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Có \(\left(x-\dfrac{1}{3}\right)^2\ge0\)
Mà \(\left(x-\dfrac{1}{3}\right)^2=-\dfrac{8}{27}< 0\) \(\Rightarrow\) Vô lý
a) 35x+8=33
<=> 5x+8=3
<=>5x=3-8=-5
<=>x=-1
b) 28-x=25
<=> 8-x=5
<=> x=8-5=3
\(3^{5x+8}=27\)
\(3^{5x+8}=3^3\)
\(5x+8=3\)
\(5x=-5\)
\(x=-1\)
\(2^{8-x}=32\)
\(2^{8-x}=2^5\)
\(8-x=5\)
\(x=3\)
=.= hok tốt!!
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{13}.3^{13}-2^{11}.3^{11}}\)
\(\Leftrightarrow\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.2^{12}-2^{11}.2^{11}}\)
\(\Leftrightarrow\frac{2^{12}.3^{10}.\left(1+5\right)}{6^{12}-6^{11}}\)
\(\Leftrightarrow\frac{2^{12}.3^{10}.6}{6^{11}.\left(6-1\right)}\)
\(\Leftrightarrow\frac{2^{12}.3^{10}.2.3}{6^{11}.\left(6-1\right)}\)
\(\Leftrightarrow\frac{2^{13}.3^{11}}{6^{11}.5}\)
\(\Leftrightarrow\frac{2^{11}.3^{11}.2^2}{6^{11}.5}\)
\(\Leftrightarrow\frac{6^{11}.4}{6^{11}.5}\Leftrightarrow\frac{4}{5}\)
ta có: \(\frac{-7}{8}=-0,875;\frac{-2}{3}=-0,6\)
\(\frac{-3}{4}=-0,75;\frac{-18}{19}=-0,94;\frac{-27}{28}=-0,96\)
\(\Rightarrow\frac{-2}{3}>\frac{-3}{4}>\frac{-7}{8}>\frac{-18}{19}>\frac{-27}{28}\)
\(\frac{3}{2^2}.\frac{2^3}{3^2}.\frac{5.3}{4^2}.......\frac{3.3333}{100^2}\)
mk ko nghĩ ra phần sau chắc là rút gọn thì phải!!?
bạn làm thử xem
\(\text{a, 3(x+1)+4x=10}\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow7x+3=10\)
\(\Rightarrow7x=10-3=7\)
\(\Rightarrow x=1\)
c, x+1/10+x+2/9=x+3/8+x+4/7
=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)
=> x+11/10 + x+11/9 = x+11/8 + x+11/7
...............
a) \(3\left(x+1\right)+4x=10\)
\(\Rightarrow3x+3+4x=10\)
\(\Rightarrow3x+4x=10-3\)
\(\Rightarrow7x=7\)
\(\Rightarrow x=7\)
(2x2 - 3)3=\(\frac{27}{8}\)
<=> (2x2 - 3)3=\(\left(\frac{3}{2}\right)^3\)
<=> 2x2 - 3 = \(\frac{3}{2}\)
<=> 2x2 =\(\frac{3}{2}+3\)
<=>2x2=\(\frac{9}{2}\)
<=> x2 = \(\frac{9}{2}:2\)
<=> x2=\(\frac{9}{4}\)
\(\Rightarrow x\in\left\{-\frac{3}{2};\frac{3}{2}\right\}\)
vậy...
\(\left(2x^2-3\right)^3=\frac{27}{8}\)
\(\Leftrightarrow2x^2-3=\frac{3}{2}\)
\(\Leftrightarrow2x^2=\frac{9}{2}\)
\(\Leftrightarrow x^2=\frac{9}{4}\)
\(\Leftrightarrow x=\pm\frac{3}{2}\)