Tìm x: \(\left(x-\frac{3}{4}\right)^3=\left(\frac{2}{3}\right)^6\) thanks!
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NT
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Những câu hỏi liên quan
31 tháng 3 2017
\(\Rightarrow\frac{3}{4}x+5-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}+3\)+3
\(\Rightarrow\left(\frac{3}{4}x-\frac{2}{3}x-\frac{1}{6}x\right)+\left(5+4-1\right)=\frac{1}{3}x+\left(4-\frac{1}{3}+3\right)\)
=>\(\frac{-1}{12}x+8=\frac{1}{3}x+\frac{20}{3}\)\(\Rightarrow\frac{-1}{12}x+8-\frac{1}{3}x=\frac{20}{3}\)
\(\Rightarrow\left(\frac{-1}{12}-\frac{1}{3}\right)x+8=\frac{20}{3}\)
\(\Rightarrow\frac{-5}{12}x+8=\frac{20}{3}\Rightarrow\frac{-5}{12}x=\frac{20}{3}-8\)
\(\Rightarrow\frac{-5}{12}x=\frac{-4}{3}\Rightarrow x=\frac{-4}{3}:\frac{-5}{12}=\frac{16}{5}\)
31 tháng 7 2016
a) \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\)\(\frac{21\left(4x+3\right)-15\left(6x-2\right)}{105}=\frac{35\left(5x+4\right)+315}{105}\)
\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+315\)
\(\Leftrightarrow84x+63-90x+30=175x+140+315\)
\(\Leftrightarrow84x-90x-175x=140+315-63-30\)
\(\Leftrightarrow-181x=362\)
\(\Leftrightarrow x=-2\)
b)\(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x+4\right)^2}{6}=0\)
\(\Leftrightarrow\)\(\frac{8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x+4\right)^2}{24}=0\)
\(\Leftrightarrow8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2+8x+16\right)=0\)
\(\Leftrightarrow8x^2-32x+32-12x^2+27+4x^2+32x+64=0\)
\(\Leftrightarrow8x^2-12x^2+4x^2-32x+32x=-64-27-32\)
\(\Leftrightarrow0x=-123\) (vô nghiệm)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)
Vậy \(x = \frac{{ - 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ - 9}}{{10}}\).
TT
3
12 tháng 5 2016
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)=\frac{4}{3}.\frac{-1}{3}=\frac{-4}{9}\)
k nha
S
24 tháng 5 2016
a)
\(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
\(\frac{9}{12}x-\frac{8}{12}x-\frac{2}{12}x=4+3-3-4+1\)
\(\left(\frac{9}{12}-\frac{8}{12}-\frac{2}{12}\right)x=1\)
\(-\frac{1}{12}x=1\)
\(x=-12\)
b)
\(\frac{x+1}{3}=\frac{x-2}{4}\)
\(\Rightarrow4\left(x+1\right)=3\left(x-2\right)\)
\(4x+4=3x-6\)
\(4x-3x=-6-4\)
\(x=-10\)
4 tháng 10 2015
đk: \(\begin{cases}x+2\ne0\\4-x>0\\6+x>0\end{cases}\)
ta có \(3\log_{\frac{1}{4}}\left(x+2\right)-3=3\log_{\frac{1}{4}}\left(4-x\right)+3\log_{\frac{1}{4}}\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right)-\log_{\frac{1}{4}}\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right).\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\frac{x+2}{4}=\left(4-x\right)\left(6+x\right)\)
giải pt tìm ra x
đối chiếu với đk của bài ta suy ra đc nghiệm của pt
\(\left(x-\frac{3}{4}\right)^3=\left(\frac{2}{3}\right)^6\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)
\(\Leftrightarrow x-\frac{3}{4}=\left(\frac{2}{3}\right)^2\)
\(\Leftrightarrow x=\frac{43}{36}\)