2^2+2^2+...+2^x=2^21
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a \(\dfrac{1}{x-y}+\dfrac{2}{x+y}+\dfrac{3x}{y^2-x^2}\)
\(=\dfrac{x+y+2x-2y-3x}{\left(x-y\right)\left(x+y\right)}=\dfrac{-y}{\left(x-y\right)\left(x+y\right)}\)
b: \(\dfrac{1}{x-2}+\dfrac{1}{x+2}-\dfrac{4x-4}{x^2-4}\)
\(=\dfrac{x+2+x-2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x+2\right)}\)
=-2/x+2
c: \(\dfrac{x+1}{x+3}-\dfrac{x-1}{3-x}+\dfrac{2x-2x^2}{x^2-9}\)
\(=\dfrac{\left(x+1\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x-6}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x+3}\)
22 + 22 + 23 + 24 + ... + 2x = 221 (1)
Đặt 22 + 23 + 24 + ... + 2x là B, ta có:
2B = 2.(22 + 23 + 24 + ... + 2x)
=> 2B = 23 + 24 + 25 + ... + 2x+1
=> 2B - B = (23 + 24 + 25 + ... + 2x+1) - (22 + 23 + 24 + ... + 2x)
=> B = 2x+1 - 22
Thay B vào (1) ta có:
22 + (2x+1 - 22) = 221
=> 2x+1 = 221
=> x + 1 = 21
=> x = 21 - 1
=> x = 20
Vậy x = 20
\(\left(\frac{21}{x}-2\right)^2-2\left(\frac{21}{x}-7\right)=x+42\)
\(\Leftrightarrow\frac{441}{x^2}-\frac{126}{x}+8=x+42\)
\(\Leftrightarrow\frac{441}{x^2}-\frac{126}{x}=x+42-8\)
\(\Leftrightarrow\frac{441}{x^2}-\frac{126}{x}=x+34\)
\(\Leftrightarrow\frac{441}{x^2}.x^2-\frac{126}{x}.x^2=x.x^2+34.x^2\)
\(\Leftrightarrow441-126x=x^3+34x^2\)
\(\Leftrightarrow x^3+34x^2=441-126x\)(chuyển vế)
\(\Leftrightarrow x^3+34x^4+126x-441=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+27x-63\right)=0\)
\(\Leftrightarrow x+7=0\)
\(\Leftrightarrow x=0-7\)
\(\Leftrightarrow x=-7\)
Vì \(x^2+27-63\ne0\)
=> x = -7
\(\left(\frac{21}{x}-2\right)^2-2\left(\frac{21}{x}-2\right)=x+42\)
\(\Leftrightarrow\frac{441}{x^2}-\frac{126}{x}+8=x+42\)
\(\Leftrightarrow\frac{441}{x^2}-\frac{126}{x}=x+42-8\)
\(\Leftrightarrow\frac{441}{x^2}-\frac{126}{x}=x+34\)
\(\Leftrightarrow\frac{441}{x^2}.x^2-\frac{126}{x}.x^2=x.x^2+34.x^2\)
\(\Leftrightarrow441-126x=x^3+34x^2\)
\(\Leftrightarrow x^3+34x^2=441-126x\)(chuyển vế nhé)
\(\Leftrightarrow x^3+34x^2+126x-441=0\)
\(\Leftrightarrow\left(x+7\right)\left(x^2+27x-63\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+7=0\\x^3+27x-63\ne0\end{cases}}\Leftrightarrow x=-7\)
=> x = -7
b: \(=\left(x^2+3x+1-3x+1\right)^2=\left(x^2+2\right)^2\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^2+y^2-xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\)
\(\Rightarrow x^2+\left(\dfrac{2}{x}\right)^2=5\)
\(\Leftrightarrow x^4-5x^2=4=0\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}\right)^2-\left(y+\dfrac{1}{y}\right)^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)\left(x+\dfrac{1}{x}-y-\dfrac{1}{y}\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=7\\x+\dfrac{1}{x}-y-\dfrac{1}{y}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=5\\y+\dfrac{1}{y}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+1=0\\y^2-2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
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