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a) \(\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}+1}}-\dfrac{\sqrt{2}}{\sqrt{\sqrt{2}-1}}=\dfrac{\sqrt{2}\left(\sqrt{\sqrt{2}-1}\right)}{\left(\sqrt{\sqrt{2}+1}\right)\left(\sqrt{\sqrt{2}-1}\right)}-\dfrac{\sqrt{2}\left(\sqrt{\sqrt{2}+1}\right)}{\left(\sqrt{\sqrt{2}+1}\right)\left(\sqrt{\sqrt{2}-1}\right)}=\dfrac{\sqrt{2}\left(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\right)}{\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\dfrac{\sqrt{2}\left(\left(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\right)\right)}{\sqrt{2-1}}=\sqrt{2}.\left(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\right)\)(1)
Đặt A=\(\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}\Leftrightarrow A^2=\sqrt{2}-1+\sqrt{2}+1-2\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=2\sqrt{2}-2\sqrt{1}=2\sqrt{2}-2\Leftrightarrow A=\pm\sqrt{2\sqrt{2}-2}\)
Ta có \(\sqrt{\sqrt{2}-1}< \sqrt{\sqrt{2}+1}\Leftrightarrow\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}< 0\Leftrightarrow A< 0\)
Vậy A=\(-\sqrt{2\sqrt{2}-2}\)
(1)\(=\sqrt{2}.\left(-\sqrt{2\sqrt{2}-2}\right)=-\sqrt{4\sqrt{2}-4}\)
b) \(\sqrt{4-2\sqrt{3}}+\sqrt{\dfrac{2}{2-\sqrt{3}}}-\sqrt{27}=\sqrt{3-2.\sqrt{3}.1+1}+\sqrt{\dfrac{2\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}-\sqrt{9.3}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\dfrac{4+2\sqrt{3}}{2^2-\left(\sqrt{3}\right)^2}}-3\sqrt{3}=\left|\sqrt{3}-1\right|+\sqrt{4+2\sqrt{3}}-3\sqrt{3}=\sqrt{3}-1-3\sqrt{3}+\sqrt{3+2\sqrt{3}+1}=-2\sqrt{3}-1+\sqrt{\left(\sqrt{3}+1\right)^2}=-2\sqrt{3}-1+\sqrt{3}+1=-\sqrt{3}\)
c) \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3x+5\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3x-3\sqrt{x}-2\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left[3\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(\sqrt{x}-1\right)\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-\left(3\sqrt{x}-2\right)}{\sqrt{x}+3}=\dfrac{2-3\sqrt{x}}{\sqrt{x}+3}\)
ĐKXĐ:......
A=\(\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{x^3-\sqrt{27}}\right)\cdot\left(\dfrac{x^2+3+\sqrt{3}x}{\sqrt{3}x}\right)\)
A=\(\dfrac{\sqrt{3}x}{\left(x-\sqrt{3}\right)\left(x^2+\sqrt{3}x+3\right)}\cdot\dfrac{x^2+\sqrt{3}x+3}{\sqrt{3}x}\)
A=\(\dfrac{1}{x-\sqrt{3}}\)
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: \(A=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{x+3+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
b: \(x=5+\sqrt{2}-4-\sqrt{2}=1\)
Khi x=1 thì \(B=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)
a) \(\sqrt{2}\cdot x-\sqrt{50}=0< =>\sqrt{2}\cdot x=\sqrt{50}\)
<=> x= 5
b) \(\sqrt{3}\cdot x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot\sqrt{4}+\sqrt{3}\cdot\sqrt{9}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot5< =>x+1=5\)
<=> x=4
c) \(\sqrt{3}\cdot x^2-\sqrt{12}=0\\ < =>x^2=\sqrt{4}=2;-2\\ < =>x=\sqrt{2};-\sqrt{2}\)
d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\\ < =>x^2=\sqrt{100}=10;-10\\ < =>x=\sqrt{10};-\sqrt{10}\)
Câu 1:
ĐK: \(4\leq x\leq 6\)
Ta thấy biểu thức vế trái luôn không âm theo tính chất căn bậc 2
Vế phải: \(x^2-10x-27=x(x-10)-27< 0-27< 0\) với mọi \(4\leq x\leq 6\), tức là biểu thức vế phải luôn âm
Do đó pt vô nghiệm
Câu 2:
\(x\geq -3; y\geq 3; z\geq 3\)
Ta có: \(\sqrt{x+3}+\sqrt{y-3}+\sqrt{z-3}=\frac{1}{2}(x+y+z)\)
\(\Leftrightarrow 2\sqrt{x+3}+2\sqrt{y-3}+2\sqrt{z-3}=x+y+z\)
\(\Leftrightarrow (x+3-2\sqrt{x+3}+1)+(y-3-2\sqrt{y-3}+1)+(z-3-2\sqrt{z-3}+1)=0\)
\(\Leftrightarrow (\sqrt{x+3}-1)^2+(\sqrt{y-3}-1)^2+(\sqrt{z-3}-1)^2=0\)
Vì \((\sqrt{x+3}-1)^2; (\sqrt{y-3}-1)^2; (\sqrt{z-3}-1)^2\) đều không âm nên để tổng của chúng bằng $0$ thì:
\((\sqrt{x+3}-1)^2=(\sqrt{y-3}-1)^2=(\sqrt{z-3}-1)^2=0\)
\(\Rightarrow x=-2; y=z=4\)
Bài 1:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
b: \(P=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{x-2\sqrt{x}}\)
\(=\dfrac{x-3\sqrt{x}}{x-2\sqrt{x}}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
c: Để \(P=\dfrac{1}{2}\) thì \(2\sqrt{x}-6=\sqrt{x}-2\)
hay x=16
`a)`\(\sqrt{3x+1}-\sqrt{x-1}=2\)
\(ĐK:x\ge1\)
\(\Leftrightarrow3x+1+x-1-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)
\(\Leftrightarrow4x-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)
\(\Leftrightarrow\sqrt{\left(3x+1\right)\left(x-1\right)}=2x-2\)
\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=4x^2-8x+4\)
\(\Leftrightarrow3x^2-3x+x-1=4x^2-8x+4\)
\(\Leftrightarrow x^2-6x+5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\) `(tm)`
Vậy \(S=\left\{5;1\right\}\)
`b)`\(\dfrac{\sqrt{x+27}+\sqrt{27-x}}{\sqrt{27+x}-\sqrt{27-x}}=\dfrac{27}{x}\)
\(ĐK:-27\le x\le27;x\ne0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}+\sqrt{27-x}\right)\left(\sqrt{x+27}-\sqrt{27-x}\right)}{\left(\sqrt{27+x}-\sqrt{27-x}\right)\left(\sqrt{27+x}-\sqrt{27-x}\right)}=\dfrac{27}{x}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}\right)^2-\left(\sqrt{27-x}\right)^2}{\left(\sqrt{27+x}-\sqrt{27-x}\right)^2}=\dfrac{27}{x}\)
\(\Leftrightarrow\dfrac{2x}{54-2\sqrt{\left(27+x\right)\left(27-x\right)}}=\dfrac{27}{x}\)
\(\Leftrightarrow\dfrac{x}{27-\sqrt{27^2-x^2}}=\dfrac{27}{x}\)
\(\Leftrightarrow x^2=27^2-\sqrt{27^2-x^2}\)
\(\Leftrightarrow27^2-x^2-\sqrt{27^2-x^2}=0\)
\(\Leftrightarrow\sqrt{27^2-x^2}\left(\sqrt{27^2-x^2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}27^2-x^2=0\\\sqrt{27^2-x^2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm27\\x=\pm\sqrt{27^2-1}\end{matrix}\right.\)
Thế vào pt ta được \(x=\pm27\) là thỏa mãn
Vậy \(S=\left\{\pm27\right\}\)
a) ĐKXĐ: \(x\ge1\)
Ta có: \(\sqrt{3x+1}-\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{3x+1}-4+2-\sqrt{x-1}=0\)
\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{4-x+1}{2+\sqrt{x-1}}=0\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{5-x}{2+\sqrt{x-1}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}\right)=0\)
+ Nếu \(x-5=0\Leftrightarrow x=5\left(tmđkxđ\right)\)
+ Nếu \(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}=0\Leftrightarrow\dfrac{3}{\sqrt{3x+1}+4}=\dfrac{1}{2+\sqrt{x-1}}\)
\(\Leftrightarrow3\sqrt{x-1}+2=\sqrt{3x+1}\Rightarrow6x-6+12\sqrt{x-1}=0\)
\(\sqrt{x-1}\left(6\sqrt{x-1}+12\right)=0\Leftrightarrow\sqrt{x-1}=0\)
\(\Leftrightarrow x=1\left(tmđkxđ\right)\)
Vậy ....