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\(a,x-\dfrac{5}{7}=\dfrac{19}{21}\\ x=\dfrac{34}{21}\\ b,\dfrac{5}{3}-\left|x-\dfrac{1}{5}\right|=\dfrac{1}{3}\\ \left|x-\dfrac{1}{5}\right|=\dfrac{4}{3}\\ TH1:x-\dfrac{1}{5}=\dfrac{4}{3}\\ x=\dfrac{23}{15}\\ TH2:x-\dfrac{1}{5}=-\dfrac{4}{3}\\ x=-\dfrac{17}{15}\\ c,x-\dfrac{2}{5}=\dfrac{1}{4}\\ x=\dfrac{13}{20}\\ d,5\sqrt{x}-30=15\\ 5\sqrt{x}=45\\ \sqrt{x}=9\\ x=9^2=81\)
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) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)
\(1.\)
\(a.\)
\(\dfrac{x}{-150}=-\dfrac{6}{x}\)
\(\Rightarrow x^2=\left(-6\right)\left(-150\right)\)
\(\Rightarrow x^2=900\)
\(\Rightarrow x=\pm30\)
\(2.\)
\(a.\) \(2x=3y;5y=7z\) và \(3x-7y+5z=30\)
Ta có : \(2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}\) \(\left(1\right)\)
\(5y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{5}\Rightarrow\dfrac{y}{14}=\dfrac{z}{10}\) \(\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}=\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
\(\Rightarrow\dfrac{x}{21}=2\Rightarrow x=42\)
\(\dfrac{y}{14}=2\Rightarrow y=28\)
\(\dfrac{z}{10}=2\Rightarrow z=20\)
Vậy : ..................
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 
\(x^2=900\Leftrightarrow x^2=30^2\Rightarrow x=30\)
Chọn A
mik nghĩ A