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\(a,5\left(x+2\right)^3+7=2\\ \Leftrightarrow5\left(x+2\right)^3=-5\\ \Leftrightarrow\left(x+2\right)^3=-1\\ \Leftrightarrow x+2=-1\\ \Leftrightarrow x=-3\\ b,14-\left|\dfrac{3}{2}x-1\right|=9\\ \Leftrightarrow\left|\dfrac{3}{2}x-1\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-1=5\\\dfrac{3}{2}x-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=6\\\dfrac{3}{2}x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow5\left(x+2\right)^3=-5\\ \Leftrightarrow\left(x+2\right)^3=-1\\ \Leftrightarrow x+2=-1\Leftrightarrow x=-3\\ b,\Leftrightarrow\left|\dfrac{3}{2}x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-1=5\\1-\dfrac{3}{2}x=5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=6\\\dfrac{3}{2}x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{8}{3}\end{matrix}\right.\)
a: \(\dfrac{x}{0.9}=\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
b: \(\dfrac{-6}{x}=\dfrac{9}{-15}\)
\(\Leftrightarrow x=10\)
c: \(\dfrac{\dfrac{14}{15}}{\dfrac{9}{10}}=\dfrac{x}{\dfrac{3}{7}}\)
\(\Leftrightarrow x=\dfrac{3}{7}\cdot\dfrac{14}{15}:\dfrac{9}{10}=\dfrac{2}{5}\cdot\dfrac{10}{9}=\dfrac{20}{45}=\dfrac{4}{9}\)
a) \(\text{}/3x-5/-\frac{1}{7}=\frac{1}{3}\) b)\(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(/3x-5/=\frac{10}{21}\) \([x.\left(\frac{3}{5}-\frac{2}{3}-1\right)]=\frac{-5}{21}.7\)
\(\Rightarrow3x-5=\frac{10}{21}hay3x-5=\frac{-10}{21}\) \(\left[x.\frac{-16}{15}\right]=\frac{-5}{3}\)
\(3x=\frac{115}{21}\) \(3x=\frac{95}{21}\) \(x=\frac{25}{16}\)
\(x=\frac{115}{63}\) \(x=\frac{95}{63}\) Vậy x = \(\frac{25}{16}\)
Vậy x \(\in\left\{\frac{115}{63};\frac{95}{63}\right\}\)
a) \(2.\left|5x-3\right|-2x=14\)
\(2\left|5x-3\right|=14+2x\)
\(\left|5x-3\right|=\frac{14+2x}{2}\)
\(\Rightarrow\orbr{\begin{cases}5x-3=\frac{-14-2x}{2}\\5x-3=\frac{14+2x}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\left(5x-3\right).2=-14-2x\\\left(5x-3\right).2=14+2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}10x-6+2x=-14\\10x-6-2x=14\end{cases}\Rightarrow\orbr{\begin{cases}12x=-14+6\\8x=14+6\end{cases}}}\Rightarrow\orbr{\begin{cases}12x=-8\\8x=20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)
vậy \(\orbr{\begin{cases}x=\frac{-2}{3}\\x=2,5\end{cases}}\)
Những câu sau tương tự nhé.
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
\(/x-\frac{1}{2}/=\frac{1}{3}\\ =>\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{1}{3}+\frac{1}{2}\\x=-\frac{1}{3}+\frac{1}{2}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)
\(a,|x-\frac{1}{2}|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}}\)
\(b,\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\)
\(\frac{28}{27}=x:\frac{3}{7}\)
\(x=\frac{4}{9}\)
Ta có: \(\frac{3}{7}.\left|3x-2\right|=\frac{9}{14}\)
\(\Rightarrow\left|3x-2\right|=\frac{9}{14}:\frac{3}{7}=\frac{3}{2}\)
\(\Rightarrow\orbr{\begin{cases}3x-2=\frac{3}{2}\\3x-2=\frac{-3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}3x=\frac{7}{2}\\3x=\frac{1}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{1}{6}\end{cases}}}\)
Vậy x = 7/6 hoặc x = 1/6
3/7.(3.x-2)=9/14 và = -9/14
3x-2= 9/14:3/7=3/2
x= (3/2+2) :3=7/6
với 3/7(3x-2)=-9/7
3x-2=-9/7 : 3/7=-3
x=(-3+2):3=-1/3
vậy x có hai giá trị 7/6 và -1/3