Rút gọng:\(\left(\frac{1}{\text{x}-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{\text{x}}+1}{\text{x}-2\sqrt{x}+1}\)
Những câu hỏi liên quan
1) B=\(\left(\frac{\sqrt{x}}{2}\text{+}\frac{1}{2\sqrt{x}}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}\text{+}1}-\frac{\sqrt{x}\text{+}1}{\sqrt{x}-1}\right)\)
\(\left(\frac{2x\text{√}x+x-\text{√}x}{x\text{√}x-1}-\frac{x+\text{√}x}{x-1}\right)\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
a) Rút gọn
b) min
\(choP=\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)a;R\text{ú}tg\text{ọ}nP....b;T\text{í}nhPkhiX=3-2\sqrt{2}c;t\text{ì}mX\text{đ}\text{ể}P=1\)
B=\(\left(\frac{x\sqrt{x}}{x\text{+}\sqrt{x}\text{+}1}-\frac{1}{x\text{+}\sqrt{x}\text{+}1}\right):\frac{2}{\sqrt{x}\text{+}1}\)
Chứng minh A<0 với mọi 0<x<1
1)intsqrt{frac{1-sqrt{x}}{1+sqrt{x}}}dx
2)intfrac{dx}{left(e^x+1right)left(x^2+1right)}
3)intfrac{1+2xsqrt{1-x^2}+2x^2}{1+x+sqrt{1+x^2}}dx
4)intfrac{sin^6x+ctext{os}^6x}{1+6^x}dx
5)int_0^{frac{pi}{2}}frac{sqrt{ctext{os}x}}{sqrt{stext{inx}}+sqrt{ctext{os}x}}dx
6)intfrac{x^4}{2^x+1}dx
7)int_0^{frac{pi^2}{4}}sinsqrt{x}dx
8)intsqrt[6]{1-ctext{os}^3x}.stext{inx}.ctext{os}^5xdx
9)intsqrt{frac{1}{4x}+frac{sqrt{x}+e^x}{sqrt{x}.e^x}}dx
10)intfrac{ctext{os}x+stext{inx}}{left(e^xstext{inx}+1right)...
Đọc tiếp
1)\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)
2)\(\int\frac{dx}{\left(e^x+1\right)\left(x^2+1\right)}\)
3)\(\int\frac{1+2x\sqrt{1-x^2}+2x^2}{1+x+\sqrt{1+x^2}}\)dx
4)\(\int\frac{sin^6x+c\text{os}^6x}{1+6^x}dx\)
5)\(\int_0^{\frac{\pi}{2}}\frac{\sqrt{c\text{os}x}}{\sqrt{s\text{inx}}+\sqrt{c\text{os}x}}dx\)
6)\(\int\frac{x^4}{2^x+1}dx\)
7)\(\int_0^{\frac{\pi^2}{4}}sin\sqrt{x}dx\)
8)\(\int\sqrt[6]{1-c\text{os}^3x}.s\text{inx}.c\text{os}^5xdx\)
9)\(\int\sqrt{\frac{1}{4x}+\frac{\sqrt{x}+e^x}{\sqrt{x}.e^x}}dx\)
10)\(\int\frac{c\text{os}x+s\text{inx}}{\left(e^xs\text{inx}+1\right)s\text{inx}}dx\)
Bài 1: cho P= \(\left(\frac{\sqrt{b}}{\sqrt{a}-\sqrt{a-b}}\text{+}\frac{\sqrt{b}}{\sqrt{a}\text{+}\sqrt{a\text{+}b}}\right)\div\left(1\text{+}\frac{\sqrt{a\text{+}b}}{\sqrt{a-b}}\right)\)
a) Rút gọn P
b) Tính P khi a=\(24-2\sqrt{15}\); b=16
Bài 2: \(x\ge0\), cm: \(x^2-3\sqrt{x}\text{+}\frac{5}{2}\)>0
1.Giải pt sau:(sqrt{2} +2)(xsqrt{2} -1)2xsqrt{2} -sqrt{2}
2.Cho pt: 2(a-1).x-a(x-1)2a+3
3.Giải pt sau:
a) frac{2}{x+frac{text{1}}{text{1}+frac{x+text{1}}{x-2}}}frac{6}{3x-text{1}}
b) frac{frac{x+text{1}}{x-text{1}}-frac{x-text{1}}{x+text{1}}}{text{1}+frac{x+text{1}}{x-text{1}}}frac{x-text{1}}{2left(x+text{1}right)}
Đọc tiếp
1.Giải pt sau:(\(\sqrt{2}\) +2)(x\(\sqrt{2}\) -1)=2x\(\sqrt{2}\) -\(\sqrt{2}\)
2.Cho pt: 2(a-1).x-a(x-1)=2a+3
3.Giải pt sau:
a) \(\frac{2}{x+\frac{\text{1}}{\text{1}+\frac{x+\text{1}}{x-2}}}=\frac{6}{3x-\text{1}}\)
b) \(\frac{\frac{x+\text{1}}{x-\text{1}}-\frac{x-\text{1}}{x+\text{1}}}{\text{1}+\frac{x+\text{1}}{x-\text{1}}}=\frac{x-\text{1}}{2\left(x+\text{1}\right)}\)
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
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Mấy bài kia sao cái phương trình dài thê,s giải sao nổi
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Cho biểu thức
A=\(\text{[}1-\frac{\sqrt{x}}{1+\sqrt{x}}\text{]}:\text{[}\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)
a, Rút gọn A
b, Tìm x để A= \(\frac{1}{2}\)
a) A= (\(\left(\frac{1+\sqrt{x}}{1+\sqrt{x}}-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x-2}\right)}+\frac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right)\)
A=\(\left(\frac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}\right)\)
A= \(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\left(\frac{1}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
A=\(\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
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b) Để A = \(\frac{1}{2}\)
thì \(\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{1}{2}\)
=> 2\(\sqrt{x}-4\)=\(\sqrt{x}+1\)
=> \(\sqrt{x}=5\)
=> x = 25
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Cho biểu thức
A= \(\text{[}1-\frac{\sqrt{x}}{1+\sqrt{x}}\text{]}:\text{[}\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\)
a, Rút gọn A
b, Tìm x để A<0
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
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