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ĐKXĐ : \(x\ge0\) ; \(x\ne4\) và \(x\ne9\)
\(A=\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2-\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}+1}{x-4}\)
Câu b : \(\dfrac{1}{A}< \dfrac{1}{5}\Leftrightarrow\dfrac{x-4}{\sqrt{x}+1}< \dfrac{1}{5}\Leftrightarrow5x-20< \sqrt{x}+1\Leftrightarrow5x-\sqrt{x}-21< 0\)Mysterious Person Tới đây làm sao nữa :(((
\(\text{a) }ĐKXĐ:x\ge0;x\ne4;x\ne9\\ A=\left(\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2-\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\\ =\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\\ =\dfrac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\\ =\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}+1}{x-4}\)
\(b\text{) }\dfrac{1}{A}\le\dfrac{1}{5}\\ \Leftrightarrow A\ge5\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{x-4}\ge5\\ \Leftrightarrow\dfrac{\sqrt{x}+1}{x-4}-5\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}+1-5\left(x-4\right)}{x-4}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}+1-5x+20}{x-4}\ge0\\ \Leftrightarrow\dfrac{\sqrt{x}-5x+21}{x-4}\ge0\\ \Leftrightarrow\dfrac{\left(\sqrt{x}-\dfrac{1+\sqrt{421}}{10}\right)\left(\sqrt{x}-\dfrac{1-\sqrt{421}}{10}\right)}{\sqrt{x}-2}\ge0\)
Rồi lập bảng xét dấu.
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
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
Lời giải:
ĐK: \(x\geq 0; x\neq 4;x\neq 9\)
a) Ta có:
\(P=\left(\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right):\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left(\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}-3)}+\frac{(\sqrt{x}+3)(\sqrt{x}-3)}{(\sqrt{x}-2)(\sqrt{x}-3)}-\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}\right):\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}+2+(x-9)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-3)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(P=\frac{\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}-3)}.\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}+1}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}+1}{x-4}\)
b)
Ta có: \(\frac{1}{P}\leq \frac{-5}{2}\)\(\Leftrightarrow \frac{x-4}{\sqrt{x}+1}\leq \frac{-5}{2}\)
\(\Leftrightarrow 2(x-4)\leq -5(\sqrt{x}+1)\)
\(\Leftrightarrow 2x+5\sqrt{x}-3\leq 0\)
\(\Leftrightarrow (2\sqrt{x}-1)(\sqrt{x}+3)\leq 0\)
\(\Rightarrow 2\sqrt{x}-1\leq 0\) (do \(\sqrt{x}+3>0\) )
\(\rightarrow x\leq \frac{1}{4}\)
Vậy \(0\leq x\leq \frac{1}{4}\)
b: \(B=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\cdot\left(2-\dfrac{\sqrt{a}\left(5-\sqrt{b}\right)}{-\left(5-\sqrt{b}\right)}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)
c: \(C=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+2\right)\left(2-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\)
=4-x
1. b) \(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(x\sqrt{\dfrac{6x}{x^2}}+\sqrt{\dfrac{6x}{9}}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\dfrac{7}{3}\sqrt{6x}:\sqrt{6x}=\dfrac{7}{3}\)
2.
P=\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)(bn có ghi sai đề ko)
a) ĐKXĐ : \(x\ge1,x\ge2,x\ge0\)
b) P=\(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{x-3\sqrt{x}-\sqrt{x}+3-2x+\sqrt{x}+4\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-2}\)
c) thay x= \(4-2\sqrt{3}\)vào P ta có :
\(\dfrac{1}{\sqrt{4-2\sqrt{3}}-2}=\dfrac{1}{\sqrt{3}-1-2}=\dfrac{1}{\sqrt{3}-3}\)
a) ĐKXĐ: : phải là 1 biểu thức có nghĩa. b) ko có x nên ko phải tìm
Ô xin lỗi bạn, do lúc trước mình ko thấy đề nên bấm bậy, xin lỗi nhiều