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\(a,\dfrac{x}{6}=\dfrac{7}{3}\Rightarrow x=\dfrac{6.7}{3}\Rightarrow x=14\)
\(b,\dfrac{20}{x}=\dfrac{-12}{15}\Rightarrow x=\dfrac{20.15}{-12}\Rightarrow x=-25\)
\(c,\dfrac{-15}{35}=\dfrac{27}{x}\Rightarrow x=\dfrac{35.27}{-15}\Rightarrow x=-63\)
\(d,\dfrac{\dfrac{4}{5}}{1\dfrac{2}{5}}=\dfrac{2\dfrac{2}{5}}{x}\Rightarrow\dfrac{\dfrac{4}{5}}{\dfrac{7}{5}}=\dfrac{\dfrac{12}{5}}{x}\Rightarrow x=\dfrac{\dfrac{7}{5}.\dfrac{12}{5}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{\dfrac{84}{25}}{\dfrac{4}{5}}\Rightarrow x=\dfrac{21}{5}\)
\(e,\dfrac{x}{1\dfrac{1}{4}}=\dfrac{5}{2}\Rightarrow\dfrac{x}{\dfrac{5}{4}}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{2}.\dfrac{5}{4}\Rightarrow x=\dfrac{25}{8}\)
\(f,\dfrac{\dfrac{1}{2}}{1\dfrac{1}{4}}=\dfrac{x}{3\dfrac{1}{3}}\Rightarrow\dfrac{\dfrac{1}{2}}{\dfrac{5}{4}}=\dfrac{x}{\dfrac{10}{3}}\Rightarrow x=\dfrac{\dfrac{10}{3}.\dfrac{1}{2}}{\dfrac{5}{4}}\Rightarrow x=\dfrac{\dfrac{5}{3}}{\dfrac{5}{4}}\Rightarrow x=\dfrac{4}{3}\)
1)
a) \(\frac{x}{6}\)= \(\frac{7}{3}\)
\(\Rightarrow\)x.3=6.7
\(\Rightarrow\)x.3=42
\(\Rightarrow\)x =42:3
\(\Rightarrow\)x =14
b) làm tương tự như câu a
c) làm tương tự như câu
d) làm tương tư như câu a nhưng hơi phúc tạp một chút là bn phải đổi ra từ hỗn số ra phân số hoặc số nguyên
e) tương tự câu d
f) làm tương tự như câu d
2)
a) 3x:\(\frac{27}{10}\)=\(\frac{1}{3}\): \(2\frac{1}{4}\)
3x: \(\frac{27}{10}\) = \(\frac{1}{3}\): \(\frac{9}{4}\)
3x: \(\frac{27}{10}\) = \(\frac{4}{27}\)
3x = \(\frac{4}{27}\). \(\frac{27}{10}\)
3x = \(\frac{2}{5}\)
x = \(\frac{2}{5}\): 3
x = \(\frac{2}{15}\)
Các câu còn lại bn làm tương tự như câu a nha
3)
Làm tương tự như bài 2 nha
mik khuyên bn nếu bn giải bài thì bn nên đổi ra cùng một kiểu số thì tốt hơn như số số thập phân thì thập phân hết ấy
Cuối cùng chúc bn học giỏi
a, \(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\Rightarrow\) \(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(\Rightarrow x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(\Rightarrow x=\dfrac{-1}{6}\)
b, \(25-\left(5-x\right)=-7\)
\(\Rightarrow\) \(5-x=25-\left(-7\right)\)
\(\Rightarrow5-x=32\)
\(\Rightarrow x=5-32\)
\(\Rightarrow x=-27\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{-5}{7}\)
d, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow\) \(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=0+\dfrac{1}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e, \(\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|-7=-3\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=-3+7\)
\(\Rightarrow\left|\dfrac{1}{2}x-\dfrac{3}{4}\right|=4\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4+\dfrac{3}{4}\\\dfrac{1}{2}x=-4+\dfrac{3}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{19}{4}\\\dfrac{1}{2}x=\dfrac{-13}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{4}:\dfrac{1}{2}\\x=\dfrac{-13}{4}:\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=\dfrac{-13}{2}\end{matrix}\right.\)
a)\(\dfrac{3}{7}+\dfrac{4}{7}x=\dfrac{1}{3}\)
\(\dfrac{4}{7}x=\dfrac{1}{3}-\dfrac{3}{7}\)
\(\dfrac{4}{7}x=\dfrac{-2}{21}\)
\(x=\dfrac{-2}{21}:\dfrac{4}{7}\)
\(x=\dfrac{-1}{6}\)
b)\(25-\left(5-x\right)=-7\)
\(5-x=25-\left(-7\right)\)
\(5-x=32\)
x= -27
c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(x=\dfrac{-5}{7}\)
d)\(2x\left(x-\dfrac{1}{7}\right)=0\)
⇒\(\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
e)\(|\dfrac{1}{2}x-\dfrac{3}{7}|-7=-3\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=-3+7\)
\(\left|\dfrac{1}{2}x-\dfrac{3}{7}\right|=4\)
⇒\(\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{3}{4}=4\\\dfrac{1}{2}x-\dfrac{3}{4}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=4\dfrac{3}{4}\Rightarrow x=9\dfrac{1}{2}=\dfrac{19}{2}\\\dfrac{1}{2}x=-3\dfrac{1}{4}\Rightarrow x=\dfrac{-13}{2}\end{matrix}\right.\)
a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)
\(x=\dfrac{-7}{10}\)
b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)
\(x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\)
\(x+\dfrac{5}{6}=\dfrac{16}{15}\)
\(x=\dfrac{16}{15}-\dfrac{5}{6}\)
\(x=\dfrac{7}{30}\)
c) \(1\dfrac{2}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)
\(\dfrac{7}{5}x=\dfrac{-43}{35}\)
\(\Rightarrow x=\dfrac{-43}{49}\)
d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)
\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)
\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{4}\)
\(x=\dfrac{-5}{12}\)
e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)
\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)
\(x+\dfrac{4}{5}=2,15-3,75\)
\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)
\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)
\(x=\dfrac{-12}{5}\)
f) \(\left(x-2\right)^2=1\)
\(\Rightarrow x=1\)
Sức chịu đựng có giới hạn -.-
- Mình tiếp tục cho Nguyễn Phương Trâm nhé.
g, \(\left(2x-1\right)^3=-27\)
\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)
\(\Rightarrow2x-1=-3\)
\(\Rightarrow2x=-2\)
=> \(x=-1\)
- Vậy x = -1
h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)
\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)
\(\Rightarrow\left(x-1\right)^2=900 \)
\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)
=> x = 31
i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)
=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{16}\)
- Vậy x=\(\dfrac{1}{16}\)
j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)
\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}\)
- Vạy x = \(\dfrac{3}{4}\)
k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)
=>\(4^x=4\)
=> x = 1
- Vậy x = 1
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
b) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\dfrac{1}{4}:x=\dfrac{7}{20}\)
\(x=\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{5}{7}\)
\(\left|x-3\right|=\dfrac{1}{2}\)
\(x-3=\pm\dfrac{1}{2}\)