2/3 + 1/3 : 3x = 20%
giúp mình với :(
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Ta có : \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)
\(\Leftrightarrow12x^2-8x+9x-6=2x-3x^2-2+3x\)
\(\Leftrightarrow12x^2-8x+9x-6-2x+3x^2+2-3x=0\)
\(\Leftrightarrow15x^2-4x-4=0\)
\(\Leftrightarrow15x^2-10x+6x-4=0\)
Lỗi :vvvv
\(\Leftrightarrow10x\left(\dfrac{3}{2}x-1\right)+4\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left(10x+4\right)\left(\dfrac{3}{2}x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...
\(\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\dfrac{36}{49}\\ \Rightarrow\left(\dfrac{1}{2}-\dfrac{x}{3}\right)^2=\left(\dfrac{6}{7}\right)^2\\ \Rightarrow\dfrac{1}{2}-\dfrac{x}{3}=\pm\dfrac{6}{7}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-\dfrac{x}{3}=\dfrac{6}{7}\\\dfrac{1}{2}-\dfrac{x}{3}=-\dfrac{6}{7}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{5}{14}\\\dfrac{x}{3}=\dfrac{19}{14}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{14}\times3\\x=\dfrac{19}{14}\times3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{15}{14}\\x=\dfrac{57}{14}\end{matrix}\right.\)
\(\left(3-\dfrac{2}{3}x\right)^3=-\dfrac{1}{64}\\ \Rightarrow\left(3-\dfrac{2}{3}x\right)^3=\left(-\dfrac{1}{4}\right)^3\\ \Rightarrow3-\dfrac{2}{3}x=-\dfrac{1}{4}\\ \Rightarrow\dfrac{2}{3}x=3-\left(-\dfrac{1}{4}\right)\\ \Rightarrow\dfrac{2}{3}x=\dfrac{13}{4}\\ \Rightarrow x=\dfrac{13}{4}:\dfrac{2}{3}\\ \Rightarrow x=\dfrac{13}{4}\times\dfrac{3}{2}\\ \Rightarrow x=\dfrac{39}{8}\)
Hic 2 câu em làm dr xong tự nhiên thử lung tung rồi lại xóa bài ;-;
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
\(\left(\dfrac{2x}{3}-\dfrac{1}{3}\right)+\left(3x-2x+1\right)=8\)
\(\Leftrightarrow\dfrac{2x-1}{3}+x-7=0\Rightarrow2x-1+3x-21=0\Leftrightarrow x=\dfrac{22}{5}\)
\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)+\left[3x-2\left(x-1\right)\right]=8\)
\(\Rightarrow\dfrac{2}{3}x-\dfrac{1}{3}+3x-2x+2=8\)
\(\Rightarrow\dfrac{5}{3}x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{5}\)
\(\frac{2}{3}+\frac{1}{3}:3x=20\%\)
\(\frac{2}{3}+\frac{1}{9}x=\frac{1}{5}\)
\(\frac{1}{9}x=\frac{1}{5}-\frac{2}{3}\)
\(\frac{1}{9}x=\frac{-7}{15}\)
\(x=\frac{-7}{15}:\frac{1}{9}\)
\(x=\frac{-21}{5}\)
\(\frac{2}{3}+\frac{1}{3}:3x=20\%\)
\(< =>\frac{1}{3}:3x=20\%-\frac{2}{3}\)
\(< =>\frac{1}{3}:3x=\frac{-7}{15}\)
\(< =>3x=\frac{-5}{7}\)
\(< =>x=\frac{-5}{21}\)
K CHO MIK NHA