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a) \(\dfrac{5}{6}:x=30:3\)
\(\Leftrightarrow\dfrac{5}{6}:x=10\)
\(\Leftrightarrow x=\dfrac{5}{6}:10\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy .......
b) \(x:2,5=0,003:0,75\)
\(\Leftrightarrow x:2,5=0,004\)
\(\Leftrightarrow x=0,004.2,5\)
\(\Leftrightarrow x=0,01\)
Vậy .......
c) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
\(\Leftrightarrow3,8:\left(2x\right)=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{3}{32}\)
\(\Leftrightarrow2x=3,8:\dfrac{3}{32}\)
\(\Leftrightarrow2x=\dfrac{698}{25}\)
\(\Leftrightarrow x=\dfrac{304}{15}\)
Vậy ...
d) \(\dfrac{2}{3}:0,4=x:\dfrac{4}{5}\)
\(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{8}{15}\)
Vậy ....
e) \(3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(\Leftrightarrow0,25:x=\dfrac{19}{5}:\dfrac{608}{15}\)
\(\Leftrightarrow0,25x=\dfrac{57}{608}\)
\(\Leftrightarrow x=\dfrac{228}{608}\)
Vậy ...
e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow xx=\left(-60\right)\left(-15\right)\)
\(\Leftrightarrow x^2=900\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=30^2\\x^2=\left(-30\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy ...
\(a,x^2=16\)
\(x^2=4^2=\left(-4\right)^2\)
\(x=2\) hoặc \(x=-2\)
\(b,x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(x=-2\)
\(c,\left(x+2\right)^2=4\)
\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)
\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)
\(d,\left(1-x\right)^3=1\)
\(1-x=1\)
\(x=0\)
e,phần này mk chưa nghĩ ra,sorry bn nha!
a)\(Từ\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)
⇒\(2\left(x-1\right)=\left(-15\right).\left(-60\right)\)
\(2\left(x-1\right)=900\)
\(\Rightarrow x-1=900:2\)
\(x-1=450\)
\(\Rightarrow x=450-1=449\)
b)\(\left|x+\dfrac{4}{5}\right|+\dfrac{3}{5}=\dfrac{2}{5}\)
\(\Rightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{2}{5}-\dfrac{3}{5}\)
\(\Rightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{-1}{5}\)
Do \(\left|x+\dfrac{4}{5}\right|\ge0\Rightarrow với\) \(\left|x+\dfrac{4}{5}\right|=\dfrac{-1}{5}\) thì x ϵ ∅
c)\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow\)\(x-\dfrac{1}{2}=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{3}\)
\(x=\dfrac{3}{6}+\dfrac{2}{6}\)
\(x=\dfrac{5}{6}\)
a) \(\frac{x-1}{-15}=\frac{-60}{x-1}\)
\(\left(x-1\right)^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=300^2\\\left(x-1\right)^2=\left(-300\right)^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-1=300\\x-1=-300\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=301\\x=-299\end{cases}}\)
b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)
\(\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)
\(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)
vì \(\left|x+\frac{4}{5}\right|\ge0\forall x\)mà \(\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)
\(\Rightarrow\)không có giá trị x nào thỏa mãn đề bài trên
c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
a) \(\Leftrightarrow\left(x-1\right)\left(x-1\right)=\left(-60\right).\left(-15\right)\)
\(\Leftrightarrow\left(x-1\right)^2=900=30^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=30\\x-1=-30\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30+1\\x=-30+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=31\\x=-29\end{cases}}}\)
Vậy x = 31 hoặc x = - 29
b) \(\left|x+\frac{4}{5}\right|+\frac{3}{5}=\frac{2}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{2}{5}-\frac{3}{5}\)
\(\Leftrightarrow\left|x+\frac{4}{5}\right|=\frac{-1}{5}\)vô lý không có giá trị tuyệt đối của số nào mà nhận giá trị âm
Vậy ko có giá trị nào của x thỏa mãn
c) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
a: =>x+1/2=5
=>x=9/2
b: =>(x-1)^2=900
=>x-1=30 hoặc x-1=-30
=>x=-29 hoặc x=31
Bài 1 :
\(\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{-3}{4}\\ \Rightarrow\dfrac{1}{2}x=\dfrac{-19}{12}\\ \Rightarrow x=\dfrac{-19}{12}\cdot2=-\dfrac{19}{6}\)
Bài 2 :
\(a)\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{30}{5}=6\\ \Rightarrow\dfrac{x}{2}=6\Rightarrow x=12\\ \dfrac{y}{3}=6\Rightarrow y=18\\ b)\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{15}{-4}\\ \Rightarrow\dfrac{x}{3}=\dfrac{-15}{4}\Rightarrow x=\dfrac{-45}{4}\\ \dfrac{y}{7}=\dfrac{-15}{4}\Rightarrow4y=-105\Rightarrow y=\dfrac{-105}{4}\)
Bài 1 :
a,\(\dfrac{1}{2}x\)+\(\dfrac{5}{6}\)=\(\dfrac{-3}{4}\)
\(\Rightarrow\)\(\dfrac{1}{2}x\)=\(\dfrac{-3}{4}\)-\(\dfrac{5}{6}\)
\(\Rightarrow\)\(\dfrac{1}{2}x\)=\(\dfrac{-18}{24}\)-\(\dfrac{20}{24}\)
\(\Rightarrow\)\(\dfrac{1}{2}x\)=\(\dfrac{-38}{24}\)
\(\Rightarrow\)\(\dfrac{1}{2}x\)=\(\dfrac{-19}{12}\)
\(\Rightarrow\)x =\(\dfrac{-19}{12}\):\(\dfrac{1}{2}\)
\(\Rightarrow\)x =\(\dfrac{-19}{12}\).2
\(\Rightarrow\)x=\(\dfrac{-19}{6}\)
Vậy x=\(\dfrac{-19}{6}\)
Bài 2:
a,x+y=30 và \(\dfrac{x}{2}=\dfrac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}\)=\(\dfrac{x+y}{2+3}\)=\(\dfrac{30}{5}\)=6
\(\dfrac{x}{2}\)=6\(\Rightarrow\)x=2.6=12
\(\dfrac{y}{3}\)=6\(\Rightarrow\)y=3.6=18
Vậy x=12,y=18
b,x-y=15 và \(\dfrac{x}{3}=\dfrac{y}{7}\)
Đặt \(\dfrac{x}{3}\),\(\dfrac{y}{7}\)=k
\(\Rightarrow\)x=3k,y=7k
Thay x=3k,y=7k vào x-y=15 ta có :
3k-7k=15
\(\Rightarrow\)-4k=15
\(\Rightarrow\)k=\(\dfrac{-15}{4}\)
x=3k\(\Rightarrow\)x=3.\(\dfrac{-15}{4}\)=\(\dfrac{-45}{4}\)
y=7k\(\Rightarrow\)y=7.\(\dfrac{-15}{4}\)=\(\dfrac{-105}{4}\)
Vậy x=\(\dfrac{-45}{4}\),y=\(\dfrac{-105}{4}\)
Nếu đúng thì tick cho mk nha 
3, Tìm x, biết
\(d,\dfrac{-16}{x}=\dfrac{x}{-4}=>x^2=\left(-16\right).\left(-4\right)=>x^2=64\)
\(=>x=8\) hay \(x=-8\)
\(e,\dfrac{x}{-2}=\dfrac{\dfrac{8}{25}}{-x}=>-x^2=-2.\dfrac{8}{5}=\dfrac{-16}{25}\)
\(=>-x^2=0,64=>x=0,8\)
\(g,\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(=>x^2=\left(-15\right).\left(-60\right)\)\(=>x^2=900=>x=30\) hay \(x=-30\)
d) \(\dfrac{-16}{x}=\dfrac{x}{-4}\)
= 16 . 4 = x.x
= 64 = \(x^2\)
= \(8^2=x^2\)
vậy x = 8
e)\(\dfrac{x}{-2}=\dfrac{8}{\dfrac{25}{-x}}\)
= -2 . \(\dfrac{8}{25}\) = -x . x
= -0,64 = \(-x^2\)
= 0,64 = \(x^2\)
0,8\(^2=x^2\)
vậy x = 0,8
g) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
= -15 . -60 = x.x
= 900 = \(x^2\)
30 \(^2=x^2\)
vậy x = 30
a) \(x^2=\left(-15\right).\left(-60\right)=900=>x=\)\(\pm\)\(30\)
b) \(-x^2=\dfrac{-16}{25}=>x^2=\dfrac{16}{25}=>x=\)\(\pm\)\(\dfrac{4}{5}\)
a)\(\dfrac{x}{-15}\)= \(-\dfrac{60}{x}\)
=> x . x = -15 . (-60)
=> \(^{x^2}\) = 900
x = 30
b) \(-\dfrac{2}{x}\) = \(-\dfrac{x}{\dfrac{8}{25}}\)
=> -2 . \(\dfrac{8}{25}\) = x . (-x)
=> \(\dfrac{-16}{25}\) = \(^{x^2}\)
=> x = \(\dfrac{4}{5}\)và \(-\dfrac{4}{5}\)
nhớ tích cho mk vs nha >_<
\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)
Chọn A
mik nghĩ A