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
Những câu hỏi liên quan
1) B=\(\left(\frac{\sqrt{x}}{2}\text{+}\frac{1}{2\sqrt{x}}\right)\left(\frac{\sqrt{x}-1}{\sqrt{x}\text{+}1}-\frac{\sqrt{x}\text{+}1}{\sqrt{x}-1}\right)\)
\(choP=\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)a;R\text{ú}tg\text{ọ}nP....b;T\text{í}nhPkhiX=3-2\sqrt{2}c;t\text{ì}mX\text{đ}\text{ể}P=1\)
Rút gọng:\(\left(\frac{1}{\text{x}-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{\text{x}}+1}{\text{x}-2\sqrt{x}+1}\)
\(\left(\frac{2x\text{√}x+x-\text{√}x}{x\text{√}x-1}-\frac{x+\text{√}x}{x-1}\right)\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
a) Rút gọn
b) min
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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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)...
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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: Tính
Asqrt{5-2text{√}6}+sqrt{5+2text{√}6}
B left(sqrt{10}+sqrt{6}right)sqrt{8-2text{√}15}
Csqrt{4+text{√}7}+sqrt{4-text{√}7}
Dleft(3+text{√}5right)left(text{√}10-text{√}2right)sqrt{3-text{√}5}
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a0
b, x-2sqrt{xy}+y left(xge0;yge0right)
c, sqrt{xy}+2text{√}x-3text{√}y-6left(xge0;yge0right)
Bài 3: Rút gọn
M left(frac{1}{text{√}x-1}-frac{1}{text{√}x}right)divleft(frac{text{√}x+1}{text{√}x-2}-frac{text{√}x+2}{text{√}x-1}right)
a, Rút gọn...
Đọc tiếp
Bài 1: Tính
A=\(\sqrt{5-2\text{√}6}+\sqrt{5+2\text{√}6}\)
B= \(\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\text{√}15}\)
C=\(\sqrt{4+\text{√}7}+\sqrt{4-\text{√}7}\)
D=\(\left(3+\text{√}5\right)\left(\text{√}10-\text{√}2\right)\sqrt{3-\text{√}5}\)
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a>=0
b, x-2\(\sqrt{xy}\)+y \(\left(x\ge0;y\ge0\right)\)
c, \(\sqrt{xy}+2\text{√}x-3\text{√}y-6\)\(\left(x\ge0;y\ge0\right)\)
Bài 3: Rút gọn
M= \(\left(\frac{1}{\text{√}x-1}-\frac{1}{\text{√}x}\right)\div\left(\frac{\text{√}x+1}{\text{√}x-2}-\frac{\text{√}x+2}{\text{√}x-1}\right)\)
a, Rút gọn M
b, Tính giá trị của M khi x=2
c, Tìm x để M>0
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
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Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
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Bài 3:
a) ĐKXĐ:\(x>0; x\neq 1; x\neq 4\)
\(M=\frac{\sqrt{x}-(\sqrt{x}-1)}{(\sqrt{x}-1)\sqrt{x}}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(x-1)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b)
Khi $x=2$ \(M=\frac{\sqrt{2}-2}{3\sqrt{2}}=\frac{1-\sqrt{2}}{3}\)
c)
Để \(M>0\leftrightarrow \frac{\sqrt{x}-2}{3\sqrt{x}}>0\leftrightarrow \sqrt{x}-2>0\leftrightarrow x>4\)
Kết hợp với ĐKXĐ suy ra $x>4$
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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
tính: \(\text{[}\sqrt{2}-1\text{]}^2-\frac{3}{2}\cdot\sqrt{\text{[}-2\text{]}^2}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}\cdot\sqrt{2}\)
Chứng minh: \(\sqrt{x}\cdot\text{[}1-\sqrt{x}\text{]}\le\frac{1}{4}v\text{ới}x\ge0\)
TÍNH : \(\left(\sqrt{2}-1\right)^2-\frac{3}{2}\sqrt{\left(-2\right)^2}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}.\sqrt{2}\)
\(=\left(\sqrt{2}-1\right)^2-\frac{3}{2}.2+\frac{4\sqrt{2}}{5}+\sqrt{\frac{36}{25}}.\sqrt{2}\)
\(=3-2\sqrt{2}-3+\frac{4\sqrt{2}}{5}+\frac{6\sqrt{2}}{5}=\frac{10\sqrt{2}}{5}-2\sqrt{2}=2\sqrt{2}-2\sqrt{2}=0\)
CHỨNG MINH :
Ta có : \(\sqrt{x}\left(1-\sqrt{x}\right)=-x+\sqrt{x}=-\left[\left(\sqrt{x}\right)^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right]+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)với mọi \(x\ge0\)
Vậy ta có điều phải chứng minh.
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