Tìm X:
X+ X: 1,4x 1,6+ X: 6x 3,6= 222.
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a) x + x : 4 *1,6 + x : 6*3,6 = 222
\(\Rightarrow x+x\cdot0,4+x.0,6=222\)
\(\Rightarrow x.\left(1+0,4+0,6\right)=222\)
\(\Rightarrow x.2=222\)
\(\Rightarrow x=222:2=111\)
b) \(\left(x+0,2\right)+\left(x+0,4\right)+...+\left(x+1\right)=9,6\)
\(\Rightarrow\left(x+x+x+x+x\right)+\left(0,2+0,4+0,6+0,8+1\right)\) =9,6
\(\Rightarrow5.x+\left[\left(1-0,2\right):0,2+1\right].\left(1+0,2\right):2\) = 9,6
\(\Rightarrow5.x+3=9,6\)
\(\Rightarrow5.x=9,6-3=6,6\)
\(\Rightarrow x=6,6:5=1,32\)
(x+0,2)+(x+0,4)+(0,6)+(0,8)+(x+1)=9,6
5x+(0,2+0,4+0,6+0,8+1)=9,6
5x+3=9,6
5x =9,6-3
5x =6,6
x =6,6:5
x =1,32
\(\Leftrightarrow\left(x+2\right)^2\cdot\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-1\end{matrix}\right.\)
1.
\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)
2.
\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt
Bài 1:
Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=6x\left(2x+1\right)\cdot2y\)
\(=12xy\left(2x+1\right)\)
Bài 2:
Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
a: \(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)
hay \(x\in\left\{0;\sqrt{3};-\sqrt{3}\right\}\)
b: \(=\dfrac{x^3-3x^2+6x-8}{x-2}=\dfrac{x^2-2x-x^2+2x+4x-8}{x-2}=x^2-x+4\)
\(6x\left(1-3x\right)+9x\left(2x-7\right)+171=0\)
\(\Leftrightarrow6x-18x^2+18x^2-63x+171=0\)
\(\Leftrightarrow-57x=-171\)
\(\Leftrightarrow x=3\)
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+1}{2015}+1\right)+\left(\frac{x+2}{2014}+1\right)-\left(\frac{x+3}{2013}+1\right)-\left(\frac{x+4}{2012}+1\right)=0\)
\(\Leftrightarrow\)\(\frac{x+2016}{2015}+\frac{x+2016}{2014}-\frac{x+2016}{2013}+\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2016=0\) ( vì \(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\ne0\) )
\(\Leftrightarrow x=-2016\)
X x 1,6 = 3,6 :0,5
X x 1,6 = 7,2
X = 7,2 : 1,6
X = 4,5
8,52 + ( 75,4 - 29,48 ) : 4
= 8,52 + 45,92 :4
= 8,52 + 11,48
= 20
X x 1,6 = 3,6 :0,5
X x 1,6 = 7,2
X= 7,2 : 1,6
X=4,5
8,52 + ( 75,4 - 29,48 ) : 4
= 8,52 + 45,92 : 4
= 8,52 +11,48
=20
X + X : 1,4 x 1,6 + X : 6 x 3,6 = 222
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