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1) \(3\left(5x-2\right)-9=6\)
\(15x-6=6+9\)
\(15x=15+6\)
\(15x=21\)
\(x=\frac{7}{5}\)
2) \(4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\left(3x-2\right)\left(4+7\right)=22\)
\(\left(3x-2\right)\cdot11=22\)
\(\left(3x-2\right)=2\)
\(3x=4\)
\(x=\frac{4}{3}\)
3) \(2x^2-6x=0\)
\(2x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy,............
\(1)3\left(5x-1\right)-9=6\)
\(\Rightarrow3\left(5x-2\right)=15\)
\(\Rightarrow5x-2=5\Rightarrow5x=7\Rightarrow x=\frac{7}{5}\)
\(2)4\left(3x-2\right)+7\left(3x-2\right)=22\)
\(\Rightarrow11\left(3x-2\right)=22\)
\(\Rightarrow33x-22=22\Rightarrow33x=44\Rightarrow x=\frac{4}{3}\)
\(3)2x^2-6x=0\)
\(\Rightarrow2x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
??????????????????????????????????????????????????????????????????????
6x+3 chia hết cho 3x+6 => 6x+3 chia hết cho 2(3x+6) => 6x+3 chia hết cho 6x+12=6x+3+9 => 9 chia hết cho 6x+3 => 6x+3 thuộc Ước của 9 ( tự lm nốt )
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
Quy đồng với mẫu số chung là 30
\(\frac{15x}{30}+\frac{30x+10}{30}+\frac{-36x-54}{30}=\frac{5}{30}\)
=> 9x - 44 = 5
=> x = \(\frac{49}{9}\)
\(3x^2+6x+6=3\)
=>\(x^2+2x+2=1\)
=>\(x^2+2x+1=0\)
=>\(\left(x+1\right)^2=0\)
=>x+1=0
=>x=-1
Có đúng là toán 6 không em?