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Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
\(a.\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)=\dfrac{x+1+\sqrt{x}}{x\sqrt{x}-1}.\dfrac{x\sqrt{x}+1-\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}=\dfrac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
\(b.ĐK:x>2\) ( thường là những bài rút gọn sẽ kèm theo ĐK nhé , mình thêm như vậy , nếu không bạn chia TH ra )
\(\dfrac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{\dfrac{1}{x^2}-\dfrac{2}{x}+1}}=\dfrac{\sqrt{x-1}-1+\sqrt{x-1}+1}{1-\dfrac{1}{x}}=\dfrac{2\sqrt{x-1}}{1-\dfrac{1}{x}}\)
\(c.\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=\dfrac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=1\)
\(d.Tuong-tự\)
bạnn giải giúp mik lun câu d lun nha?!:)))))cảm ơn nhiw!:))))))
\(\sqrt{28-6\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=3\sqrt{3}-1\)
\(\sqrt{6-\sqrt{20}}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-1\)
\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\left(x+2\right)}}\)
\(=\sqrt{\left(\sqrt{x+2}+\sqrt{x+1}\right)^2}\)
\(=\sqrt{x+2}+\sqrt{x+1}\)
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
\(=\sqrt{\left(x-1\right)-2\sqrt{\left(x-1\right)\left(x+3\right)}+\left(x+3\right)}\)
\(=\sqrt{\left(\sqrt{x+3}-\sqrt{x-1}\right)^2}\)
\(=\left|\sqrt{x+3}-\sqrt{x-1}\right|\)
\(\sqrt{21-6\sqrt{6}}+\sqrt{21+6\sqrt{6}}\)
\(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)
\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\)
\(=6\sqrt{2}\)
\(M=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)\(\left[\dfrac{\left(\sqrt{x}+1\right)-\left(3-\sqrt{x}\right)}{\sqrt{x}+1}\right]\)
\(=\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right]\times\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}\times2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
\(A=\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\)
\(=\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}}{2x-1}-1\)
\(=\dfrac{2x\sqrt{2}+2\sqrt{2x}-1+2x-2x+1}{2x-1}=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}\)
\(B=\left(1+\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=1+\dfrac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{2x-1}\)
\(=1+\dfrac{-2\sqrt{x}-1-2x}{2x-1}\)
\(=\dfrac{2x-1-2\sqrt{x}-1-2x}{2x-1}=\dfrac{-2-2\sqrt{x}}{2x-1}\)
\(P=A:B=\dfrac{2x\sqrt{x}+2\sqrt{2x}}{2x-1}:\dfrac{-2\sqrt{x}-2}{2x-1}\)
\(=\dfrac{2\sqrt{x}\left(x+\sqrt{2}\right)}{2x-1}\cdot\dfrac{2x-1}{-2\left(\sqrt{x}+1\right)}=\dfrac{-\sqrt{x}\left(x+\sqrt{2}\right)}{\sqrt{x}+1}\)
b: Thay \(\sqrt{x}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\) vào P, ta được:
\(P=\left[-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\left(\dfrac{3+2\sqrt{2}}{2}+\sqrt{2}\right)\right]:\left[\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{2}+1\right]\)
\(=\left[\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)}{2}\cdot\dfrac{3+4\sqrt{2}}{2}\right]:\left[\dfrac{2+\sqrt{2}+2}{2}\right]\)
\(=\dfrac{-\sqrt{2}\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{4}\cdot\dfrac{2}{4+\sqrt{2}}\)
\(=\dfrac{-\left(\sqrt{2}+1\right)\left(4\sqrt{2}+3\right)}{2\cdot\left(2\sqrt{2}+1\right)}=\dfrac{-\left(4\sqrt{2}+3\right)}{3\cdot\left(3+\sqrt{2}\right)}\)
ĐKXĐ : \(0\le x\le1\) ; \(\sqrt{x}+\sqrt{1-x}>0\)
Đặt \(\sqrt{x}=a;\sqrt{1-x}=b\left(a;b\ge0\right)\)
=> a2 + b2 = 1 (1)
Khi đó phương trình <=> \(\dfrac{2a^3}{a+b}+ab=1\)
<=> 2a3 + ab(a + b) = a + b
<=> 2a3 + a2b + ab2 - (a + b) = 0
<=> (a3 + a2b + ab2 + b3) - (a + b) + (a3 - b3) = 0
<=> (a + b)(a2 + b2 - 1) + a3 - b3 = 0 (2)
Kết hợp (1);(2) được a3 - b3 = 0
<=> a = b
<=> \(\sqrt{x}=\sqrt{1-x}\Leftrightarrow x=\dfrac{1}{2}\)(tm)
Vậy tập nghiệm \(S=\left\{\dfrac{1}{2}\right\}\)