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\(3x.\left(x-\frac{2}{3}\right)=0\)
\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)
\(3x=0\Rightarrow x=0\)
\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)
Vậy..
a) \(\frac{6x-5}{-7}=\frac{5x-3}{-5}\)
=> -5(6x - 5) = -7(5x - 3)
=> -30x + 25 = -35x + 21
=> -30x + 25 + 35x - 21 = 0
=> (-30x + 35x) + (25 - 21) = 0
=> 5x + 4 = 0
=> 5x = -4
=> x = -4/5
b) \(\frac{12-7x}{-13}=\frac{4-3x}{-5}\)
=> -5(12 - 7x) = -13(4 - 3x)
=> -60 + 35x = -52 + 39x
=> -60 + 35x + 52 - 39x = 0
=> (-60 + 52) + (35x - 39x) = 0
=> -8 - 4x = 0
=> -8 = 4x
=> x = -2
c) \(\frac{2x+4}{7}=\frac{4x-2}{15}\)
=> 15(2x + 4) = 7(4x - 2)
=> 30x + 60 = 28x - 14
=> 30x + 60 - 28x + 14 = 0
=> 2x + 74 = 0
=> 2x = -74
=> x = -37
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
\(\frac{x}{2}-\left(\frac{3}{5}x-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
\(\Rightarrow\frac{5x-6x+26+14+7x}{10}=0\Rightarrow6x+40=0\Rightarrow x=-\frac{20}{3}\)
X=3/7:5/7-3/11:5/11+3/13:5/13
+
1/2:5/4-1/3+1/4:5/6
=3/5-3/5+3/5 + 2/5-1/3+3/10
=3/5 + 11/10
=17/10
\(A=\frac{x^2-10x+36}{x-5}=\frac{x^2-10x+25+9}{x-5}\) \(=\frac{\left(x-5\right)^2+9}{x-5}=x-5+\frac{9}{x-5}\)
để \(A\in Z\)
<=> \(\frac{9}{x-5}\in Z\)mà \(x\in Z\)
=> \(x-5\inƯ\left(9\right)\)
=> \(x-5\in\left(1;-1;3;-3;9;-9\right)\)
=> \(x\in\left(6;4;8;2;14;-4\right)\)
học tốt
\(\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10x}\right)\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10x}\)\(\Leftrightarrow\frac{x}{2}-\frac{3x}{5}+\frac{7}{10x}=-\frac{7}{5}-\frac{13}{5}\)
\(\Leftrightarrow\frac{5x^2}{10x}-\frac{6x^2}{10x}+\frac{7}{10x}=-\frac{20}{5}\)\(\Leftrightarrow\frac{5x^2-6x^2+7}{10x}=-\frac{20}{5}\)
\(\Leftrightarrow\frac{-x^2+7}{10x}=-4\)\(\Leftrightarrow x^2-7=40x\)\(\Leftrightarrow x^2-40x-7=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot20+20^2-20^2-7=0\)\(\Leftrightarrow x^2-2\cdot x\cdot20+20^2=400+7\)
\(\Leftrightarrow\left(x-20\right)^2=407\)\(\Leftrightarrow x-20=\pm\sqrt{407}\)\(\Leftrightarrow x\in\left\{20-\sqrt{407};20+\sqrt{407}\right\}\) (TMĐKXĐ)
Vậy \(S=\left\{20-\sqrt{407};20+\sqrt{407}\right\}\)