Tìm giá trị của x biết:
a) x - 3/5 = 1 - 4/5 + 1/6
b) ( - 10/4 ) + 1/4 = 3/4
c) -3/2 + ( x - 1/2 ) = 1/2
d) x - 2/7 = 7/2 1
đ) ( 1/6 + 1/10 - 1/15 ) + x = 0
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C = 59/10 : 3/2 - ( 5/3 x 9/2 x 14/3) : 7/4
= 59/15 - 35 : 7/4
= 59/15 - 20
= -241/15
\(a,\dfrac{-1}{8}=\dfrac{3}{x}\\ \dfrac{3}{-24}=\dfrac{3}{x}\\ x=-24\\ b,\dfrac{x}{3}=\dfrac{3}{x}\\ x.x=3.3\\ x^2=9\\ x=\pm3\\ c,\dfrac{3}{4}.x=1\dfrac{1}{2}\\ \dfrac{3}{4}.x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:\dfrac{3}{4}\\ x=2\\ d,x-\dfrac{3}{10}=\dfrac{7}{15}:\dfrac{3}{5}\\ x-\dfrac{3}{10}=\dfrac{7}{9}\\ x=\dfrac{7}{9}+\dfrac{3}{10}\\ x=\dfrac{97}{90}\\ e,\dfrac{-4}{7}-x=\dfrac{-8}{3}.\dfrac{3}{7}\\ \dfrac{-4}{7}-x=\dfrac{-8}{7}\\ x=\dfrac{-4}{7}+\dfrac{8}{7}\\ x=\dfrac{4}{7}\\ \)
Bài 3:
a: =>3x^2-6x-x-3x^2=14
=>-7x=14
=>x=-2
b: \(\Leftrightarrow2x^2+10x-x-5-2x^2-9x-x-4.5=3.5\)
=>-x-9,5=3,5
=>-x=12
=>x=-12
c: =>\(3x-3x^2+9x=36\)
=>-3x^2+12x-36=0
=>x^2-6x+12=0(loại)
d: \(\Leftrightarrow3x^2-3x+x-1+4x-3x^2=5\)
=>2x=6
=>x=3
Câu 4 :
\(x^2+y^2-2\left(x-y-1\right)=0\)
\(\Leftrightarrow x^2+y^2-2x+2y+2=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+1\right)^2=0\)
Do \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+1\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Vậy \(x=1;y=-1\)
a; \(x\) - \(\dfrac{3}{5}\) = 1 - \(\dfrac{4}{5}\) + \(\dfrac{1}{6}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{30}{30}\) - \(\dfrac{24}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{6}{30}\) + \(\dfrac{5}{30}\)
\(x\) - \(\dfrac{3}{5}\) = \(\dfrac{11}{30}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{3}{5}\)
\(x\) = \(\dfrac{11}{30}\) + \(\dfrac{18}{30}\)
\(x\) = \(\dfrac{29}{30}\)
Vậy \(x\) = \(\dfrac{29}{30}\)
b; (- \(\dfrac{10}{4}\)) + \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) thế \(x\) của em đâu nhỉ???
c; - \(\dfrac{3}{2}\) + (\(x\) - \(\dfrac{1}{2}\)) = \(\dfrac{1}{2}\)
\(x\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\)
\(x\) - \(\dfrac{1}{2}\) = 2
\(x\) = 2 + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{4}{2}\) + \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)