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a) x2 + x = 0
=> x( x+ 1 ) = 0
=> x = 0
hoặc x = -1
b) b, (x-1)x+2 = (x-1)x+4
=> x + 2 = x + 4
=> 0x = 2 ( ktm)
Vậy ko có giá trị x nào thoả mãn đk
d) Ta có: x-1/x+5 = 6/7
=>(x-1).7 = (x+5).6
=>7x-7 = 6x+ 30
=> 7x-6x = 7+30
=> x = 37
Vậy x = 37
e, x2/ 6= 24/25
=> x2 . 25 = 6 . 24
⇒x2.25=144⇒x2.25=144
⇒x2=144÷25⇒x2=144÷25
⇒x2=5,76=2,42=(−2,42)⇒x2=5,76=2,42=(−2,42)
⇒x∈{2,4;−2,4}⇒x∈{2,4;−2,4}
Vậy x∈{2,4;−2,4}
c: \(=\dfrac{7}{23}\cdot\dfrac{-24-45}{18}=\dfrac{7}{23}\cdot\dfrac{-69}{18}=\dfrac{7}{18}\cdot\left(-3\right)=-\dfrac{7}{6}\)
d: \(=\dfrac{7}{5}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{7}{5}\cdot10=14\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{9^5}=3^{40}-1\)
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)