³√x-20 + √x+15 =7
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a) \(\frac{7}{12}+\frac{x}{15}=\frac{1}{20}\)
=> \(\frac{35}{60}+\frac{4x}{60}=\frac{3}{60}\)
=> 35 + 4x = 3
=> 4x = -32
=> x = -8
b) \(\frac{-7}{x}+\frac{8}{15}=\frac{-1}{20}\)(ĐK \(x\ne0\))
=> \(\frac{-7}{x}=\frac{-1}{20}-\frac{8}{15}\)
=> \(\frac{-7}{x}=\frac{-3}{60}-\frac{32}{60}\)
=> \(\frac{-7}{x}=\frac{-35}{60}\)
=> \(\frac{-7}{x}=\frac{-7}{12}\)
=> x = 12(TM)
tìm x,biết
5-x-16=40+x
4x-10=15-x
15-x=4x-5
x-15=6+4x
-12+x=5x-20
7x-4=20+3x
5x-7=-21-2x
x+15=20-4x
17-x=7-6x
1) 3x - 6= 5x + 2
5x - 3x = -6 - 2
2x = -8
x = -4
2) 15 - x = 4x - 5
4x + x = 15 + 5
5x = 20
x = 4
Tương tự như trên
a) x+15 = 20-4x
=> x=1
b) 17-x=7-6x
=> x=-2
c) -12+x=5x-20
=> x=2
d) 4x-5=15-x
=> x=4
e) 9x-7=20-6x
=>x= \(\frac{9}{5}\)
g)2.(x-10)=3.(x-20)=x-4
=> x thuộc ∅
h) (x^2+2).(x-3) <0
=> x=3,...
a) \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)
\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)
\(=\frac{1}{2}+\frac{1}{3}\)
\(=\frac{5}{6}\)
b) \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)
\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)
\(=1\times\frac{1}{2}=\frac{1}{2}\)
c) \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)
\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)
\(=\frac{247}{420}+\frac{72}{420}\)
\(=\frac{319}{420}\)
a, \(x=-\frac{21}{20}-\frac{-7}{15}\)
\(x=\frac{-7}{12}\)
b, \(\frac{7}{2}-x=\frac{-21}{20}:\frac{5}{4}\)
\(\frac{7}{2}-x=\frac{-21}{25}\)
\(x=\frac{7}{2}-\frac{-21}{25}\)
\(x=\frac{217}{50}\)
\(\sqrt[3]{x}-20+\sqrt{x}+15=7\)
\(\sqrt[3]{x}-20+15+\sqrt{x}=7\)
\(\sqrt[3]{x}-5+\sqrt{x}=7\)
\(\sqrt[3]{x}+\sqrt{x}=7+5\)
\(\sqrt[3]{x}+\sqrt{x}=12\)
còn lại mình chịu
\(\sqrt[3]{x}+\sqrt{x}=12=8+4\)
\(\sqrt[3]{x}=8\) và \(\sqrt{x}=4\)
Vậy x = 2