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2 tháng 9 2018
a) \(\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}\)
\(=\left(\frac{3}{4}\right)^{45}:\left(\frac{3}{4}\right)^{20}\)
\(=\left(\frac{3}{4}\right)^{25}\)
b) \(\frac{125^{100}.2^{160}}{5^{298}.4^{80}}\)
\(=\frac{5^{300}.2^{160}}{5^{298}.2^{160}}\)
\(=5^2=25\)
2 tháng 9 2018
a) \(\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}\)
\(\Leftrightarrow\frac{\left(\frac{3}{4}\right)^{45}}{\frac{3^{10}}{2^{10}}}=\frac{\frac{3^{45}}{4^{45}}}{\frac{3^{10}}{2^{10}}}=\frac{3^{45}.2^{10}}{4^{45}}=\frac{3^{35}.2^{10}}{2^{90}}=\frac{3^{35}}{2^{80}}\)
\(\Rightarrow\left(\frac{3}{4}\right)^{45}:\left(\frac{9}{6}\right)^{10}=\frac{3^{35}}{2^{80}}\)
B
21 tháng 12 2016
a)\(\Rightarrow\frac{3}{2.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}\)
\(\Rightarrow\frac{3x-x+6}{2x.\left(x+3\right)}\)
\(\Rightarrow\frac{2x+6}{2x.\left(x+3\right)}=\frac{2.\left(x+3\right)}{2x.\left(x+3\right)}=\frac{2}{2x}=\frac{1}{x}\)
NN
22 tháng 12 2016
b
=\(\frac{96x^4-75y^7}{40x^3y^3}\)
c, phan tich ra:
=\(\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}=\frac{x+2}{6}\)
=
4 tháng 12 2016
a)\(dk,x\ne7;x\ne0\)
\(\frac{4x+13}{5x\left(x-7\right)}-\frac{x-48}{5x\left(7-x\right)}=\frac{4x+13}{5x\left(x-7\right)}+\frac{x-48}{5x\left(x-7\right)}=\frac{\left(4x+13\right)+\left(x-48\right)}{5x\left(x-7\right)}\\ \)
\(=\frac{5x-35}{5x\left(x-7\right)}=\frac{5\left(x-7\right)}{5x\left(x-7\right)}=\frac{1}{x}\)
b)
\(\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{1-\left(5x\right)^2}=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(1+5x\right)}\)
\(\frac{1+5x}{x\left(1-5x\right)\left(1+5x\right)}+\frac{x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-15x+5x+1}{x\left(1-5x\right)\left(1+5x\right)}=\frac{25x^2-10x+1}{x\left(1-5x\right)\left(1+5x\right)}\)
EC
5 tháng 9 2016
a)\(\left(\frac{1}{2}x-1\right)\left(2x-3\right)=x^2-\frac{3}{2}x-2x+3=x^2-\frac{7}{2}x+3\)
b)\(\left(x-7\right)\left(x-5\right)=x^2-5x-7x+5=x^2-12x+5\)
c)\(\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)\left(4x-1\right)=\left(x^2-\frac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\frac{1}{4}\)
M
2
H9
HT.Phong (9A5)
CTVHS
2 tháng 8 2023
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1+x-18+x+2}{x-5}\)
\(=\dfrac{3x-15}{x-5}\)
\(=\dfrac{3\left(x-5\right)}{x-5}\)
\(=\dfrac{3}{1}\)
\(=3\)
2 tháng 8 2023
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)
22 tháng 4 2020
\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
\(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(x\left(x+1\right)+x\left(x-3\right)=4x\)
\(x^2+x+x^2-3x=4x\)
\(2x^2-2x=4x\)
\(2x^2-2x-4x=0\)
\(2x\left(x-3\right)=0\)
\(2x=0\Leftrightarrow x=0\)
hoặc
\(x-3=0\Leftrightarrow x=3\)
NN
22 tháng 4 2020
b) \(ĐKXĐ:x\ne\pm4\)
\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)
\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)
\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)
\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)
\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )
Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)
\(A=\frac{125^{100}}{5^{298}}\cdot\frac{2^{160}}{4^{80}}=>A=\frac{\left(5^3\right)^{100}}{5^{298}}\cdot\frac{2^{160}}{\left(2^2\right)^{80}}\)
\(=>A=\frac{5^{300}}{5^{298}}\cdot\frac{2^{160}}{2^{160}}=>A=5^2\cdot1=>A=25\)
\(A=\frac{125^{100}}{5^{298}}.\frac{2^{160}}{4^{80}}\)
\(=\frac{\left(5^3\right)^{100}}{5^{298}}.\frac{2^{160}}{\left(2^2\right)^{80}}\)
\(=\frac{5^{300}}{5^{298}}.\frac{2^{160}}{2^{160}}\)
\(=5^2.1=25\)
Vậy \(A=25\)