tìm x biết
(3x-2)\(^5\) =-243
\(\frac{1}{9}.27^x=3^x\)
\(\frac{1}{3^2}.3^4.3^n=3^7\)
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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: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a)\(\left(-3\right)^{x+3}=-\frac{1}{27}\)
\(\left(-3\right)^{x+3}=\left(-\frac{1}{3}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-\frac{3^0}{3^1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3^{-1}\right)^3\)
\(\left(-3\right)^{x+3}=\left(-3\right)^{-3}\)
\(\Rightarrow x+3=-3\)
\(\Rightarrow x=-6\)
b)\(\left(-6\right)^{2x+2}=\frac{1}{36}\)
\(\left(-6\right)^{2x+2}=\left(-\frac{1}{6}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-\frac{6^0}{6^1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6^{-1}\right)^2\)
\(\left(-6\right)^{2x+2}=\left(-6\right)^{-2}\)
\(\Rightarrow2x+2=-2\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\)
c)\(\left(-3\right)^{x+5}=\frac{1}{81}\)
\(\left(-3\right)^{x+5}=\left(-\frac{1}{3}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-\frac{3^0}{3^1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3^{-1}\right)^4\)
\(\left(-3\right)^{x+5}=\left(-3\right)^{-4}\)
\(\Rightarrow x+5=-4\)
\(\Rightarrow x=-9\)
d)\(\left(\frac{1}{9}\right)^x=\left(\frac{1}{27}\right)^6\)
\(\left[\left(\frac{1}{3}\right)^2\right]^x=\left[\left(\frac{1}{3}\right)^3\right]^6\)
\(\left(\frac{1}{3}\right)^{2x}=\left(\frac{1}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
e)\(\left(\frac{4}{9}\right)^x=\left(\frac{8}{27}\right)^6\)
\(\left[\left(\frac{2}{3}\right)^2\right]^x=\left[\left(\frac{2}{3}\right)^3\right]^6\)
\(\left(\frac{2}{3}\right)^{2x}=\left(\frac{2}{3}\right)^{18}\)
\(\Rightarrow2x=18\)
\(\Rightarrow x=9\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
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.
a) ...
(3x-2)^5=(-3)^5
=) 3x-2=(-3)
3x=(-1)
x=(-1/3)
DUYỆT CHO MÌH ĐI RỒI MÌH LẠI GIẢI TIẾP CHO
\(\left(3x-2\right)^5=-243\)
=> \(\left(3x-2\right)^5=\left(-3\right)^5\)
=> 3x - 2 = -3
=> 3x = -3 + 2
=> 3x = -1
=> x = -1/3
\(\frac{1}{9}.27^x=3^x\)
=> \(3^{-2}.\left(3^3\right)^x=3^x\)
=> 3-2.33x=3x
=> 33x-2=3x
=> 3x - 2 = x
=> 3x - x = 2
=> 2x = 2
=> x = 1
\(\frac{1}{3^2}.3^4.3^n=3^7\)
=> \(3^{-2}.3^4.3^n=3^7\)
=> 3n+4-2=37
=> n + 4 - 2 = 7
=> n = 7 + 2 - 4
=> n = 5