\(\Leftrightarrow-x^2=-9\cdot\dfrac{4}{49}=\dfrac{-36}{49}\\ \Leftrightarrow x^2=\dfrac{36}{49}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{7}\\x=-\dfrac{6}{7}\end{matrix}\right.\)
\(\Leftrightarrow-x^2=-9\cdot\dfrac{4}{49}=\dfrac{-36}{49}\\ \Leftrightarrow x^2=\dfrac{36}{49}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{7}\\x=-\dfrac{6}{7}\end{matrix}\right.\)
Tìm x,y,z biết:
a) \(\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=9\)
b) \(\dfrac{x-3}{12}=\dfrac{-3}{3-x}\)
c) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và \(x-y-z=-49\)
\(x\) x {\(\dfrac{1}{4}\) + \(\dfrac{1}{5}\)} - {\(\dfrac{1}{7}\) + \(\dfrac{1}{8}\)}
2 : \(x\) = \(x\) : \(\dfrac{8}{49}\)
a, (\(\dfrac{1}{2}x-\dfrac{1}{3}\))2 - \(\dfrac{4}{25}=0\) b , (\(1-\dfrac{1}{4}x\) )-\(\dfrac{121}{49}=0\)
bài 7 tìm những giá trị nhuyên dương X thỏa mãn
3) \(\dfrac{-5}{11}\)<\(\dfrac{9}{x}\)<\(\dfrac{-5}{12}\)
4) \(\dfrac{-11}{13}\)<\(\dfrac{9}{x}\)<\(\dfrac{-11}{15}\)
5) \(\dfrac{-4}{5}\)<\(\dfrac{9}{x}\)<\(\dfrac{-4}{7}\)
nhanh cần gấp nhé
a,-12:(3/4-5/6)^2
,b,10.\(\sqrt{0.01}.\sqrt{\dfrac{16}{9}+3\sqrt{49}-\dfrac{1}{6}\sqrt{4}}\)
c,x/6=y/3=z/2 và x-2y+4z=8
d,|1/4+x|-1/3=2/5
Bài 4: tìm x:
a) \(\dfrac{4}{3}\) + (1,25 - x) = 2,25
b) \(\dfrac{17}{6}\) - (x - \(\dfrac{7}{6}\) ) = \(\dfrac{7}{4}\)
c) 4 - (2x + 1) = 3 - \(\dfrac{1}{3}\)
bài 15:
a) (\(\dfrac{-2}{3}\))9 : x = (\(\dfrac{-2}{3}\))
b) x : (\(\dfrac{4}{9}\))5 = (\(\dfrac{4}{9}\))4
c) (x + 4)3 = -125
d) (10 - 5x)3 = 64
e) (4x + 5)2 = 81
Bài 16:
a) 4 - \(1\dfrac{2}{5}\) - \(\dfrac{8}{3}\)
b) -0,6 - \(\dfrac{-4}{9}\) - \(\dfrac{16}{15}\)
c) \(-\dfrac{15}{4}\) . (\(\dfrac{-7}{15}\)) . (\(-2\dfrac{2}{5}\)
Gi ải gấp giúp mình ạ, mình rất cần gấp
\(5x-9=5+3x;2^3+0,5x=1,5;\left(5x+1\right)^2=\dfrac{36}{49};\left(\dfrac{-3}{81}\right)^x=-27;2^{x-1}=16\)
a) \(\dfrac{3}{4}=\dfrac{3x}{20}\) b) \(\dfrac{1,2}{x+3}=\dfrac{5}{4}\) c) \(\dfrac{x^2}{32}=\dfrac{9}{8}\)
\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)