Cho A =2x^3 ; B =-3x^4 . Biết đơn thức A và B cùng dấu. So sánh x với 0
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\(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\)
=>\(\dfrac{5}{6}x=-\dfrac{4}{15}+\dfrac{2}{5}=\dfrac{2}{15}\)
=>\(x=\dfrac{2}{15}:\dfrac{5}{6}=\dfrac{2}{15}\cdot\dfrac{6}{5}=\dfrac{12}{75}=\dfrac{4}{25}\)
\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)
<=> \(\left[{}\begin{matrix}x+\dfrac{5}{3}=0\\x-\dfrac{5}{4}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\\ TH1:x+\dfrac{5}{3}=0\\ =>x=\dfrac{-5}{3}\\ TH2:x-\dfrac{5}{4}=0\\ =>x=\dfrac{5}{4}\)
Vậy: ...
\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1.5+\dfrac{-3}{5}:x\right)=0\left(x\ne0\right)\\ TH1:\dfrac{3}{4}x-\dfrac{9}{16}=0\\ =>\dfrac{3}{4}x=\dfrac{9}{16}\\ =>x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{3}{4}\left(tm\right)\\ TH2:1,5+\dfrac{-3}{5}:x=0\\ =>\dfrac{3}{5}:x=\dfrac{3}{2}\\ =>x=\dfrac{3}{5}:\dfrac{3}{2}=\dfrac{2}{5}\left(tm\right)\)
\(\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{3}{4}-2\dfrac{1}{4}:1,\left(3\right)\)
\(=\dfrac{3}{64}-\dfrac{9}{4}:\dfrac{4}{3}\)
\(=\dfrac{3}{64}-\dfrac{27}{16}=\dfrac{3}{64}-\dfrac{108}{64}=-\dfrac{105}{64}\)
\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\\ \Rightarrow\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\\\Rightarrow\dfrac{7}{4}:x=\dfrac{1}{6} \\ \Rightarrow x=\dfrac{7}{4}:\dfrac{1}{6}\\ \Rightarrow x=\dfrac{21}{2}\)
Vậy \(x=\dfrac{21}{2}\)
\(\dfrac{2}{3}+\dfrac{7}{4}:x=\dfrac{5}{6}\)
=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{2}{3}\)
=> \(\dfrac{7}{4}:x=\dfrac{5}{6}-\dfrac{4}{6}\)
=> \(\dfrac{7}{4}:x=\dfrac{1}{6}\)
=> \(x=\dfrac{7}{4}:\dfrac{1}{6}\)
=> x = \(\dfrac{7}{4}.6\)
=> \(x=\dfrac{21}{2}\)
Vậy ...
\(\dfrac{5^4.18^4}{125.9^5.16}\\ =\dfrac{5^4.\left(2.3^2\right)^4}{5^3.\left(3^2\right)^5.2^4}\\ =\dfrac{5^4.2^4.3^8}{5^3.2^4.3^{10}}\\ =\dfrac{5}{3^2}\\ =\dfrac{5}{9}\)
26.
\(a.\left(\dfrac{3}{7}\right)^5\cdot x=\left(\dfrac{3}{7}\right)^7\\ =>x=\left(\dfrac{3}{7}\right)^7:\left(\dfrac{3}{7}\right)^5\\ =>x=\left(\dfrac{3}{7}\right)^{7-5}\\ =>x=\left(\dfrac{3}{7}\right)^2\\ =>x=\dfrac{9}{49}\\ b.\left(0,09\right)^3x=-\left(0,09\right)^2\\ =>\left[\left(0,3\right)^2\right]^3\cdot x=-\left[\left(0,3\right)^2\right]^2\\ =>x=-\left(0,3\right)^4:\left(0,3\right)^6\\ =>x=-\left(0,3\right)^{-2}\\ =>x=-\left(\dfrac{10}{3}\right)^2\\ =>x=-\dfrac{100}{9}\)
27:
a: Vì \(0< \dfrac{1}{2}< 1\)
và 40<50
nên \(\left(\dfrac{1}{2}\right)^{40}>\left(\dfrac{1}{2}\right)^{50}\)
b: \(243^3=\left(3^5\right)^3=3^{15};125^5=\left(5^3\right)^5=5^{15}\)
mà 3<5
nên \(243^3< 125^5\)
A và B cùng dấu nên AB>0
=>\(2x^3\cdot\left(-3\right)x^4>0\)
=>\(x^7< 0\)
=>x<0
có nhân 2 với -3 k ạ