tim x biet
a;\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
b; \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
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`2x-15 = 17``
`=> 2x = 17 + 15`
`=> 2x = 32`
`=> X = 32 : 2`
`=> x = 16`
`156 - (x + 61) = 82`
`=> x + 61 = 156 - 82`
`=> x + 61 = 74`
`=> x = 13`
`2x - 138 = 2^3 . 3^2`
`=>2x - 138 = 72`
`=> 2x = 210`
`=> x = 105`
bài 2:
`23-3x = 8`
`=> 3x = 23 - 8`
`=> 3x = 15`
`=> x = 5`
`(x-35) - 120 = 0`
`=>(x-35) = 120`
`=> x = 120 +35`
`=> x = 155`
`3^x + 2 = 29`
`=> 3^x = 27`
`=> 3^x = 3^3`
`=> x = 3`
\(x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a) \(\left|x\right|=x\\ \left|-\dfrac{4}{7}\right|=\dfrac{4}{7}\)
\(\dfrac{1}{2}=\dfrac{x}{-4}\\ \Rightarrow x=\dfrac{1.\left(-4\right)}{2}=\dfrac{-4}{2}=-2\)
a/ \(x^2-25=0\)
\(\Rightarrow\left(x+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\Rightarrow x=-5\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
b/ \(x\left(x+7\right)+x+7=0\)
\(x\left(x+7\right)+\left(x+7\right)=0\)
\(\left(x+7\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
a) x-12=-15
x=-15+12
x=-3
b) 123-5(x+4)=38
5(x+4)=123-38
5(x+4)=85
x+4=85:5=17
x=17-4
x=13
\(a.x-12=-15\) \(b.123-5.\left(x+4\right)=38\)
\(x=\left(-15\right)+12\) \(5.\left(x+4\right)=123-38\)
\(x=-3\) \(5.\left(x+4\right)=85\)
\(x+4=85:5=17\)
\(x=17-4=13\)
a) (x - 6)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)
b) \(\left|x\right|=3\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
a) Ta có: \(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)
\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Leftrightarrow3x\left(5x+1\right)+2\left(5x+1\right)=5x\left(3x-1\right)+7\left(3x-1\right)\)
\(\Leftrightarrow15x^2+3x+10x+2=15x^2-5x+21x-7\)
\(\Leftrightarrow15x^2-15x^2+3x+10x+5x-21x=-7-2\)
\(\Leftrightarrow-3x=-9\)
\(\Leftrightarrow x=3\)
Vậy x = 3
b) Ta có: \(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x\left(x+3\right)+\left(x+3\right)=2x\left(0,5x+2\right)+\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+3x+x+3=x^2+4x+0,5x+2\)
\(\Leftrightarrow x^2-x^2+3x+x-4x-0,5x=2-3\)
\(\Leftrightarrow-0,5x=-1\Leftrightarrow x=2\)
Vậy x = 2
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