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a: \(=\dfrac{-3}{5}\cdot\dfrac{5}{7}+\dfrac{-3}{5}\cdot\dfrac{3}{7}+\dfrac{-3}{5}\cdot\dfrac{6}{7}\)
\(=\dfrac{-3}{5}\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)=\dfrac{-3}{5}\cdot2=-\dfrac{6}{5}\)
b: \(=\dfrac{3}{13}\cdot\dfrac{6}{11}+\dfrac{3}{13}\cdot\dfrac{5}{11}-\dfrac{2}{13}=\dfrac{3}{13}-\dfrac{2}{13}=\dfrac{1}{13}\)
c: =>1/2x+1+3/8=7/16
=>1/2x=-15/16
=>x=-15/8
d: =>5/2x-1/3=1/6*(-9)/2=-9/12=-3/4
=>5/2x=-3/4+1/3=-9/12+4/12=-5/12
=>x=-1/6
1; 5.22 + (\(x\) + 3) = 52
5.4 + (\(x\) + 3) = 25
20 + (\(x\) + 3) = 25
\(x\) + 3 = 25 - 20
\(x+3\) = 5
\(x\) = 5 - 3
\(x\) = 2
Vậy \(x=2\)
2; 23 + (\(x\) - 32) = 53 - 43
8 + (\(x\) - 9) = 125 - 64
8 + (\(x\) - 9) = 61
\(x\) - 9 = 61 - 8
\(x\) - 9 = 53
\(x\) = 53 + 9
\(x\) = 62
Vậy \(x\) = 62
làm ơn giúp mình ,bạn nào làm nhanh đúng mình chọn cho nhanh nha mọi người
1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
1: =>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
2: =>7/6x=5/2:3,75=2/3
=>x=2/3:7/6=2/3*6/7=12/21=4/7
3: =>2x-3=0 hoặc 6-2x=0
=>x=3 hoặc x=3/2
4: =>-5x-1-1/2x+1/3=3/2x-5/6
=>-11/2x-3/2x=-5/6-1/3+1
=>-7x=-1/6
=>x=1/42
Bài 1:
a) Ta có: \(x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2;-2\right\}\)
b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)
\(\Leftrightarrow2x-3-3x-x+5=40\)
\(\Leftrightarrow-2x+2=40\)
\(\Leftrightarrow-2x=38\)
hay x=-19
Vậy: x=-19
Bài 2:
a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)
\(=-45\cdot\left(12+34+54\right)\)
\(=-45\cdot100\)
\(=-4500\)
b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)
\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)
\(=43\cdot57-33\cdot57\)
\(=57\cdot\left(43-33\right)\)
\(=57\cdot10=570\)
\(-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow-5x-\frac{1}{5}-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\left(-5x-\frac{1}{2}x\right)+\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\left(\frac{-10}{2}x-\frac{1}{2}x\right)+\left(\frac{5}{15}-\frac{3}{15}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\frac{-11}{2}x+\frac{2}{15}=\frac{3}{2}x-\frac{5}{6}\)
\(\Leftrightarrow\frac{-11}{2}x-\frac{3}{2}x=-\frac{5}{6}-\frac{2}{15}\)
\(\Leftrightarrow\frac{-14}{2}x=-\frac{25}{30}-\frac{4}{30}\)
\(\Leftrightarrow-7x=-\frac{29}{30}\)
\(\Leftrightarrow x=-\frac{29}{30}\times\frac{-1}{7}\)
\(\Leftrightarrow x=\frac{29}{210}\)
\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\Leftrightarrow3x-\frac{3}{2}-5x-3=\frac{1}{5}-x\)
\(\Leftrightarrow\left(3x-5x\right)-\left(\frac{3}{2}+3\right)=\frac{1}{5}-x\)
\(\Leftrightarrow-2x-\left(\frac{3}{2}+\frac{6}{2}\right)=\frac{1}{5}-x\)
\(\Leftrightarrow-2x-\frac{9}{2}=\frac{1}{5}-x\)
\(\Leftrightarrow-2x+x=\frac{1}{5}+\frac{9}{2}\)
\(\Leftrightarrow-x=\frac{2}{10}+\frac{45}{10}\)
\(\Leftrightarrow-x=\frac{47}{10}\)
\(\Leftrightarrow x=\frac{-47}{10}\)
\(6^2-\left(x+3\right)=45\)
\(36-\left(x+3\right)=45\)
\(x+3=35-45\)
\(x+3=-10\)
\(x=-13\)
lớp 6 mà bạn