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\(a.A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}=\dfrac{x}{\sqrt{x}-1}-\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x-1}\) \(b.x=3+2\sqrt{2}\left(TM\right)\)
Khi đó , ta có : \(A=\sqrt{3+2\sqrt{2}-1}=\sqrt{2+2\sqrt{2}}\)
Bài 1:
a) 4x-\(\sqrt{9x^2-12x+4}\)
= 4x-\(\sqrt{\left(3x-2\right)^2}\)
= 4x-\(|3x-2|\)
= 4x-3x+2
= x+2
b) Thay x=\(\dfrac{2}{7}\)vào biểu thức A, ta có:
A= \(\dfrac{2}{7}+\dfrac{1}{2}\)= \(\dfrac{11}{14}\)
Bài 2:
a) \(\sqrt{x^2+2x+1}=\sqrt{x+1}\)
\(\Leftrightarrow\)\(\left(\sqrt{x^2+2x+1}\right)^2=\left(\sqrt{x+1}\right)^2\)
\(\Leftrightarrow\)x2+2x+1=x+1
\(\Leftrightarrow\)x2+2x+1-x-1=0
\(\Leftrightarrow\)x2-x=0
\(\Leftrightarrow\)x(x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)
....... Tới đây được chưa bạn?
đề bài sai ở câu b ạ
chỗ \(x+4+2\sqrt{3}\) sửa lại là \(x=4+2\sqrt{3}\)
Mấy bài này rất dài , đăng từ từ thôi nhé bạn .
\(1.\dfrac{\sqrt{30}-\sqrt{2}}{\sqrt{8}-\sqrt{15}}-\sqrt{8-\sqrt{49+8\sqrt{3}}}=\dfrac{\sqrt{60}-\sqrt{4}}{\sqrt{16-2\sqrt{15}}}-\sqrt{8-\sqrt{48+2.4\sqrt{3}+1}}=\dfrac{2\left(\sqrt{15}-1\right)}{\sqrt{\left(\sqrt{15}-1\right)^2}}-\sqrt{8-|4\sqrt{3}+1|}=2-\sqrt{4-2.2\sqrt{3}+3}=2-|2-\sqrt{3}|=\sqrt{3}\)
\(2.\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{\sqrt{4}+\sqrt{4+2\sqrt{3}}}+\dfrac{2\sqrt{2}-\sqrt{6}}{\sqrt{4}-\sqrt{4-2\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+|\sqrt{3}+1|}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-|\sqrt{3}-1|}=\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}+\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}=\dfrac{12\sqrt{2}-2\sqrt{18}}{9-3}=\dfrac{12\sqrt{2}-6\sqrt{2}}{6}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
\(3.\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{2}{4+\sqrt{5+2\sqrt{5}+1}}+\dfrac{2}{4-\sqrt{5-2\sqrt{5}+1}}=\dfrac{2}{4+|\sqrt{5}+1|}+\dfrac{2}{4-|\sqrt{5}-1|}=\dfrac{2}{\sqrt{5}+5}+\dfrac{2}{5-\sqrt{5}}=\dfrac{10-2\sqrt{5}+10+2\sqrt{5}}{20}=\dfrac{20}{20}=1\)
a)
= \(\sqrt{18-6\sqrt{6}+3}\)
= \(\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
= \(\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
= \(|3\sqrt{2}-\sqrt{3}|\)
= \(3\sqrt{2}-\sqrt{3}\)
b)
= \(\sqrt{\frac{7}{2}-\sqrt{7}+\frac{1}{2}}\)
= \(\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2+2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\left(\sqrt{\frac{1}{2}}\right)^2}\)
= \(\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)
= \(|\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}|\)
= \(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\)
c)
= \(\sqrt{3+2\sqrt{3}+1}\)
= \(\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}\)
= \(\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
d)
Đặt t = \(\sqrt{x-1}\left(ĐK:t\ge0\right)\)
= \(\sqrt{t^2+1-2t}\)
= \(\sqrt{\left(t+1\right)^2}\)
\(=t+1\)
= \(\sqrt{x-1}+1\)
\(\sqrt{21-6\sqrt{6}}=\sqrt{18-2\sqrt{9}\sqrt{6}+3}=\sqrt{\left(\sqrt{18}\right)^2-2\sqrt{18}\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{18}+\sqrt{3}\right)^2}=\sqrt{18}+\sqrt{3}=\sqrt{3}+3\sqrt{2}\)
\(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2}\sqrt{4-\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}-1}{\sqrt{2}}=\frac{\sqrt{14}-\sqrt{2}}{2}\)
\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
Với \(x\ge1\)thì \(\sqrt{x-2\sqrt{x-1}}=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}\sqrt{1}+\left(\sqrt{1}\right)^2}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\sqrt{x-1}-1\)
T đã tốn mấy phút cuộc đời viết lời giải cho bạn r, tiếc j mấy giây mà bấm k cho t ik =))
a)P= \(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}+\dfrac{4\sqrt{a}-4}{4-a}\)
=\(\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)-4\sqrt{a}+4}{a-4}\)
=\(\dfrac{a+3\sqrt{a}+2\sqrt{a}+6-a+\sqrt{a}+2\sqrt{a}-2-4\sqrt{a}+4}{a-4}\)
=\(\dfrac{4\sqrt{a}+8}{a-4}=\dfrac{4\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}=\dfrac{4}{\sqrt{a}-2}\)
b) thay a=\(6-2\sqrt{5}\)vào P ta có :
\(\dfrac{4}{\sqrt{6-2\sqrt{5}}-2}=\dfrac{4}{\sqrt{\left(\sqrt{5}-1\right)^2}-2}=\dfrac{4}{\sqrt{5}-3}\)
\(a.P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}=\sqrt{a}+2+\sqrt{a}+2=2\sqrt{a}+4\) \(b.P=a+1\)
⇔ \(2\sqrt{a}+4=a+1\)
⇔ \(a-2\sqrt{a}-3=0\)
⇔ \(a+\sqrt{a}-3\sqrt{a}-3=0\)
⇔ \(\sqrt{a}\left(\sqrt{a}+1\right)-3\left(\sqrt{a}+1\right)=0\)
⇔ \(a=9\left(TM\right)\)
KL.............