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Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)
\(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1-\left(x+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+1-x-2-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+0-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left[-\left(\sqrt{x}-1\right)\right]}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(-1\right)}{x+\sqrt{x}+1}\\ =-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
Bài 3:
\(P=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{\left(2x+\sqrt{x}\right)\sqrt{x}}{x}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+2\left(\sqrt{x}+1\right)\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x\left(2\sqrt{x}+1\right)}{x}+2\sqrt{x}+2\)
\(=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}+1\\ =\dfrac{x-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{2x+1}{x+\sqrt{x}+1}\)
a: \(A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}\cdot\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+2\sqrt{x}+2\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\)
\(=\left(x-\sqrt{x}\right)\left(2\sqrt{x}+1\right)+2\sqrt{x}+2\)
\(=2x\sqrt{x}+x-2x-\sqrt{x}+2\sqrt{x}+2\)
\(=2x\sqrt{x}-x+\sqrt{x}+2\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}=-\sqrt{x}+1\)
c: \(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3x+8\sqrt{x}+5-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}+8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
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
\(Q=\frac{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\cdot\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(Q=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(Q=x+1\)
Không thể tìm được GTLN hay GTNN của Q.
b)
\(\frac{3x+3}{\sqrt{x}}=3\sqrt{x}+\frac{3}{\sqrt{x}}\)
Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)
Vậy x=1, x=9 là các giá trị cần tìm
a: \(A=\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{4+\sqrt{3}}{5-2\sqrt{3}}}\)
\(=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: \(B=\dfrac{x\sqrt{x}-2x+28}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}-\dfrac{x-16}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4\sqrt{x}-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)
a: \(=\dfrac{2x+1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
c: \(=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}+1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}-x+\sqrt{x}+1}{x-1}=\dfrac{-x+\sqrt{x}+2}{x-1}\)
\(=\dfrac{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{x-1}=\dfrac{-\sqrt{x}+2}{\sqrt{x}-1}\)
Lời giải:
a) ĐK: \(x\geq 0; x\neq 1\)
\(A=\left(\frac{x+2}{(\sqrt{x})^3-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}-1}{2}\)
\(=\left(\frac{x+2}{(\sqrt{x}-1)(x+\sqrt{x}+1)}+\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\right):\frac{\sqrt{x}-1}{2}\)
\(=\frac{x+1-2\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\frac{2}{\sqrt{x}-1}\)
\(=\frac{2(\sqrt{x}-1)^2}{(\sqrt{x}-1)^2(x+\sqrt{x}+1)}=\frac{2}{x+\sqrt{x}+1}\)
----------------------------
\(B=\frac{2\sqrt{x}}{x+\sqrt{x}+2\sqrt{x}+2}+\frac{5\sqrt{x}+1}{x+\sqrt{x}+3\sqrt{x}+3}+\frac{\sqrt{x}+10}{x+2\sqrt{x}+3\sqrt{x}+6}\)
\(=\frac{2\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}+2)}+\frac{5\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}+3)}+\frac{\sqrt{x}+10}{(\sqrt{x}+2)(\sqrt{x}+3)}\)
\(=\frac{2\sqrt{x}(\sqrt{x}+3)+(5\sqrt{x}+1)(\sqrt{x}+2)+(\sqrt{x}+10)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}+2)(\sqrt{x}+3)}\)
\(=\frac{8x+28\sqrt{x}+12}{(\sqrt{x}+1)(\sqrt{x}+2)(\sqrt{x}+3)}=\frac{4(2\sqrt{x}+1)(\sqrt{x}+3)}{(\sqrt{x}+1)(\sqrt{x}+2)(\sqrt{x}+3)}\)
\(=\frac{4(2\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}+2)}\)
ĐKXĐ:...
\(A=\left(\frac{\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a}+1}\right).\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}.\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}=\frac{1}{a}\)
\(C=\left(\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}-1}\)
\(=\left(\frac{\left(\sqrt{x}+1\right)}{-\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)}.\frac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)}\)
\(=\left(-1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\right).\sqrt{x}=\left(\frac{-x-\sqrt{x}-1+x+\sqrt{x}}{x+\sqrt{x}+1}\right)\sqrt{x}=\frac{-\sqrt{x}}{x+\sqrt{x}+1}\)
Sửa đề: