GIA TRI BIEU THUC
\(\left(\dfrac{1}{2}\right)^2.\left(\dfrac{2}{3}\right)^2.\left(\dfrac{3}{4}\right)^2.............\left(\dfrac{9}{10}\right)^2\)
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29 tháng 6 2023
\(=\left(\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{3}{4}\right):\left(4+\dfrac{3}{4}-3-\dfrac{1}{2}\right)\)
\(=\dfrac{4}{3}:\left(1+\dfrac{1}{4}\right)=\dfrac{4}{3}:\dfrac{5}{4}=\dfrac{16}{15}\)
27 tháng 11 2022
\(P=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3+x\right)\left(2x+3-x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(x-5\right)\cdot5\left(x+1\right)}-\dfrac{3\left(x+1\right)\left(x+3\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{5\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{\left(x+5\right)}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{5x^2+30x+45+x^2+10x+25-x^2-2x-1}{\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{5x^2+38x+69}{\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{5x^2+38x+69}{x^2+6x+5}\)
Để P là số nguyên thì \(5x^2+30x+25+8x+34⋮x^2+6x+5\)
=>\(8x+34⋮x^2+6x+5\)
=>\(\left\{{}\begin{matrix}8x+34⋮x+1\\8x+34⋮x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+8+26⋮x+1\\8x+40-6⋮x+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\\x+5\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\end{matrix}\right.\)
=>\(x\in\left\{-2;1\right\}\)
22 tháng 12 2021
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
NN
13 tháng 12 2017
a) ĐKXĐ: \(a\ne0\) ; \(a\ne3\) ; \(a\ne-3\)
b) \(P=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\left(\dfrac{a^2-9}{a^2-9}-\dfrac{6a-18}{a^2-9}\right)\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{\left(a^2-9\right)-\left(6a-18\right)}{a^2-9}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{a^2-9-6a+18}{a^2-9}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{a^2-6a+9}{a^2-9}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\dfrac{\left(a-3\right)^2}{\left(a-3\right)\left(a+3\right)}\)
\(\Leftrightarrow P=\dfrac{a+3}{2a}.\dfrac{a-3}{a+3}\)
\(\Leftrightarrow P=\dfrac{\left(a+3\right)\left(a-3\right)}{2a\left(a+3\right)}\)
\(\Leftrightarrow P=\dfrac{a-3}{2a}\)
NN
13 tháng 12 2017
( ko biết đúng hay ko)
c) \(P=\dfrac{a-3}{2a}=0\)
\(\Leftrightarrow a-3=0\)
\(\Leftrightarrow a=3\left(loai\right)\) ( không thỏa mãn điều kiện )
\(P=\dfrac{a-3}{2a}=1\)
\(\Leftrightarrow a-3=2a\)
\(\Leftrightarrow a-3-2a=0\)
\(\Leftrightarrow-a-3=0\)
\(\Leftrightarrow-a=3\)
\(\Leftrightarrow a=-3\left(loai\right)\) ( không thỏa mãn điều kiện )
4 tháng 11 2022
a: \(A=\left(\dfrac{6x+4}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\dfrac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
\(=\dfrac{6x+4-3x+2\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\cdot\left(1-\sqrt{3x}\right)^2\)
\(=\dfrac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\)
b: Để A là số nguyên thì \(3x-2\sqrt{3x}+1⋮\sqrt{3x}-2\)
=>\(\sqrt{3x}-2\in\left\{1;-1;3;-3\right\}\)
=>\(3x\in\left\{9;1;25\right\}\)
hay x=3
16 tháng 12 2022
a: \(P=\dfrac{a+3}{a}\cdot\dfrac{a^2-9-6a+18}{\left(a-3\right)\left(a+3\right)}\)
\(=\dfrac{\left(a-3\right)^2}{a\left(a-3\right)}=\dfrac{a-3}{a}\)
b: Để P=-2 thì -2a=a-3
=>-3a=-3
=>a=1
c: Để P nguyên thì a-3 chia hết cho a
=>-3 chia hết cho a
mà a<>0; a<>3; a<>-3
nên \(a\in\left\{1;-1\right\}\)
14 tháng 8 2021
a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
\(\left(\dfrac{1}{2}\right)^2.\left(\dfrac{2}{3}\right)^2.\left(\dfrac{3}{4}\right)^2.....\left(\dfrac{9}{10}\right)^2\)
\(=\left(\dfrac{1}{1}\right)^2.\left(\dfrac{1}{1}\right)^2.\left(\dfrac{1}{1}\right)^2....\left(\dfrac{1}{10}\right)^2\)
\(=\dfrac{1^2}{10^2}=\dfrac{1}{100}\)
đơn giản vậy cơ à?? :D