Giúp mình với, minhd đabg cần gấp lắm!!
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1: \(A=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
2: căn x-2<căn x+1
=>A<1
3: A=1/4
=>căn x-2/căn x+1=1/4
=>4 căn x-8=căn x+1
=>3 căn x=9
=>x=9
\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)
\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)
Rút gọn các thừa số ở tử và mẫu ta được:
\(B=\frac{1}{2004}\)
Đ/S:\(\frac{1}{2004}\)
Ta có:
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}....\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)
Đơn giản hết sẽ là:
\(=\frac{1}{2004}\)
1 do you do
2 had done - went
3 went - had read
4 will attend
5 hadn't worn
6 to be
7 weren't sleeping - were playing
8 to be
9 had lived - moved
10 locking
11 had work - retired
12 told - had learned
13 won't call
14 had met
do you do
had done-went
went- had read
will attend
hadn't worn
to be
a: Xét tứ giác AEHF có
\(\widehat{AEH}=\widehat{AFH}=\widehat{FAE}=90^0\)
Do đó: AEHF là hình chữ nhật
Coi như bài toán đã cho là x;y;z hết từ điều kiện đến biểu thức (lẫn lộn abc với xyz)
Đặt \(\left(x^3;y^3;z^3\right)=\left(a^2;b^2;c^2\right)\Rightarrow abc=1\)
Ta có: \(Q=\dfrac{1}{a^2+b^2+b^2+1+2}+\dfrac{1}{b^2+c^2+c^2+1+2}+\dfrac{1}{c^2+a^2+a^2+1+2}\)
\(Q\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ca+2a+2}\)
\(Q\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{ab.bc+abc+ab}+\dfrac{b}{cab+ab+b}\right)\)
\(Q\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{ab}{b+1+ab}+\dfrac{b}{1+ab+b}\right)=\dfrac{1}{2}\)
\(a,\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\\ =\sqrt{1}=1\)
\(b,\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\left(2\sqrt{2}+1\right)}}\\ =\sqrt{13+30\sqrt{3+2\sqrt{2}}}\\ =\sqrt{13+30\left(\sqrt{2}+1\right)}\\ =\sqrt{43+30\sqrt{2}}\\ =5+3\sqrt{2}\)
\(c,\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\\ =\sqrt{1+\sqrt{3+\left(2\sqrt{3}+1\right)}}+\sqrt{1-\sqrt{3-\left(2\sqrt{3}-1\right)}}\\ =\sqrt{1+\sqrt{4+2\sqrt{3}}}+\sqrt{1-\sqrt{4-2\sqrt{3}}}\\ =\sqrt{1+\left(\sqrt{3}+1\right)}+\sqrt{1-\left(\sqrt{3}-1\right)}\\ =\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
Đặt \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(A^2=4+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=4+2=6\\ A=\sqrt{6}\)
Vậy gt biểu thức là \(\sqrt{6}\)
\(d,\sqrt{5-\sqrt{13+4\sqrt{3}}}+\sqrt{3+\sqrt{13+4\sqrt{3}}}\\ =\sqrt{5-\left(2\sqrt{3}+1\right)}+\sqrt{3+\left(2\sqrt{3}+1\right)}\\ =\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\\ =\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
\(4,\)
\(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\\ A^2=2-\sqrt{3}+2+\sqrt{3}+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ A^2=4+2=6\\ A=\sqrt{6}\\ B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\\ B^2=4-\sqrt{7}+4+\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\\ B^2=8-2\sqrt{9}=8-6=2\\ B=\sqrt{2}\)
\(5,\\ a,\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\\ =\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\\ b,\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\\ =\left(\sqrt{5}-\sqrt{2}\right)-\left(\sqrt{2}+\sqrt{5}\right)=-2\sqrt{2}\\ c,\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\left(2\sqrt{5}-2\right)+\left(\sqrt{5}-2\right)\\ =3\sqrt{5}\\ d,\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\left(3-2\sqrt{2}\right)+\left(2\sqrt{2}+1\right)\\ =4\)