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1)
2x.(x-2) - x.(2x+1) = 3
=> 2x2 - 4x - 2x2 - x = 3
=> (2x2 - 2x2 ) - (4x+x) = 3
=> -5x = 3
=> x = \(\dfrac{-3}{5}\)
2) (2x-1).(x-2) - (x+3).(2x-7) = 3
=> 2x2 - 4x - x + 2 - 2x2 + 7x - 6x + 21 = 3
=> (2x2 - 2x2) - (4x + 6x + x - 7x) + 2 + 21 = 3
=> -4x = -20
=> x = -20 : (-4)
=> x = 5
3) (x - 5).(-x + 4) - (x - 1).(x + 3) = -2x2
=> Bạn tách tương tự như mấy câu 2 nhé! Nếu không làm được thì bảo mình
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a)Ta có:
\(\left\{{}\begin{matrix}x+y=\frac{1}{3}\\y+z=\frac{-1}{4}\\z+x=\frac{1}{5}\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{3}+\frac{-1}{4}+\frac{1}{5}\)
\(\Rightarrow2\left(x+y+z\right)=\frac{17}{60}\)
\(\Rightarrow x+y+z=\frac{17}{60}:2=\frac{17}{120}\)
\(\Rightarrow\left\{{}\begin{matrix}z=\frac{-23}{120}\\x=\frac{47}{120}\\y=\frac{-7}{120}\end{matrix}\right.\)
b)Ta có:
\(\left\{{}\begin{matrix}xy=\frac{3}{5}\\yz=\frac{4}{5}\\zx=\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow xyyzzx=\frac{3}{5}.\frac{4}{5}.\frac{3}{4}=\frac{9}{25}\)
\(\Rightarrow\left(xyz\right)^2=\frac{9}{25}\Rightarrow\left[{}\begin{matrix}xyz=\frac{3}{5}\\xyz=-\frac{3}{5}\end{matrix}\right.\)
TH1: \(xyz=\frac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}z=1\\x=\frac{3}{4}\\y=\frac{4}{5}\end{matrix}\right.\)
TH2:
\(xyz=-\frac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}z=-1\\x=-\frac{3}{4}\\y=-\frac{4}{5}\end{matrix}\right.\)
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a,
\(x:\frac{3}{4}+\frac{1}{4}=-\frac{2}{3}\)
\(x:\frac{3}{4}=-\frac{2}{3}-\frac{1}{4}=-\frac{11}{12}\)
\(x=-\frac{11}{12}.\frac{3}{4}=-\frac{11}{16}\)
a)\(x:\frac{3}{4}+\frac{1}{4}=-\frac{2}{3}\)
\(x:\frac{3}{4}=-\frac{2}{3}-\frac{1}{4}\)
\(x:\frac{3}{4}=-\frac{11}{12}\)
\(x=-\frac{11}{16}\)
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ở hàng thứ 3 tính cả đề, ở phân số thứ 2 trên tử là số 3 ak bn???
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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 \(x=\dfrac{5}{2}\)
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a, 11/12 - ( 2/5 + x ) = 2/3
<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)
b, 2x . ( x - 1/7 ) = 0
<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
vậy x={\(0;\frac{1}{7}\)}
c, 3/4 + 1/4 : x = 2/5
<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)
vậy x=-5/7
a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)
\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)
\(\Leftrightarrow x=-\frac{3}{20}\)
b) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow4x=\frac{-20}{7}\)
\(\Leftrightarrow x=-\frac{5}{7}\)