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a: \(\Leftrightarrow x\cdot\dfrac{1}{2}-\dfrac{3}{5}x+\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7}{10}x\)
=>3/5x=-4
hay x=-4:3/5=-20/3
b: \(\Leftrightarrow4x-6-9=5-3x-2\)
=>4x-15=-3x+3
=>7x=18
hay x=18/7
a) Ta có:
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
b)Ta có:
\(\frac{5}{8x-2}=\frac{-4}{7-x}\)
\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)
\(\Rightarrow35-5x=-32x+8\)
\(\Rightarrow-5x+32x=8-35\)
\(\Rightarrow27x=-27\)
\(\Rightarrow x=-1\)
c) Ta có:
\(\frac{4}{3}=\frac{2x-1}{x}\)
\(\Rightarrow4x=3\left(2x-1\right)\)
\(\Rightarrow4x=6x-3\)
\(\Rightarrow3=6x-4x=2x\)
\(\Rightarrow x=\frac{3}{2}\)
d)Ta có:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)
\(\Rightarrow8x-4=9x+3\)
\(\Rightarrow8x-9x=3+4\)
\(\Rightarrow-x=7\Rightarrow x=-7\)
e)Ta có:
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)
\(\Rightarrow12x+4=7x+14\)
\(\Rightarrow12x-7x=14-4\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
f)Ta có:
\(\frac{-3}{x+1}=\frac{4}{2-2x}\)
\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)
\(\Rightarrow-6+6x=4x+4\)
\(\Rightarrow6x-4x=4+6\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
\(x⋮12;25;30\Rightarrow x\in BC\left(12;25;30\right)\)
\(12=2^2.3;25=5^2;30=2.3.5\)
\(BCNN\left(12;25;30\right)=2^2.3.5^2=300\)
\(BC\left(12;25;30\right)=B\left(300\right)\)
\(B\left(300\right)=\left\{0;300;600;....\right\}\)
\(0\le x\le500\Rightarrow x=300\)
\(\left(3x-2^4\right).7^3=2.7^4\)
\(\left(3x-16\right)=2.7\)
\(3x-16=14\)
\(3x=30\)
\(x=10\)
\(\left|x-5\right|=16+2.\left(-3\right)\)
\(\left|x-5\right|=10\)
\(\Leftrightarrow x-5=10\Rightarrow x=15\)
\(\Leftrightarrow x-5=-10\Rightarrow x=-5\)
Để x2 - 3x + 7 \(⋮\)x - 3
=> x(x - 3) + 7 \(⋮\) x - 3
Vì x(x - 3) \(⋮\)x - 3
=> 7 \(⋮\)x - 3
=> \(x-3\inƯ\left(7\right)\)
=> x - 3 \(\in\left\{1;7;-1;-7\right\}\)
=> x \(\in\left\{4;10;2;-4\right\}\)
Vậy x \(\in\left\{4;10;2;-4\right\}\)
Để \(x^2-3x+7⋮x-3\)
Thì \(x\left(x-3\right)+7⋮x-3\)
Vì \(x\left(x-3\right)⋮x-3\)
\(\Rightarrow7⋮x-3\)
\(\Rightarrow x-3\inƯ\left(7\right)\)
\(\Rightarrow x-3\in\left\{1;7;-1;-7\right\}\)
\(\Rightarrow x\in\left\{4;10;2;-4\right\}\)
Vậy ...