Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(P=\left(3-\dfrac{3}{\sqrt{x}-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3\left(\sqrt{x}-1\right)-3}{\sqrt{x}-1}\right):\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+2}\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-6}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)\)
\(=3\sqrt{x}-6\)
b) \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
\(\Leftrightarrow3\sqrt{x}-6=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\) (1)
ĐKXĐ: \(x>0\)
\(\left(1\right)\Leftrightarrow3x-6\sqrt{x}=4\sqrt{x}-1\)
\(\Leftrightarrow3x-6\sqrt{x}-4\sqrt{x}+1=0\)
\(\Leftrightarrow3x-10\sqrt{x}+1=0\) (2)
Đặt \(t=\sqrt{x}\ge0\)
\(\left(2\right)\Leftrightarrow3t^2-10t+1=0\)
\(\Delta'=25-4=22\)
Phương trình có hai nghiệm phân biệt:
\(t_1=\dfrac{5+\sqrt{22}}{3}\) (nhận)
\(t_2=\dfrac{5-\sqrt{22}}{3}\) (nhận)
Với \(t=\dfrac{5+\sqrt{22}}{3}\) \(\Leftrightarrow\sqrt{x}=\dfrac{5+\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47+10\sqrt{22}}{9}\) (nhận)
Với \(t=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow\sqrt{x}=\dfrac{5-\sqrt{22}}{3}\Leftrightarrow x=\dfrac{47-10\sqrt{22}}{9}\) (nhận)
Vậy \(x=\dfrac{47+10\sqrt{22}}{9};x=\dfrac{47-10\sqrt{22}}{9}\) thì \(P=\dfrac{4\sqrt{x}-1}{\sqrt{x}}\)
a: \(P=\dfrac{3\sqrt{x}-3-3}{\sqrt{x}-1}:\dfrac{x+2-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=3\sqrt{x}-6\)
b: P=(4căn x-1)/căn x
=>3x-6căn x-4căn x+1=0
=>3x-10căn x+1=0
=>x=(47+10căn 22)/9 hoặc x=(47-10căn 22)/9
\(a)E=\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-1\right):\left(\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+2}\right):\dfrac{4-x+x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}:\dfrac{9-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-2}{\sqrt{x}+2}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{9-x}\\ =\dfrac{-2\left(\sqrt{x}-3\right)}{9-x}=\dfrac{2\left(\sqrt{x}-3\right)}{x-9}\\ =\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2}{\sqrt{x}+3}\)
\(b)\)E dương
\(\Leftrightarrow E>0\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+3}>0\\ \Leftrightarrow\sqrt{x}+3>0\left(Vì.2>0\right)\\ \Leftrightarrow\sqrt{x}>-3\forall x\in R\\ \Rightarrow x\ge0\)
Kết hợp đk
\(x\ge0;x\ne4;x\ne9\)
Vậy \(x\ge0;x\ne4;x\ne9\) thì E dương
\(a,A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\left(dk:x\ge0,x\ne4\right)\\ =\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x-4+10-x}\)
\(=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\sqrt{x}-2}.\dfrac{1}{6}\\ =\dfrac{-6}{\left(\sqrt{x}-2\right).6}\\
=-\dfrac{1}{\sqrt{x}-2}\)
\(b,A>0\Leftrightarrow-\dfrac{1}{\sqrt{x}-2}>0\Leftrightarrow\sqrt{x}-2< 0\\
\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với \(dk:x\ge0,x\ne4\), ta kết luận \(0\le x< 4\)
\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)
Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)
Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(1-3x\right)^2}=\left|x-1\right|+\left|1-3x\right|\)
\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)
\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)
\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)
\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)
Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)
\(=2\sqrt{x-4}\)
Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
a) \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=-\dfrac{3}{\sqrt{x}+3}.\dfrac{\sqrt{x}+3}{-\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
b) \(\sqrt{x}=\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
Thế vào B \(\Rightarrow B=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)
a) Ta có: \(B=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\dfrac{x-3\sqrt{x}-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-x+4\sqrt{x}-4}\)
\(=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
a: Sửa đề: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\dfrac{2}{x^2-2x+1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{1}{2}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}< 1\)
=>\(0< =x< 1\)
c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
Câu 1:
a) ĐKXĐ: \(x>0;x\ne9\)
Với x=36 (thỏa mãn ĐKXĐ) thì A có giá trị :
\(A=\dfrac{\sqrt{36}+2}{1+\sqrt{36}}=\dfrac{6+2}{1+6}=\dfrac{8}{7}\)
b) Ta có:
\(B=\left(\dfrac{2\sqrt{x}}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
c) Ta có:
\(P=A\cdot B=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)
Vì x là số nguyên lớn hơn 0 nên
\(x\ge1\Rightarrow\sqrt{x}\ge1\Rightarrow\sqrt{x}+1\ge2>0\Rightarrow P\le1+\dfrac{3}{2}=\dfrac{5}{2}\)
Dấu bằng xảy ra khi x=1;
Gọi số sản phẩm dự định của xí nghiệp A và B lần lượt là x,y \(\left(x,y\in N;0< x,y< 720\right)\)
Vì tổng sản phẩm dự định là 720 nên ta có phương trình: \(x+y=720\left(1\right)\)
Vì thực tế , xí nghiệp A hoàn thành vượt mức 12% nên số sản phẩm xí nghiệp A thực tế là : \(112\%x=\dfrac{28}{25}x\)
Xí nghiệp B hoàn thành vượt mức 10% nên số sản phẩm xí nghiệp B thực tế là : \(110\%y=\dfrac{11}{10}y\)
Vì tổng số sản phẩm thực tế là 800 nên ta có phương trình: \(\dfrac{28}{25}x+\dfrac{11}{10}y=800\Leftrightarrow56x+55y=40000\left(2\right)\)
Từ (1)(2) ta có hệ phương trình:
\(\left\{{}\begin{matrix}x+y=720\\56x+55y=40000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=720\\55\cdot720+x=40000\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=400\\y=320\end{matrix}\right.\left(t.m\right)\)
Vậy số sản phẩm 2 xí nghiệp làm theo kế hoạch lần lượt là 400 và 320 sản phẩm
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
b) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 1\)
hay x<1
Kết hợp ĐKXĐ, ta được: 0<x<1
a: \(2\sqrt{x^2}=2\left|x\right|=-2x\)
b: \(=\dfrac{1}{2}\cdot\left| x^5\right|=-\dfrac{1}{2}x^5\)
c: \(=\left|\left(a-5\right)^2\right|=\left(a-5\right)^2\)
d: \(=\left|8a\right|+2a=8a+2a=10a\)
e: \(=\left|3a^3\right|-6a^3=-3a^3\)