(10mủ 5-10x-4)+2mủ5 =89968
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3x(8^2-2(2^5-1))=2022
=>3x(64-2*31)=2022
=>3x=1011
=>x=337
3x[8² - 2(2⁵ - 1)] = 2022
3x[64 - 2(32 - 1)] = 2022
3x(64 - 2.31) = 2022
3x(64 - 62) = 2022
3x.2 = 2022
6x = 2022
x = 2022 : 6
x = 337
\(\frac{3}{10}\)x \((\frac{-5}{9}+\frac{3}{5})\)\(+\frac{3}{10}x\)\((\frac{-4}{9}+\frac{2}{5})\)= \(\frac{3}{10}x\)\((\frac{-2}{45})\)+ \(\frac{3}{10}x\) \(\times\frac{2}{45}\)= \(\frac{3}{10}x\left(\frac{-2}{45}+\frac{2}{45}\right)\)= \(\frac{3}{10}x\times0\)= 0
Chúc bạn học tốt !
= 6x2 + 21x -2x - 7 - 6x2 + 5x + 6x - 5 - 10x - 12
= 20x - 19
\(a,=x\left(x^2-10x+25\right)=x\left(x-5\right)^2\\ b,=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\\ c,=\left(x-5\right)^2\\ d,=\left(x-8\right)\left(x+8\right)\)
Ta có: \(\frac{\left(x^2\right)^2-10x^2+9}{x^4+6x^3+9x^2+2x^3+12x^2+18x+x^2+6x+9}\)
= \(\frac{\left(x^2-1\right)\left(x^2-3\right)}{x^2\left(x^2+6x+9\right)+2x\left(x^2+6x+9\right)+\left(x^2+6x+9\right)}\)
= \(\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}{\left(x^2+6x+9\right)\left(x^2+2x+1\right)}\)
= \(\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)^2.\left(x+1\right)^2}\)
= \(\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)\left(x+1\right)\left(x+1\right)}\)
= \(\frac{\left(x-1\right)\left(x-3\right)}{\left(x+1\right)\left(x+3\right)}\)
x=(căn bậc hai(127)+5^(3/2))^(1/3)/(4^(1/3)*căn bậc hai(5))-4^(1/3)/(2*căn bậc hai(5)*(căn bậc hai(127)+5^(3/2))^(1/3))
; x = -((căn bậc hai(127)+5^(3/2))^(2/3)*(căn bậc hai(3)*i+1)+2^(1/3)*căn bậc hai(3)*i-2^(1/3))/(2^(5/3)*căn bậc hai(5)*(căn bậc hai(127)+5^(3/2))^(1/3))
;x = ((căn bậc hai(127)+5^(3/2))^(2/3)*(căn bậc hai(3)*i-1)+2^(1/3)*căn bậc hai(3)*i+2^(1/3))/(2^(5/3)*căn bậc hai(5)*(căn bậc hai(127)+5^(3/2))^(1/3));
mủ sao?????
(105 - 10x -4) + 25 = 89968