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a) \(|x+4|=\frac{7}{3}\) \(\Rightarrow x+4=\pm\left(\frac{7}{3}\right)\)
TH1: \(x+4=\frac{7}{3}\)
\(x=\frac{7}{3}-4=-\frac{5}{3}\)
TH2: \(x+4=-\frac{7}{3}\)
\(x=-\frac{7}{3}-4=-\frac{19}{3}\)
a) 2|2/3 - x| = 1/2
|2/3 - x| = 1/4
|2/3 - x| = 1/4 hoặc |2/3 - x| = -1/4
Xét 2 TH...
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)
\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)
Vậy A = \(\frac{1}{2}\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
Vậy B = \(\frac{1}{2}\)
a) Vì : \(\left|3x+1\right|\ge0\forall x\)
\(\Rightarrow\left|3x+1\right|+3\ge3\)
Dấu " = " xảy ra khi :
\(\left|3x+1\right|=0\)
=> x = \(-\frac{1}{3}\)
Vậy MinA = 3 <=> x = \(-\frac{1}{3}\)
Mấy con sau cũng làm tương tự nha
ta có |x| luôn luôn lớn hơn hoặc bằng 0
a) Để A đạt GTNN thì |3x+1|=0=>A=3
=>x=1/3
b) Để B đạt GTNN thì |-x+4|=0
=>B=-1/3=>x=4
c)Để C đạt GTNN thì |2-3/2x|=0
=>C=-2/7=>x=4/3
d) Để D đạt GTNN thì |8-3/2x|=0
=>D=-8/9=>x=16/3
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
a.
\(\left|x\right|+\frac{7}{4}=\frac{19}{8}\)
\(\left|x\right|=\frac{19}{8}-\frac{7}{4}\)
\(\left|x\right|=\frac{19-14}{8}\)
\(\left|x\right|=\frac{5}{8}\)
\(x=\pm\frac{5}{8}\)
Vậy \(x=\frac{5}{8}\) hoặc \(x=-\frac{5}{8}\)
b.
\(\left|x\right|-\frac{4}{9}=\frac{2}{3}\)
\(\left|x\right|=\frac{2}{3}+\frac{4}{9}\)
\(\left|x\right|=\frac{6+4}{9}\)
\(\left|x\right|=\frac{10}{9}\)
\(x=\pm\frac{10}{9}\)
Vậy \(x=\frac{10}{9}\) hoặc \(x=-\frac{10}{9}\)
c.
\(\frac{8}{3}-\left|x\right|=-\frac{1}{2}\)
\(\left|x\right|=\frac{8}{3}+\frac{1}{2}\)
\(\left|x\right|=\frac{16+3}{6}\)
\(\left|x\right|=\frac{19}{6}\)
\(x=\pm\frac{19}{6}\)
Vậy \(x=\frac{19}{6}\) hoặc \(x=-\frac{19}{6}\)
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Cảm ơn bạn nhiều nhé.