
`=>8x-7,6x=98,8-96`
`=>0,4x=2,8`
`=>x=7`
Vậy `x=7`
cho \(x-\dfrac{1}{2}x^2=0\)
\(=>x\left(1-\dfrac{1}{2}x\right)=0\)
\(=>\left[{}\begin{matrix}x=0\\\dfrac{1}{2}x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1:\dfrac{1}{2}=2\end{matrix}\right.\)
\(Q\left(x\right)=x^4+3x^2+1\)
vì \(\left\{{}\begin{matrix}x^4\ge0\\3x^2\ge0\end{matrix}\right.\)
\(=>x^4+3x^2\ge0\)
mà 1 > 0
\(=>x^4+3x^2+1\ge0\)
hay Q(x ) ko có nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
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`@` `\text {Ans}`
`\downarrow`
Ta có:
\(5^{333}=\left(5^3\right)^{111}=125^{111}\)
\(11^{222}=\left(11^2\right)^{111}=121^{111}\)
Vì `125 > 121 =>`\(125^{111}>121^{111}\)
`=>`\(5^{333}>11^{222}\)
Vậy, \(5^{333}>11^{222}\)
_____
`@` So sánh lũy thừa cùng cơ số:
Nếu `m > n =>`\(a^m>a^n\left(m,n\ne0,a>1\right)\)
`@` So sánh lũy thừa cùng số mũ:
Nếu `a > b =>`\(a^m>b^m\left(a,b>1,m\ne0\right)\)
`@` `\text {Kaizuu lv uuu}`
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