CÂU 2: Tìm x, biêt
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\(x=\dfrac{1}{2}+\left(\dfrac{7}{12}:\dfrac{21}{18}\right)=\dfrac{1}{2}+\left(\dfrac{7}{12}\times\dfrac{18}{21}\right)=\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{2}{2}=1\)
X-7/12 : 21/18 = 1/2
X-7/12 =1/2 x21/18
X-7/12 =7/12
X = 7/12 + 7/12
X = 7/6
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\(\frac{x}{1}\)= \(\frac{3}{x+2}\)
=> x (x+2) = 3
=> x+2 = 3
=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left|4x-1\right|=\left|x+2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=x+2\\4x-1=-x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=3\\5x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Số số hạng là :
\(\left(x-2\right):2+1=x:2-1+1=x:2\) (số hạng)
Tổng trên là :
\(\left(x+2\right).x:2:2=42\)
\(\left(x+2\right).x=42.2.2\)
\(\left(x+2\right).x=168\)
\(\left(x+2\right).x=14.12\)
Vậy x = 14
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Với mọi \(x\in R\)ta có:
\(\left|x\right|+\left|x+1\right|+\left|x+2\right|+\left|x+3\right|+\left|x+4\right|\ge0\Leftrightarrow6x\ge0\Leftrightarrow x\ge0\)
Với \(x\ge0\)thì: \(\left|x\right|=x;\left|x+1\right|=x+1;\left|x+2\right|=x+2;\left|x+3\right|=x+3;\left|x+4\right|=x+4\)
\(pt\Leftrightarrow5x+10=6x\Leftrightarrow x=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x/5 + x2 + x/5:1/2=13
x/5 + x.2 + x/5 x 2 = 13
x/5.(1 + 2 ) + 2x = 13
x/5 .3 +2x = 13
3x/5 + 2x = 13
3x + 10x = 65
13x = 65
x = 65 : 13
x = 5
tìm x nguyên biêt : \(\left(x-3\right)^2\) + \(\left(x-2\right)^2\) + \(\left|x-1\right|\)+ x = 2019
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(\left(x-5\right)^2-1=80\)
\(\left(x-5\right)^2=80+1\)
\(\left(x-5\right)^2=81\)
\(\left(x-5\right)^2=9^2\)
\(\left(x-5\right)=9\)
\(x=9+5\)
\(x=14\)
Vậy ...
\(\left(x-5\right)^2-1=80\)
\(\left(x-5\right)^2=80+1\)\(=81\)
\(\left(x-5\right)^2=9^2\)
\(\left(x-5\right)=9\)
\(x=9+5=14\)
tui học lớp 7 nàk kết bạn hong