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=> 12y + (1,2 + 1,5 + 1,8 + ... + 4,2 + 4,5) = 61,8
=> 12y + 34,2 = 61,8
=> 12y = 61,8 - 34,2 = 27,6
=> y = 2,3
a) y*(3,6+6,4)=18,9
y*10=18,9
y=1,89
b)(9,4+ 5,3):y=14
14,7:y=14
y=1,05
a) 2y - 12y = 0
\(\Rightarrow\) y ( 2-12) = 0
\(\Rightarrow\) y . (-10) =0
\(\Rightarrow\) y = 0 : (-10) = 0
b) (y-7)(y-8) = 0
\(\Rightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}\Rightarrow}\orbr{\begin{cases}y=7\\y=8\end{cases}}}\)
c) x + x.2+x.3+x.4+...+x.10 = 165
\(\Rightarrow\) x ( 1+2+3+.....+8+9+10) = 165
\(\Rightarrow\)x . \(\frac{\left(1+10\right).10}{2}\)=165
\(\Rightarrow\) x . 55 = 165
\(\Rightarrow x=\frac{165}{55}=3\)
Can you k for me ,Lê Thị Kim Chi!
a) \(2y-12y=0\)
\(\Leftrightarrow-10y=0\)
\(\Leftrightarrow y=0:\left(-10\right)\)
\(\Leftrightarrow y=0\)
b) \(\left(y-7\right)\left(y-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y-7=0\\y-8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}y=0+7\\y=0+8\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}y=7\\y=8\end{cases}}\)
c) \(x+x.2+x.3+......+x.10=165\)
\(\Leftrightarrow x.\left(1+2+3+.....+10\right)=165\)
\(\Leftrightarrow x.55=165\)
\(\Leftrightarrow x=165:55\)
\(\Leftrightarrow x=3\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+1\right|\ge0;\left|x+1,2\right|\ge0;\left|x+1,3\right|\ge0;\left|x+1,4\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\)
mà \(\left|x+1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
nên \(5x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+1,2\right|=x+1,2\\\left|x+1,3\right|=x+1,3\\\left|x+1,4\right|=x+1,4\end{matrix}\right.\)
\(\Rightarrow x+1+x+1,2+x+1,3+x+1,4=5x\)
\(\Rightarrow4x+4,9=5x\Rightarrow x=4,9\)
Chúc bạn học tốt!!!
\(\left|x+1\right|\ge0\) \(\left|x+1,2\right|\ge0\)
\(\left|x+1,3\right|\ge0\) \(\left|x+1,4\right|\ge0\)
\(\Leftrightarrow\left|x+1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\)
\(\Leftrightarrow x+1+x+1,2+x=1,3+x+1,4=5x\)
\(4x+4,9=5x\)
\(\Leftrightarrow x=4,9\)
Ta có: \(1,2\left(\dfrac{2,4y-0,23}{y}-0,05\right)=1,44\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}-0,05=1,2\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}=1,25\)
\(\Rightarrow2,4y-0,23=1,25y\)
\(\Rightarrow2,4y-1,25y=0,23\)
\(\Rightarrow1,15y=0,23\)
\(\Rightarrow y=0,2=\dfrac{1}{5}\)
Vậy \(y=\dfrac{1}{5}.\)
\(1,2.\left(\dfrac{2,4y-0,23}{y}-0,05\right)=1,44\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}-0,05=1,2\)
\(\Rightarrow\dfrac{2,4y-0,23}{y}=1,25\)
\(\Rightarrow2,4y-0,23=1,25y\)
\(\Rightarrow2,4y-1,25y=0,23\)
\(\Rightarrow1,15y=0,23\)
\(\Rightarrow y=\dfrac{0,23}{1,15}=0,2\)
Vậy y = 0,2
y= 1,2 hay y= -1,2