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Để A <\(\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\left(\text{đ}k\text{x}\text{đ}:x\ne0\right)\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow2\sqrt{x}-2-\sqrt{x}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\left(TM\right)\)
Vậy x<4 và x≠0 thì ta có A<\(\dfrac{1}{2}\)
a, Mk làm hơi tắt chút bạn thông cảm nha . mk vội ý mà
\(A=\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right).\left(x-3\sqrt{x}+2\right)\)
\(A=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
bạn ơi có chép sai đề bài rồi không đáng lẽ phải chia cho \(\dfrac{\sqrt{x}}{\sqrt{x}-3}\)thì mới rút gọn được .
a ,
\(A=\left(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\right)\left(\dfrac{3}{\sqrt{1-a^2}}+1\right)\)
\(A=\left(\dfrac{3}{\sqrt{1+a}}+\dfrac{\sqrt{1-a^2}}{\sqrt{1+a}}\right)\left(\dfrac{3}{\sqrt{1-a^2}}+\dfrac{\sqrt{1-a^2}}{\sqrt{1-a^2}}\right)\)
\(A=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}:\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(A=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}.\dfrac{\sqrt{1-a^2}}{3+\sqrt{1-a^2}}\)
\(A=\sqrt{1-a}\)
b, thay \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}\)ta có :
\(A=\sqrt{1-\dfrac{\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\dfrac{2+\sqrt{3}-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\dfrac{2}{2+\sqrt{3}}}=\sqrt{\dfrac{4-2\sqrt{3}}{4-3}}=\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
CHo 16.8 lít khí CO2 ( đktc) tác dụng với NaOH 2M thu được dung dịch a . tính khối lượng khi a khô cạn
\(P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x-1}\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
mình chỉ rút gọn thôi , bạn tự so sánh nha
Ta có :
\(x=\dfrac{\sqrt[3]{10+6\sqrt{3}}\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)
\(\Leftrightarrow x=\dfrac{\sqrt[3]{3\sqrt{3}+9+3\sqrt{3}+1}\left(\sqrt{3}-1\right)}{\sqrt{5+2\sqrt{5}+1}-\sqrt{5}}\)
\(\Leftrightarrow x=\dfrac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(\sqrt{5}+1\right)^2-5}}\)
\(\Leftrightarrow x=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}\)
\(\Leftrightarrow x=\dfrac{3-1}{1}=2\)
thay x=2 vào biểu thức P ta có :
\(P=\left(2^3-4.2+1\right)^{2015}\)
\(P=1^{2015}=1\)
Nhớ like đúng cho mk nha mọi người
\(a,\sqrt{3-\sqrt{5}}:\sqrt{2}\)
=\(\dfrac{\sqrt{3-\sqrt{5}}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6-2\sqrt{5}}}{2}=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{2}=\dfrac{\sqrt{5}-1}{2}\)
y2+3xy-12x2
= (y2+3xy+\(\dfrac{9}{4}x^2\) )-\(\dfrac{57}{4}x^2\)
= \(\left(y+\dfrac{3}{2}x\right)^2-\dfrac{57}{4}x^2\)
=\(\left(y+\dfrac{3}{2}-\dfrac{\sqrt{57}}{2}\right)\left(y+\dfrac{3}{2}+\dfrac{\sqrt{57}}{2}\right)\)
a, 4x2-25+9+12x=0
⇔4x2+12x-16=0
⇔(4x2-4x )-( 16x-16)=0
⇔4x(x-1)-16(x-1 )=0
⇔(x-1 )( 4x-16)=0
⇔4(x-1 )(x-4 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{1;4\right\}\)
b,16-25+k2-8k=0
⇔k2-8k-9=0
⇔(k2+k )-(9k+9 )=0
⇔k(k+1 )-9(k+1 )=0
⇔(k-9 )(k+1 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}k-9=0\\k+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}k=9\\k=-1\end{matrix}\right.\)
Vậy \(S=\left\{-1;9\right\}\)