B=\(\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
a)Rút GọnB
b)Tìm x để B>2
c)Tìm MaxB
a/ \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{x-4}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-2}-1\right)\)
=> \(B=\left(\frac{1}{\sqrt{x}+2}+\frac{7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
=> \(B=\frac{\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{1}{\sqrt{x}-2}\)
=> \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ B>2 <=> \(\frac{\sqrt{x}+5}{\sqrt{x}+2}>2\) <=> \(\sqrt{x}+5>2\sqrt{x}+4\)
<=> \(1>\sqrt{x}\)=> \(-1\le x\le1\)
c/ \(B=\frac{\sqrt{x}+5}{\sqrt{x}+2}=\frac{\sqrt{x}+2+3}{\sqrt{x}+2}=1+\frac{3}{\sqrt{x}+2}\)
Để Bmax thì \(\sqrt{x}+2\) đạt giá trị nhỏ nhất . Do \(\sqrt{x}+2\ge2\)=> Đạt nhỏ nhất khi x=0
Khí đó giá trị lớn nhất của B là: \(1+\frac{3}{2}=\frac{5}{2}\)Đạt được khi x=0
P=\(\frac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\frac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\frac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
a.Rút gọn P
b.Tìm x,y nguyên để P=2
Rút gọn biểu thức
M=\(\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a.Rút gọn M
b.Tìm x để 3M=4-x
a, ĐKXĐ: \(x>0;x\ne1;x\ne4\)
\(M=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
\(=\frac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-1-x+2}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{2}{2\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\)
=\(\frac{\left(x+1\right).\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x-2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x+2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+4x+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(x+4\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
LƯU Ý: CAP NÀY CHỈ LÀ CAP NHÁP
\(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-3}{x-\sqrt{x}-2}\right):\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{2}{\sqrt{x}-2}\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x+\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-\sqrt{x}+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-x+3+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{x-\sqrt{x}+2\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+2}\)
#)Hỏi j đi bn, bn ph hỏi cái j chứ làm lun rùi còn để cộng đồng ngắm ak ???
Bó cả tay lẫn chân !!! Bất lực như gặp cực hình !
Chắc là bạn ấy hỏi bạn ấy làm có đúng ko ha gì đó ?
\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}}{1-x}\right)\)
a.Rút gọn
b. Tìm x để P <0 (cứu mình câu này với T_T)
c. Tìm GTNN của P (câu này tớ cũng hông biết T_T)
Kamsamita.
a: \(P=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}}{x-1}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x}\)
b: Để P<0 thì \(\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)< 0\)
=>1<x<4
Cho B=\(\left(\frac{2x+1}{\sqrt{x^3-1}}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
a.Rút gọn B
b.Tìm x để B=3
Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
