a) (6-2x)^ 9=(6-2x)^31
giải giùm mình nha
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\(f\left(3\right)=3a-3=9\)
\(3a=12\Rightarrow a=4\)
\(f\left(5\right)=5a-3=11\)
\(5a=14\Rightarrow a=\dfrac{14}{5}\)
\(f\left(-1\right)=-a-3=6\)
\(-a=9\Rightarrow a=9\)
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\(a,\frac{7}{3}x+5=\frac{9}{6}x-7\)
\(\frac{7}{3}x-\frac{9}{6}x=-7-5\)
\(\left(\frac{7}{3}-\frac{9}{6}\right).x=-12\)
\(\frac{5}{6}.x=-12\)
\(x=\left(-12\right):\frac{5}{6}\)
\(x=-\frac{72}{5}\)
\(b,3x+\frac{6}{4}=5-\frac{7x}{6}\)
\(3x+\frac{7x}{6}=5-\frac{6}{4}\)
\(\left(3+\frac{7}{6}\right).x=\frac{7}{2}\)
\(\frac{25}{6}.x=\frac{7}{2}\)
\(x=\frac{7}{2}:\frac{25}{6}\)
\(x=\frac{21}{25}\)
\(c,\frac{12}{5}x+6=\frac{17}{3}x+12\)
\(\frac{12}{5}x-\frac{17}{3}x=12-6\)
\(-\frac{49}{15}x=6\)
\(x=6:\left(-\frac{49}{15}\right)\)
\(x=-\frac{90}{49}\)
\(d,6\left(3x+7\right)=12\left(2x-4\right)\)
\(18x+42=24x-48\)
\(18x-24x=-48-42\)
\(-6x=-90\)
\(x=15\)
\(e,13\left(2x-9\right)=8\left(3x-13\right)\)
\(26x-117=24x-104\)
\(26x-24x=-104+117\)
\(2x=13\)
\(x=\frac{13}{2}\)
\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)
\(2\cdot\left(2x-6\right)+\left(x-1\right)=2\)
\(\Leftrightarrow4x-12+x-1-2=0\)
\(\Leftrightarrow5x-15=0\)
\(\Leftrightarrow5x=15\)
\(\Leftrightarrow x=3\)
Nhớ ghi dấu ngoặc tránh giải sai.
\(a.\) \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}\)
Ta có:
\(2x+6=2\left(x+3\right)\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
nên \(MTC:\) \(2\left(x-3\right)\left(x+3\right)\)
Do đó: \(\frac{x+4}{2x+6}+\frac{3}{x^2-9}=\frac{x+4}{2\left(x+3\right)}+\frac{3}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+4\right)\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}+\frac{2.3}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2+x-12+6}{2\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x^2-2x+3x-6}{2\left(x-3\right)\left(x+3\right)}=\frac{x\left(x-2\right)+3\left(x-2\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-2\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{x-2}{2\left(x-3\right)}\)
a) x^4 + 2^3-x -2
=x^4 - x^3 + 3x^3 - 3x^2 + 3x^2 - 3x + 2x-2
=x^3.(x-1) + 3x^2.(x-1) + 3x.(x-1)+2.(x-1)
=(x-1).( x^3+ 3x^2 + 3x+2)
=(X+1).(X^3 + 2X^2 + X^2 +2X +X+2)
=(X+1).(X+2).(X^2 +X + 1)
\(A=\frac{2x-2}{3+2x}\)hay \(A=\frac{2x-2}{2x+3}\)
\(A=\frac{2x-2}{2x+3}=\frac{2x+3-5}{2x+3}=1-\frac{5}{2x+3}\)
A là số nguyên thì 2x + 3 là ước nguyên của 5
\(2x+3=1\Rightarrow x=-1\)
\(2x+3=-1\Rightarrow x=-2\)
\(2x+3=5\Rightarrow x=1\)
\(2x+3=-5\Rightarrow x=-4\)
Vậy \(x\in\){ -1 ; -2 ; 1; ;-4}
Ai thấy đúng thì ủng hộ, tháy sai thì góp ý nha !!!