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Cho \(5\sqrt{x}7\) mk viet nham
Sua lai thanh \(5\sqrt{x}-7\)
a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)
=>10 căn x+5-5 chia hết cho 2 căn x+1
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
hay \(x\in\varnothing\)
\(ĐKXĐ:x\ge0,x\ne1\)
= \(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
= \(\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\) (1)
b/ Ta có: \(x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
Thay \(x=\left(\sqrt{3}-1\right)^2\) vào (1) ta được:
\(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\left(\sqrt{3}-1\right)^2+\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)= \(\dfrac{\sqrt{3}-1}{4-2\sqrt{3}+\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{4-\sqrt{3}}\) = \(\dfrac{\left(\sqrt{3}-1\right)\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}=\dfrac{3\sqrt{3}-1}{13}\)
Vậy giá trị của A khi \(x=4-2\sqrt{3}\) là \(\dfrac{3\sqrt{3}-1}{13}\)
\(p=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
=\(\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)
=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
=\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
học tốt nhé anh trai
Câu a : \(A=\left(\dfrac{1}{x+\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}+1\)
\(=\dfrac{1}{\sqrt{x}}\times\dfrac{x+2\sqrt{x}+1}{\sqrt{x}-1}+1\)
\(=\dfrac{x+2\sqrt{x}+1}{x-\sqrt{x}}+1\)
\(=\dfrac{x+2\sqrt{x}+1}{x-\sqrt{x}}+\dfrac{x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{2x+\sqrt{x}+1}{x-\sqrt{x}}\)
Câu b : Thay \(x=1\dfrac{1}{3}=\dfrac{4}{3}\) vào A ta được :
\(A=\dfrac{2.\dfrac{4}{3}+\sqrt{\dfrac{4}{3}}+1}{\dfrac{4}{3}-\sqrt{\dfrac{4}{3}}}=\dfrac{\dfrac{8}{3}+\dfrac{2\sqrt{3}}{3}+\dfrac{3}{3}}{\dfrac{4}{3}-\dfrac{2\sqrt{3}}{3}}=\dfrac{\dfrac{11+2\sqrt{3}}{3}}{\dfrac{4-2\sqrt{3}}{3}}=\dfrac{11+2\sqrt{3}}{4-2\sqrt{3}}\)
Chúc bạn học tốt
Bn ơi nếu như mk bấm máy tính thì nó ra là \(\dfrac{28+15\sqrt{3}}{2}\)
a: \(=\dfrac{-a-4\sqrt{a}-4+a-2\sqrt{a}-4a-2\sqrt{a}+4}{a-4}:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\)
\(=\dfrac{-4a-8\sqrt{a}}{a-4}\cdot\dfrac{\sqrt{a}\left(2-\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\dfrac{-4\sqrt{a}\left(\sqrt{a}+2\right)}{a-4}\cdot\sqrt{a}=-\dfrac{4a}{\sqrt{a}-2}\)
b: Để \(A=\sqrt{a}+2\) thì \(-4a=a-4\)
=>-4a-a=-4
=>-5a=-4
=>a=4/5
b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)
\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)
\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)
\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)
\(VT=0=VP\)
\(P=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-2\sqrt{a}-1+1=a+\sqrt{a}-2\sqrt{a}=a-\sqrt{a}\)
\(A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}\)\(B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}\)
\(C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a\)\(D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
\(P=\dfrac{\left(\sqrt{a+1}+1\right)\left(\sqrt{a+1}+2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}+\dfrac{2\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}-\dfrac{2+5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{a+3+3\sqrt{a+1}}{a-3}+\dfrac{2a+2-4\sqrt{a+1}}{a-3}-\dfrac{2+5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{a+3+3\sqrt{a+1}+2a+2-4\sqrt{a+1}-2-5\sqrt{a+1}}{a-3}\)
\(P=\dfrac{3a+3-6\sqrt{a+1}}{a-3}\)
Có thể dừng ở đây hoặc nếu thích thì làm tiếp như sau (chưa chắc gọn hơn):
\(P=\dfrac{3\left(a+1\right)-6\sqrt{a+1}}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}=\dfrac{3\sqrt{a+1}\left(\sqrt{a+1}-2\right)}{\left(\sqrt{a+1}-2\right)\left(\sqrt{a+1}+2\right)}\)
\(P=\dfrac{3\sqrt{a+1}}{\sqrt{a+1}-2}\)