bài 1:Tìm x,biết:
a)(3x)2:33=\(\frac{1}{243}\)
b)(5.x=1)2=\(\frac{36}{49}\)
c)2.(\(\frac{3}{4}\)- 5.x)=\(\frac{4}{5}\)- 3.x
Tìm x, biết:
a)\(\frac{2}{9}:x + \frac{5}{6} = 0,5;\)
b)\(\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3};\)
c)\(1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75;\)
d)\(\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\).
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 \(x = \frac{{ - 9}}{{10}}\).
Tìm x, biết :
a, \(60\%x+0,4x+x:3=2\)
b, \(\left|2x-5\right|-7=\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right)...\left(\frac{1}{49}-\frac{1}{2015^2}\right)\)
c, \(\frac{x+1}{1}+\frac{2x+3}{3}+\frac{3x+5}{5}+...+\frac{20x+39}{39}=22+\frac{4}{3}+\frac{6}{5}+...+\frac{40}{39}\)
a. 60%x + 0,4x + x : 3 = 2
0.6x + 0,4x + x : 3 = 2
x(0,6 + 0,4 : 3 ) = 2
\(x.\frac{1}{3}=2=>x=2:\frac{1}{3}=\frac{1}{6}\)
câu B tự làm nha .
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
Tìm x biết:
a) \(x:1\frac{2}{7} = - 3,5\)
b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)
a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)
\(\begin{array}{l}x:1\frac{2}{7} = - 3,5\\x:\frac{9}{7} = - \frac{7}{2}\\x = - \frac{7}{2}.\frac{9}{7}\\x = - \frac{9}{2}\end{array}\)
b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)
\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)
Tìm x, biết:
a)\(x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\)
b)\(\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3};\)
c)\(\frac{2}{5}:x = \frac{1}{{16}}:0,125\)
d)\( - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\)
a)
\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)
Vậy \(x = \frac{{ - 3}}{2}\).
b)
\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)
Vậy \(x = \frac{{ - 5}}{6}\).
c)
\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)
Vậy \(x = \frac{4}{5}\)
d)
\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)
Vậy \(x = \frac{{ - 2}}{5}\).
Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.
bài 1: Tìm x,y biết rằng:
\(x+(-\frac{31}{12})^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x biết:
a.\(5^x.\left(5^3\right)^2=625\) b.\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\) c.\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d.\(172x^2-7^9:98^3=2^{-3}\)
Bài 3: Tìm x \(\varepsilon\)N biết:
a.\(8< 2^x\le2^9\times2^{-5}\) b.\(27< 81^3:3^x< 243\) \(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}\times\left(-\frac{2}{5}\right)^2\)c.
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
Tìm x, biết:
a)\(x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\);
b)\(3,7 - x = \frac{7}{{10}};\)
c)\(x.\frac{3}{2} = 2,4\);
d)\(3,2:x = - \frac{6}{{11}}\).
a)
\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{15}}\).
b)
\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)
Vậy \(x = 3\).
c)
\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)
Vậy \(x = \frac{8}{5}\)
d)
\(\begin{array}{l}3,2:x = - \frac{6}{{11}}\\\frac{{16}}{5}:x = - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 88}}{{15}}\).
Đề bài
Giải mỗi bất phương trình sau:
a) \({3^x} > \frac{1}{{243}}\)
b) \({\left( {\frac{2}{3}} \right)^{3x - 7}} \le \frac{3}{2}\)
c) \({4^{x + 3}} \ge {32^x}\)
d) \(\log (x - 1) < 0\)
e) \({\log _{\frac{1}{5}}}(2x - 1) \ge {\log _{\frac{1}{5}}}(x + 3)\)
f) \(\ln (x + 3) \ge \ln (2x - 8)\)
\(a,3^x>\dfrac{1}{243}\\ \Leftrightarrow3^x>3^{-5}\\ \Leftrightarrow x>-5\\ b,\left(\dfrac{2}{3}\right)^{3x-7}\le\dfrac{3}{2}\\ \Leftrightarrow3x-7\le1\\ \Leftrightarrow3x\le8\\ \Leftrightarrow x\le\dfrac{8}{3}\\ c,4^{x+3}\ge32^x\\ \Leftrightarrow2^{2x+6}\ge2^{5x}\\ \Leftrightarrow2x+6\ge5x\\ \Leftrightarrow3x\le6\\ \Leftrightarrow x\le2\)
d, Điều kiện: x > 1
\(log\left(x-1\right)< 0\\ \Leftrightarrow x-1< 1\\ \Leftrightarrow1< x< 2\)
e, Điều kiện: \(x>\dfrac{1}{2}\)
\(log_{\dfrac{1}{5}}\left(2x-1\right)\ge log_{\dfrac{1}{5}}\left(x+3\right)\\ \Leftrightarrow2x-1\ge x+3\\ \Leftrightarrow x\ge4\)
f, Điều kiện: x > 4
\(ln\left(x+3\right)\ge ln\left(2x-8\right)\\ \Leftrightarrow x+3\ge2x-8\\\Leftrightarrow4< x\le11\)
Tìm x biết
a) ( 2x - 4 ) 2 =\(\frac{36}{49}\)
b) ( 3x - 5 ) 2 =\(\frac{36}{25}\)
c) \(|1-x|+0,73=3\)
d) \(|x+\frac{3}{4}|-5=-2\)
a, (2x - 4)^2 = 36/49
=> 2x - 4 = 6/7 hoặc 2x - 4 = -6/7
=> 2x = 34/7 hoặc x = 22/7
=> x = 34/14 hoặc x = 22/14
b, tương tự a
c, |1 - x| + 0,73 = 3
=> |1 - x| = 2,23
=> 1 - x = 2,23 hoặc 1 - x = -2,23
=> x = -1,23 hoặc x = 3,23
d, tương tự c
a) \(\left(2x-4\right)^2=\frac{36}{49}=\frac{6^2}{7^2}=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow2x-4=\frac{6}{7}\Rightarrow2x=\frac{34}{7}\Rightarrow x=\frac{17}{7}\)
b) \(\left(3x-5\right)^2=\frac{36}{25}=\frac{6^2}{5^2}=\left(\frac{6}{5}\right)^2\)
\(\Rightarrow3x-5=\frac{6}{5}\Rightarrow3x=\frac{31}{5}\Rightarrow x=\frac{31}{15}\)
c)\(\left|1-x\right|+0,73=3\Rightarrow\left|1-x\right|=2,27\)
\(\orbr{\begin{cases}TH1.1-x=2,27\Rightarrow x=-1,27\\TH2.1-x=-2,27\Rightarrow x=3,27\end{cases}}\)
Vậy, x=......
d) \(\left|x+\frac{3}{4}\right|-5=-2\Rightarrow\left|x+\frac{3}{4}\right|=3\)
\(\orbr{\begin{cases}TH1.x+\frac{3}{4}=3\Rightarrow x=\frac{9}{4}\\TH2.x+\frac{3}{4}=-3\Rightarrow x=-3,75\end{cases}}\)
Vậy, x=.......
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