Tìm x:
a,27\(^{x+1}\)\(-10^{3x-1}=\frac{71}{81}\)
b,\(2^{x+3}-16^9\times8^7\times4^5=0\)
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\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
=-1+(4-7)+(10-13)+(16-19)+...+(94-97)+(100-103)
=-1+(-3)+(-3)+(-3)+...+(-3)+(-3)(17 cặp)
=-1+(-3)*17
=-1+(-51)
=-52
a) \(T=\frac{9^{14}\times25^6\times8^7}{18^{12}\times625^3\times24^3}\)
\(=\frac{\left(3^2\right)^{14}\times25^6\times\left(2^3\right)^7}{\left(2\times3^2\right)^{12}\times\left(25^2\right)^3\times\left(3\times2^3\right)^3}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{12}\times3^{24}\times25^6\times3^3\times2^9}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{\left(2^{12}\times2^9\right)\times\left(3^{24}\times3^3\right)\times25^6}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{21}\times3^{27}\times25^6}=3\)
b) \(A=\frac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^9\times6^{19}-7\times2^{29}\times27^6}\)
\(=\frac{5\times\left(2^2\right)^{15}\times\left(3^2\right)^9-2^2\times3^{20}\times\left(2^3\right)^9}{5\times2^9\times\left(2\times3\right)^{19}-7\times2^{29}\times\left(3^3\right)^6}\)
\(=\frac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^9\times2^{19}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{28}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{28}\times3^{18}\times\left(5\times3-7\times2\right)}\)
\(=\frac{2\times\left(10-9\right)}{15-14}=\frac{2\times1}{1}=2\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9