\(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)

\(\dfrac{5}{3}:\dfrac{x}{4}=20\)

\(\dfrac{x}{4}=\dfrac{5}{3}:20\)

\(\dfrac{x}{4}=\dfrac{1}{12}\)

\(x=\dfrac{1}{3}\)

Giải:

a) \(\left(\dfrac{1}{3}.x\right):\dfrac{2}{3}=1\dfrac{3}{4}:\dfrac{2}{5}\)

\(\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{7}{4}:\dfrac{2}{5}\)

\(\Leftrightarrow\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{35}{8}\)

\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{8}.\dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{1}{3}x=\dfrac{35}{12}\)

\(\Leftrightarrow x=\dfrac{35}{12}:\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{35}{4}\)

Vậy \(x=\dfrac{35}{4}\).

b) \(4,5:0,3=2,25\left(0,1.x\right)\)

\(\Leftrightarrow15=2,25\left(0,1.x\right)\)

\(\Leftrightarrow2,25\left(0,1.x\right)=15\)

\(\Leftrightarrow0,1.x=\dfrac{15}{2,25}\)

\(\Leftrightarrow0,1.x=\dfrac{20}{3}\)

\(\Leftrightarrow x=\dfrac{20}{3}:0,1\)

\(\Leftrightarrow x=\dfrac{200}{3}\)

Vậy \(x=\dfrac{200}{3}\).

c) \(8:\left(\dfrac{1}{4}.x\right)=2:0,02\)

\(\Leftrightarrow8:\left(\dfrac{1}{4}.x\right)=100\)

\(\Leftrightarrow\dfrac{1}{4}.x=\dfrac{2}{25}\)

\(\Leftrightarrow x=\dfrac{2}{25}:\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Vậy \(x=\dfrac{8}{25}\).

d) \(3:2\dfrac{1}{4}=\dfrac{3}{4}:\left(6.x\right)\)

\(\Leftrightarrow3:\dfrac{9}{4}=\dfrac{3}{4}:\left(6.x\right)\)

\(\Leftrightarrow\dfrac{4}{3}=\dfrac{3}{4}:\left(6.x\right)\)

\(\Leftrightarrow\dfrac{4}{3}:\left(6.x\right)=\dfrac{3}{4}\)

\(\Leftrightarrow6.x=\dfrac{4}{3}:\dfrac{3}{4}\)

\(\Leftrightarrow6.x=\dfrac{16}{9}\)

\(\Leftrightarrow x=\dfrac{16}{9}:6\)

\(\Leftrightarrow x=\dfrac{8}{27}\)

Vậy \(x=\dfrac{8}{27}\).

Chúc bạn học tốt!!!

\(B=\dfrac{\dfrac{2}{10}-\dfrac{3}{8}+\dfrac{5}{11}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\)\(-\dfrac{1}{3}\)

\(B=\dfrac{-2}{3}-\dfrac{1}{3}=-1\)

6)a) \(\left|\dfrac{5}{3}:x\right|=\left|\dfrac{-1}{6}\right|\)

⇒ \(\left|\dfrac{5}{3}:x\right|=\dfrac{1}{6}\)

⇒ \(\dfrac{5}{3}:x=\dfrac{1}{6}\) hoặc \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)

*TH1 : \(\dfrac{5}{3}:x=\dfrac{1}{6}\)

⇒ \(x=\dfrac{5}{3}:\dfrac{1}{6}=10\)

*TH2 : \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)

⇒ \(x=\dfrac{5}{3}:\dfrac{-1}{6}=-10\)

Vậy \(x\) ∈ \(\left\{10;-10\right\}\)

\(b,\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|\dfrac{-3}{4}\right|\)

⇒ \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)

⇒\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)

⇒ \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\) hoặc \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)

TH1 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\)

⇒ \(\dfrac{3}{4}x=\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)

⇒\(x=\dfrac{9}{4}:\dfrac{3}{4}=3\)

TH2 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)

⇒ \(\dfrac{3}{4}x=\dfrac{-3}{2}+\dfrac{3}{4}=\dfrac{-3}{4}\)

⇒ \(x=\dfrac{-3}{4}:\dfrac{3}{4}=-1\)

Vậy \(x\) ∈ \(\left\{3;1\right\}\)

1: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{0,3}=\dfrac{y}{0.2}=\dfrac{z}{0.1}=\dfrac{x-y}{0.3-0.2}=\dfrac{1}{0.1}=10\)

Do đó: x=3; y=2; z=1

ai giúp với mình sắp đi học rồi

1/

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{x-y}{5-6}=\dfrac{36}{-1}=-36\)

\(\Rightarrow\left\{{}\begin{matrix}x=-36\cdot5=-180\\y=-36\cdot6=-216\\z=-36\cdot4=-144\end{matrix}\right.\)

2/

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{y+z}{3+4}=\dfrac{28}{7}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=4\cdot3=12\\z=4\cdot4=16\end{matrix}\right.\)

3/

\(\dfrac{x}{1,2}=\dfrac{y}{1,3}\Leftrightarrow\dfrac{2x}{2,4}=\dfrac{y}{1,3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{2,4}=\dfrac{y}{1,3}=\dfrac{2x-y}{2,4-1,3}=\dfrac{5,5}{1,1}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5\cdot2,4}{2}=6\\y=5\cdot1,3=6,5\\z=5\cdot1,4=7\end{matrix}\right.\)

4/

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{x-y}{0,5-0,3}=\dfrac{1}{0,2}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\cdot0,5=2,5\\y=5\cdot0,3=1,5\\z=5\cdot0,2=1\end{matrix}\right.\)

5/

\(z=\dfrac{x}{0,3}\Leftrightarrow z=\dfrac{3x}{0,9}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(z=\dfrac{3x}{0,9}=\dfrac{z-3x}{1-0,9}=\dfrac{1}{0,1}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot0,9}{3}=3\\y=10\cdot0,7=7\\z=10\end{matrix}\right.\)

4\(\dfrac{1}{3}\):\(\dfrac{x}{4}\)=6:0,3

=>\(\dfrac{13}{3}\):\(\dfrac{x}{4}\)=20

=>\(\dfrac{x}{4}\)=\(\dfrac{13}{3}\):20=\(\dfrac{13}{60}\)

=>x\(\in\varnothing\)

\(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)

\(\dfrac{13}{3}:\dfrac{x}{4}=20\)

\(\dfrac{x}{4}=20.\dfrac{13}{3}\)

\(\dfrac{x}{4}=\dfrac{260}{3}\)

\(\dfrac{3x}{12}=\dfrac{1040}{12}\)

\(3x=1040\)

\(x=1040:3\)

\(x=\dfrac{1040}{3}\)

\(4\dfrac{1}{3}:\dfrac{x}{4}=6:0.3\\ \dfrac{13}{3}:\dfrac{x}{4}=20\\ x:4=\dfrac{13}{3}:20\\ x=\dfrac{13}{60}\cdot4\\ x=\dfrac{13}{15}\) tất cả sai hết

a) \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)

\(\dfrac{13}{3}:\dfrac{x}{4}=20\)

\(\dfrac{x}{4}=\dfrac{13}{3}:20\)

\(\dfrac{x}{4}=\dfrac{13}{3}\cdot\dfrac{1}{20}\)

\(\dfrac{x}{4}=\dfrac{13}{60}\)

\(x=\dfrac{13}{60}\cdot4\)

Vậy \(x=\dfrac{13}{15}\)

b)\(2^3:4.2^{\left(x+1\right)}=64\)

\(8:4.2^{\left(x+1\right)}=64\)

\(2.2^{\left(x+1\right)}=64\)

\(2\cdot2^x.2=64\)

\(4.2^x=64\)

\(2^x=64:4\)

\(2^x=16\)

\(2^x=2^4\)

Vậy \(x=4\)

Tìm x

a) x= \(\frac{13}{15}\)

b) x= 4

Chúc bạn học tốt

a, x = \(\dfrac{13}{15}\)

b, x =4

Lời giải:

$(0,3-x).2\frac{1}{4}=\frac{-2}{5}$

$(0,3-x).\frac{9}{4}=\frac{-2}{5}$

$0,3-x=\frac{-2}{5}: \frac{9}{4}=\frac{-8}{45}$

$x=0,3-\frac{-8}{45}=\frac{43}{90}$