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\(a,x^2-11x+30=0\\ \Leftrightarrow x^2-5x-6x+30=0\\ \Leftrightarrow x\left(x-5\right)-6\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
\(b,\Delta=\left(-8\right)^2-4.3\left(-5\right)=64+60=124\)
\(x_1=\dfrac{8+\sqrt{124}}{2.3}=\dfrac{8+2\sqrt{31}}{6}=\dfrac{4+\sqrt{31}}{3}\)
\(x_1=\dfrac{8-\sqrt{124}}{2.3}=\dfrac{8-2\sqrt{31}}{6}=\dfrac{4-\sqrt{31}}{3}\)
a) \(\frac{6^7}{4^3\cdot9^2}=\frac{2^7\cdot3^7}{2^6\cdot3^4}=2\cdot3^3=2\cdot27=54\)
b) \(\frac{12^3\cdot15^3}{4^3\cdot25^2\cdot9^2}=\frac{2^6\cdot3^3\cdot3^3\cdot5^3}{2^6\cdot5^4\cdot3^4}=\frac{3^2}{5}=1,8\)
c) \(\frac{2^{11}+3\cdot2^{10}}{10\cdot4^5}=\frac{2^{10}\left(2+3\right)}{2\cdot5\cdot2^{10}}=\frac{1}{2}=0,5\)
d) \(\frac{3^8\cdot2-3^6}{2\cdot17\cdot3^7}=\frac{3^6\left(3^2\cdot2-1\right)}{2\cdot17\cdot3^7}=\frac{1}{2\cdot3}=\frac{1}{6}\)
a) \(\left(x+5\right).6-7=29\)
\(\Rightarrow\left(x+5\right).6=29+7\)
\(\Rightarrow\left(x+5\right).6=36\)
\(\Rightarrow\left(x+5\right)=36:6=6\)
\(\Rightarrow x=6-5=1\)
b) \(5x-3x=12-3.2\)
\(\Rightarrow2x=6\Rightarrow x=6:2=3\)
c) \(\dfrac{1}{4}.3< x< \dfrac{51}{50}.4\)
\(\Rightarrow\dfrac{3}{4}< x< \dfrac{102}{25}\)
\(3x^2-5=11\)
\(3x^2=11+5\)
\(3x^2=16\)
\(3x^2=4^2\)
\(\Rightarrow3x=4\)
\(x=4\div3\)
\(x=1,333....\)
Nếu thế thì không có số nào thỏa mãn đề bài.