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5x-6x-9x=-100
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\(6,\\ a,P=9\left(x^2-2\cdot\dfrac{1}{9}x+\dfrac{1}{81}\right)+\dfrac{26}{9}=9\left(x-\dfrac{1}{9}\right)^2+\dfrac{26}{9}\ge\dfrac{26}{9}\\ P_{min}=\dfrac{26}{9}\Leftrightarrow x-\dfrac{1}{9}=0\Leftrightarrow x=\dfrac{1}{9}\\ b,Q=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\\ Q_{min}=\dfrac{1}{4}\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\\ c,R=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\\ R_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)
Câu 2:
\(1.f\left(x\right)=2x^3+x^2-3x+9.\\ g\left(x\right)=-2x^3-x^2+3.\)
Câu 3:
1. Ta có: \(AB=AC\) (\(\Delta ABC\) cân).
Mà \(AC=IC\left(gt\right). \)
\(\Rightarrow AB=IC.\)
Ta có: \(\widehat{ABC}=\widehat{ACB}\) (\(\Delta ABC\) cân).
Mà \(\widehat{ACB}=\widehat{ICE}\) (đối đỉnh).
\(\Rightarrow\widehat{ABC}=\widehat{ICE}.\)
Hay \(\widehat{ABD}=\widehat{ICE}.\)
Xét \(\Delta ABD\) và \(\Delta ICE:\)
BD = CE (gt).
\(\widehat{ABD}=\widehat{ICE}\left(cmt\right).\)
AB = IC (cmt).
\(\Rightarrow\Delta ABD=\Delta ICE\left(c-g-c\right).\)
2. Xét \(\Delta BDM\) và \(\Delta CEN:\)
\(\widehat{MBD}=\widehat{NCE}\left(\widehat{ABD}=\widehat{ICE}\right).\)
\(BD=CE\left(gt\right).\)
\(\widehat{BDM}=\widehat{CEN}\left(=90^o\right).\)
\(\Rightarrow\Delta BDM=\Delta CEN\left(g-c-g\right).\)
\(\Rightarrow BM=CN\) (2 cạnh tương ứng).
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A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+.....+ \(\dfrac{1}{134}\)+ \(\dfrac{1}{268}\)
A \(\times\) 2 = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) +.....+ \(\dfrac{1}{134}\)
A \(\times\) 2 - A = 1 - \(\dfrac{1}{268}\)
A \(\times\) ( 2 - 1) = \(\dfrac{267}{268}\)
A = \(\dfrac{267}{268}\)
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x3 + 3x = 12.11
=> x(3+3) = 12.11
=> 6x = 12.11
=> 6x - 12.11 = 0
=> 6( x - 2.11) = 0
=> 6 (x- 22) = 0
=> x-22=0
=> x = 22
vậy x = 22
sửa
(5/2*x-3)/15=3/10
5/2*x-3=3/10*15
5/2*x-3=9/2
5/2*x=9/2+3=9/2+6/2
5/2*x=15/2
x=15/2:5/2
x=3
\(5x-6x-9x=-100\)
\(\Rightarrow x\left(5-6-9\right)=-100\)
\(\Rightarrow x.-10=-100\)
\(x=-100:-10\)
\(x=10\)
=> (5-6-9)x=-100
=>-10x=-100
=>x= -100 : -10
=> x = 10
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