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3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
\(\dfrac{6}{x}=\dfrac{24}{x-27}\)
Theo định nghĩa phân số bằng nhau, ta có:
\(6\cdot\left(x-27\right)=24x\\ 6x-162=24x\\ -162=24x-6x\\ -162=18x\\ x=\left(-162\right):18\\ x=-9\)
Theo bài ra ta có:
6.(x-27)=x.24
6x-162=x.24
-162=24x-6x
-162=18x
x =(-162):18
x = -9
tick mk nha
b) \(\dfrac{x}{27}=\dfrac{3}{x}\)
\(\Rightarrow x.x=27.3\)
\(\Rightarrow x^2=81\)
\(\Rightarrow x^2=9^2\)
\(\Rightarrow x=9\)
Vậy x=9
bài này áp dụng quy tắc nhân chéo nha :vv
a) \(\dfrac{x}{5}=\dfrac{2}{5}\Leftrightarrow5x=2.5=10\Leftrightarrow x=\dfrac{10}{5}=2\)
b) \(\dfrac{3}{8}=\dfrac{6}{x}\Leftrightarrow3x=6.8=48\Leftrightarrow x=\dfrac{48}{3}=16\)
c)\(\dfrac{1}{9}=\dfrac{x}{27}\Leftrightarrow9x=27\Leftrightarrow x=\dfrac{27}{9}=3\)
d)\(\dfrac{4}{x}=\dfrac{8}{6}\Leftrightarrow8x=4.6=24\Leftrightarrow x=\dfrac{24}{8}=3\)
e) \(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+4x=-6+20\\ \Leftrightarrow7x=14\Leftrightarrow x=\dfrac{14}{7}=2\)
g) \(\dfrac{x}{-2}=\dfrac{-8}{x}\Leftrightarrow x^2=\left(-2\right)\left(-8\right)=16\\ \Rightarrow x=\pm4\)
a,Vì \(\dfrac{5}{6}\)=\(\dfrac{x}{24}\) nên ta có: 5.24:6= 20 \(\Rightarrow\)x =20
Mấy câu sau làm tương tự như vậy.
b,x =21
c,x =9
d,x = -5
a) \(\left(3\dfrac{1}{3}-x\right)1\dfrac{1}{6}=\dfrac{7}{24}\)
\(\Leftrightarrow\left(\dfrac{10}{3}-x\right)\dfrac{7}{6}=\dfrac{7}{24}\)
\(\Leftrightarrow\dfrac{10}{3}-x=\dfrac{7}{24}:\dfrac{7}{6}\)
\(\Leftrightarrow\dfrac{10}{3}-x=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{10}{3}-\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{37}{12}\).
b) \(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)
\(\Leftrightarrow\left(4,5-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)
\(\Leftrightarrow4,5-2x=\dfrac{4}{3}.\dfrac{3}{4}\)
\(\Leftrightarrow4,5-2x=1\)
\(\Leftrightarrow2x=4,5-1\)
\(\Leftrightarrow2x=3,5\)
\(\Leftrightarrow x=\dfrac{35}{2}\).
\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)
\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)
\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)
\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)
\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)
\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)
B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)
B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)
B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)
B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)
B = 12 . \(\dfrac{2}{13}\)
B = \(\dfrac{24}{13}\)
a: \(\Leftrightarrow-\dfrac{720}{150}=-4.8< x< \dfrac{-63}{210}=-0.3\)
mà x là số nguyên
nen \(x\in\left\{-4;-3;-2;-1\right\}\)
b: \(\Leftrightarrow-\dfrac{125}{27}< x< \dfrac{120}{210}=\dfrac{4}{7}\)
mà x là số nguyên
nên \(x\in\left\{-4;-3;-2;-1;0\right\}\)
Ta có :
\(-\dfrac{24}{-6}=\dfrac{x}{3}\)
\(\Rightarrow x=\dfrac{-24\cdot3}{-6}=12\)
=> TA CÓ :
\(\dfrac{12}{3}=\dfrac{4}{y^2}\)
\(\Rightarrow y^2=\dfrac{4\cdot3}{12}=1\)
\(\Rightarrow y=\pm1\)
=> Ta có :
\(\dfrac{4}{1}=\dfrac{z^3}{-2}\)
\(\Rightarrow z^3=\dfrac{4\cdot\left(-2\right)}{1}=-8\)
\(\Rightarrow z=-2\)
Vậy x= 12 ; y = \(\pm1\) ;z=-2
\(\dfrac{6}{x}=\dfrac{24}{x-7}\left(x\ne0;x\ne27\right)\)
suy ra
`6(x-27)=24x`
`=>6x-162-24x=0`
`=>-18x=162`
`=>x=-9(tm)`
\(\dfrac{6}{x}=\dfrac{24}{x-27}\left(dkxd:x\ne0,x\ne27\right)\)
\(\Leftrightarrow\dfrac{6}{x}-\dfrac{24}{x-27}=0\)
\(\Leftrightarrow\dfrac{6\left(x-27\right)-24x}{x\left(x-27\right)}=0\)
\(\Leftrightarrow6x-162-24x=0\)
\(\Leftrightarrow-18x=162\)
\(\Leftrightarrow x=--9\)