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1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)

\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)
=> \(x=56\)
2) => 18x = 18
=> x = 1
3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)
=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)
=> \(x=\dfrac{-1}{2}\)
4) 45%.x =\(\dfrac{3}{5}\)
=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)
=> \(x=\dfrac{4}{3}\)

\(x-\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
\(x=\frac{3}{6}+\left(\frac{1}{9}-\frac{3}{5}\right)\)
\(x=\frac{3}{6}+\left(-\frac{22}{45}\right)\)
\(x=\frac{1}{90}\)
\(\frac{-12}{25}.\frac{3}{4}-x+\frac{6}{-11}-\frac{5}{6}=0\)
\(\frac{-9}{25}-x+\left(\frac{-91}{66}\right)=0\)
\(\frac{-9}{25}-x=0+\left(\frac{-91}{66}\right)\)
\(\frac{-9}{25}-x=\frac{-91}{66}\)
\(x=\left(\frac{-9}{25}\right)-\left(\frac{-91}{66}\right)\)
\(x=\frac{1681}{1650}\)

\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_

Bài 6:
a: \(x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14-3}{21}=\dfrac{-17}{21}\)
d: \(x=\dfrac{9}{10}\cdot\dfrac{-5}{9}=\dfrac{-1}{2}\)
e: \(\Leftrightarrow x\cdot\dfrac{1}{3}=\dfrac{14}{21}-\dfrac{3}{21}=\dfrac{11}{21}\)
=>x=11/7

câu a : \(\frac{1}{7}=\frac{8}{-x}\Rightarrow\frac{8}{56}=\frac{8}{-x}\)
\(\Rightarrow-x=56\)
\(\Rightarrow x=-56\)
câu b
\(\left(x-2\frac{1}{4}\right):\left(-\frac{5}{6}\right)=3\)
\(\Rightarrow x-2\frac{1}{4}=3.\left(-\frac{5}{6}\right)\)
\(\Rightarrow x-2\frac{1}{4}=\frac{-15}{6}\)
đến đây thực hiện tìm x dễ rồi

g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
Phương trình:
\(\frac{1}{9} + \left(\left(\right. x - \frac{1}{3} \left.\right)\right)^{2} = \frac{5}{6}\)
Bước 1: Trừ \(\frac{1}{9}\) hai vế:
\(\left(\left(\right. x - \frac{1}{3} \left.\right)\right)^{2} = \frac{5}{6} - \frac{1}{9}\)
Quy đồng mẫu:
Vậy:
\(\left(\left(\right. x - \frac{1}{3} \left.\right)\right)^{2} = \frac{15}{18} - \frac{2}{18} = \frac{13}{18}\)
Bước 2: Lấy căn hai vế:
\(x - \frac{1}{3} = \pm \sqrt{\frac{13}{18}}\)
Bước 3: Giải ra \(x\):
\(x = \frac{1}{3} \pm \sqrt{\frac{13}{18}}\)
Kết quả:
Đây là nghiệm dưới dạng căn thức. Nếu bạn muốn gần đúng:
Vậy:
sao em có toán 6 ?🤔