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\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)
b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)
\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)
c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)
\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)
\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)
\(=\left(\dfrac{3}{7}\right)^{15}\)
\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)
(5x + 1)2 = 36/49
=> (5x + 1)2 = (6/7)2
=> \(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{cases}}\)
Làm từ phần b nha
b) \(\left(x-\frac{1}{9}\right)^3=\frac{2}{3}^6\)
\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{1}{3}\right)^6\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1^6}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{3^6}\)
\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{729}\)
\(\Rightarrow x-\frac{2}{9}=\frac{1}{9}\)
\(x=\frac{1}{9}+\frac{2}{9}\)
\(x=\frac{3}{9}=\frac{1}{3}\)
c) Sai đề rồi, xem lại đi
d) \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4< 0\)
\(\Rightarrow\frac{10000y^4-4000y^3+600y^3-40y+10000x^2+122501-70000x}{10000}< 0\)
=> Sai \(\forall y\inℝ\)
2.(x-3)+3x+0.5=\(\dfrac{3}{4}\)
4x+2+4x=272
(1,2-5x).(2\(\dfrac{1}{8}\) +1/2 x)=0
GIÚP MÌNH VỚI !!!!
\(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\\ \Leftrightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\\ \Leftrightarrow x\left(2+3\right)=\dfrac{3}{4}-\dfrac{1}{2}+6\\ \Leftrightarrow5x=\dfrac{25}{4}\\ \Leftrightarrow x=\dfrac{25}{4}:5=\dfrac{5}{4}\\ ---\\ 4^{x+2}+4^x=272\\ \Leftrightarrow4^x\left(4^2+1\right)=272\\ \Leftrightarrow4^x.17=272\\ \Leftrightarrow4^x=\dfrac{272}{17}=16=4^2\\ Vậy:x=2\\ ----\\ \left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1,2-5x=0\\2,125+0,5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1,2\\0,5x=-2,125\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}=0,24\\x=\dfrac{-2,125}{0,5}=-4,25\end{matrix}\right.\)
a) \(2\left(x-3\right)+3x+0,5=\dfrac{3}{4}\)
\(\Rightarrow2x-6+3x+\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow5x-6=\dfrac{3}{4}-\dfrac{1}{2}\)
\(\Rightarrow5x-6=\dfrac{1}{4}\)
\(\Rightarrow5x=\dfrac{1}{4}+6\)
\(\Rightarrow5x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:5\)
\(\Rightarrow x=\dfrac{5}{4}\)
b) \(4^{x+2}+4^x=272\)
\(\Rightarrow4^x\cdot4^2+4^x\cdot1=272\)
\(\Rightarrow4^x\cdot\left(16+1\right)=272\)
\(\Rightarrow4^x\cdot17=272\)
\(\Rightarrow4^x=16\)
\(\Rightarrow4^x=4^2\)
\(\Rightarrow x=2\)
c) \(\left(1,2-5x\right)\left(2\dfrac{1}{8}+\dfrac{1}{2}x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}1,2-5x=0\\\dfrac{15}{8}+\dfrac{1}{2}x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=1,2\\\dfrac{1}{2}x=-\dfrac{15}{8}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1,2}{5}\\x=-\dfrac{15}{8}:\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{25}\\x=-\dfrac{15}{4}\end{matrix}\right.\)
Câu 1 :
a) \(\left(\dfrac{-1}{3}\right)^3.x=\dfrac{1}{81}\)
\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)^3\Rightarrow x=\dfrac{-1}{3}\)
b) \(\left(5.x\right)^3=-64\)
\(\left(5.x\right)^3=\left(-4\right)^3\Rightarrow5x=-4\Rightarrow x=\dfrac{-4}{5}\)
c) \(\left(2x-3\right)^2-9=0\)
\(\left(2x-3\right)^2=9=\left(\pm3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
d) \(\left(5X+1\right)^2=\dfrac{36}{49}=\left(\pm\dfrac{6}{7}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}5X+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{-1}{35}\\\dfrac{-13}{35}\end{matrix}\right.\)
Câu 2: mik chỉ nêu đáp án thôi nhé :
a) \(x=0\)
\(y=\dfrac{1}{2}\) hoặc \(y=\dfrac{-1}{2}\)
b) x =10 còn y giống câu a
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
a)Ta có:
\(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)
\(\Rightarrow x-3,5=y-\dfrac{1}{10}=0\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}=0,1\end{matrix}\right.\)
b) Ta có:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
b: ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)