Tìm x biết:
a) 2x-1 = 16
b) \(\frac{\left(-3\right)^x}{81}=-27\)
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15 tháng 11 2021
a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)
\(\Rightarrow2x^2+x-6-4x^2+22x-10=-16\)
\(\Rightarrow2x^2-23x=0\Rightarrow x\left(2x-23\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)
b) \(7x^2-7=x^2-2x+1\)
\(\Rightarrow7\left(x^2-1\right)-\left(x^2-2x+1\right)=0\)
\(\Rightarrow7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)\left(7x+7-x+1\right)=0\Rightarrow2\left(x-1\right)\left(3x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)
MH
15 tháng 11 2021
a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)
\(2x^2+x-6-4x^2+22x-10=-16\)
\(-2x^2+23x-16=-16\)
\(23x-2x^2=0\)
\(x\left(23-2x\right)=0\)
⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)
b) \(7x^2-7=x^2-2x+1\)
\(7\left(x^2-1\right)=\left(x-1\right)^2\)
\(7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)
\(\left(7x+7\right)\left(x-1\right)-\left(x-1\right)^2=0\)
\(\left(x-1\right)\left(7x+7-x+1\right)=0\)
\(\left(x-1\right)\left(6x+8\right)=0\)
⇔ \(\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)
DA
1
28 tháng 9 2021
\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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.
4 tháng 10 2021
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
28 tháng 5 2019
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
Chắc Sai kết quả chứ công thức đúng nha!!!...
Fighting!!!...
28 tháng 5 2019
Đặt:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)
Giải phương trình:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
11 tháng 11 2021
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)
\(\Rightarrow3x-6x^2+6x+14=29\)
\(\Rightarrow-6x^2+9x-15=0\)
\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)
\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)
Vậy \(S=\varnothing\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
6 tháng 12 2023
\(a,1-3\left|2x-3\right|=-\dfrac{1}{2}\\ 3\left|2x-3\right|=1+\dfrac{1}{2}\\ 3\left|2x-3\right|=\dfrac{3}{2}\\ \left|2x-3\right|=\dfrac{3}{2}:3\\ \left|2x-3\right|=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{9}{2}\\2x-3=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{15}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy `x in {15/4;-3/4}`
\(b,\left(\left|x\right|-0,2\right)\left(x^3-8\right)=0\\ \left(\left|x\right|-0,2\right)\left(x-2\right)\left(x^2+2x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|-0,2=0\\x-2=0\\x^2+2x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left|x\right|=0,2\\x=2\\\left(x+1\right)^2+3=0\left(lọai\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,2\\x=-0,2\\x=2\end{matrix}\right.\)
Vậy `x in {+-0,2;2}`
30 tháng 8 2016
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\right)=\frac{51}{81}\)
\(x\times4+\frac{40}{81}=\frac{51}{81}\)
\(x\times4=\frac{51}{81}-\frac{40}{81}\)
\(x\times4=\frac{11}{81}\)
\(\Rightarrow x=\frac{11}{81}\div4=\frac{11}{81}\times\frac{1}{4}\)
\(\Rightarrow x=\frac{11}{324}\)
[ 61 + ( 53 - x ) ] . 17 = 1785
61 + ( 53 - x ) = 1785 : 17
61 + ( 53 - x ) = 105
( 53 - x ) = 105 - 61
53 - x = 44
=> x = 53 - 44
=> x = 9
\(2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(\Rightarrow x-1=4\)
\(x=4+1\)
\(x=5\)
Vậy \(x=5\)
b) \(\frac{\left(-3\right)^x}{81}=-27\)
\(\left(-3\right)^x=-27.81\)
\(\left(-3\right)^x=-2187\)
\(\left(-3\right)^x=\left(-3\right)^7\)
\(\Rightarrow x=7\)
Vậy \(x=7\)