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Bài làm:
Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)
\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)
\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)
\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)
\(\Rightarrow x=-36\)
tick cho
a/ \(\left|3x-1\right|=\left|5-2x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5-2x\\3x-1=-5+2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2x=5+1\\3x-2x=-5+1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}5x=6\\x=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-4\end{matrix}\right.\)
Vậy ......
b/ \(\left|x+2\right|-\left|x+7\right|=0\)
\(\Leftrightarrow\left|x+2\right|=\left|x+7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x+7\\x+2=-x-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x=7-2\\x+x=-7-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\2x=-9\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{9}{2}\)
Vậy ...............
c/ \(\left|2x-1\right|+x=2\)
\(\Leftrightarrow\left|2x-1\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2-x\\2x-1=-2+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+x=2+1\\2x-x=-2-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=3\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy ..
a: \(\Leftrightarrow\left|x-3\right|=12-5x-8=-5x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(-5x+4\right)^2=\left(x-3\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(5x-4-x+3\right)\left(5x-4+x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{4}{5}\\\left(4x-1\right)\left(6x-7\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\)
b: \(\left(\sqrt{x}+3\right)^{10}=1024\cdot125^2\cdot25^2\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)^{10}=2^{10}\cdot5^6\cdot5^4=10^{10}\)
\(\Leftrightarrow\sqrt{x}+3=10\)
hay x=49
c: \(\dfrac{3-0.2x}{5}=\dfrac{7}{15}+1.4x\)
\(\Leftrightarrow\dfrac{9-0.6x}{15}=\dfrac{7}{15}+\dfrac{21x}{15}\)
=>21x+7=9-0,6x
=>21,6x=-2
hay x=-5/54
d: \(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{5^9\cdot7^9\left(4\cdot7-5^2\right)}{5^9\cdot7^9\cdot4}\)
\(\Leftrightarrow\left(\dfrac{4}{3}\right)^{3x}=\dfrac{28-25}{4}=\dfrac{3}{4}\)
=>3x=-1
hay x=-1/3
bài 4:
a, 3105 + 4105 = 2735 + 6435 chia hết cho 91 ( vì 27+64=91)
mà 91 chia hết cho 13 nên 3105 + 4105 chia hết cho 13
b, 62n+1 + 5n+2 = 62n . 6 + 5n . 25 = 36n . 6 + 5n .25
36 đồng dư với 5 ( mod 31)
=> 36n đồng dư với 5n ( mod 31)
=> 36n .6 + 5n .25 đồng dư với 5n . 6 + 5n . 25 = 5n . (6+25) = 31. 5n đồng dư với 0 ( mod 32)
Vậy 62n+1 + 5n+3 chia hết cho 31
1/
a/ \(\left|2x-1\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=2x\\2x-1=-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-2x=1\left(loại\right)\\2x+2x=1\end{matrix}\right.\)
\(\Leftrightarrow4x=1\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy ......
b/ \(\left|x-3\right|-\left|4-x\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|4-x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-x\\x-3=-4+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+x=4+3\\x-x=-4+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\0x=-1\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
Vậy ....
- Ta có : \(5\left(x+2\right)^3+7=2\)
=> \(5\left(x^3+6x^2+12x+8\right)+7=2\)
=> \(5x^3+30x^2+60x+40+7=2\)
=> \(5x^3+30x^2+60x+40+7-2=0\)
=> \(5x^3+30x^2+60x+45=0\)
=> \(5x^3+15x^2+15x^2+45x+15x+45=0\)
=> \(5x^2\left(x+3\right)+15x\left(x+3\right)+15\left(x+3\right)=0\)
=> \(\left(5x^2+15x+15\right)\left(x+3\right)=0\)
=> \(\left[{}\begin{matrix}x+3=0\\5x^2+15x+15=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\x^2+3x+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\x^2+2x.\frac{3}{2}+\frac{9}{4}+\frac{3}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\\left(x+\frac{3}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-3\\\left(x+\frac{3}{2}\right)^2=-\frac{3}{4}\left(VL\right)\end{matrix}\right.\)
=> \(x=-3\)
Vậy phương trình có tập nghiệm là \(S=\left\{-3\right\}\)
5(x + 2)3 + 7 = 2
(x + 2)3 = -1
=> x + 2 = -1
<=> x = -3