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https://hoc24.vn/cau-hoi/.1145652136620
b) \(x=25\Rightarrow P=\dfrac{\sqrt{25}+1}{\sqrt{25}-3}=\dfrac{6}{2}=3\)
c) \(A< 1\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}< 1\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\Rightarrow\dfrac{4}{\sqrt{x}-3}< 0\)
mà \(4>0\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0\le x< 9,x\ne4\)
\(a,=6\sqrt{3}-10\sqrt{3}+15\sqrt{3}=11\sqrt{3}\\ b,=2\sqrt{5}-\sqrt{5}+70\sqrt{5}=71\sqrt{5}\\ c,=\dfrac{\sin43^0}{\sin43^0}+1=1+1=2\\ d,Sửa:\dfrac{\tan32^0}{\cot68^0}-\cos30^0-\dfrac{\sin18^0}{\sin82^0}=\dfrac{\tan32^0}{\tan32^0}-\dfrac{\sqrt{3}}{2}-\dfrac{\sin18^0}{\cos18^0}=1-1-\dfrac{\sqrt{3}}{2}=-\dfrac{\sqrt{3}}{2}\)
\(a)\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\left(\sqrt{7}+\sqrt{5}\right)^2+\left(\sqrt{7}+\sqrt{5}\right)^2}{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)}\)
\(=\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\)
\(=\frac{24}{2}\)
\(=12\)
\(b)\frac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\left(2+\sqrt{8}-\sqrt{6}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{2}+2-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(2+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}\right)}{2+\sqrt{2}-\sqrt{3}}\)
\(=1+\sqrt{2}\)
\(c)A=\left(\sqrt{3}+1\right)\sqrt{\frac{14-6\sqrt{3}}{5+\sqrt{3}}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{70-14\sqrt{3}-30\sqrt{3}+18}{25-3}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{88-44\sqrt{3}}{22}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\frac{44\left(2-\sqrt{3}\right)}{22}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{2\left(2-\sqrt{3}\right)}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{4+2\sqrt{3}}\)
\(A=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(A=3-1=2\)
P/s: nếu đề là vậy thì t ra kết quả như vậy ạ, nhưng lần sau khi đăng câu hỏi bạn nên viết rõ hơn ra nhé
a) \(\sqrt{33-12\sqrt{6}}+\sqrt{15+6\sqrt{6}}=\sqrt{24-2.2\sqrt{6}.3+9}+\sqrt{6+2.\sqrt{6}.3+9}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}+3\right)^2}=\left|2\sqrt{6}-3\right|+\left|\sqrt{6}+3\right|=2\sqrt{6}-3+\sqrt{6}+3=3\sqrt{6}\)
b) \(\dfrac{\sqrt{99}}{\sqrt{11}}+\dfrac{\sqrt{28}}{\sqrt{7}}-\sqrt{\sqrt{81}}=\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-\sqrt{9}=\sqrt{9}+\sqrt{4}-\sqrt{9}=\sqrt{4}=2\)
a) \(\sqrt{33-12\sqrt{6}}\) + \(\sqrt{15+6\sqrt{6}}\)
= \(\sqrt{9-2.3.2\sqrt{6}+24}\)+\(\sqrt{9+2.3\sqrt{6}+6}\)
= \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)+\(\sqrt{\left(3+\sqrt{6}\right)^2}\)
=\(\left|3-2\sqrt{6}\right|+\left|3+\sqrt{6}\right|\)
=\(2\sqrt{6}-3+3+\sqrt{6}\)
=\(\sqrt{6}\)
b)\(\dfrac{\sqrt{99}}{\sqrt{11}}\)+\(\dfrac{\sqrt{28}}{\sqrt{7}}\)\(-\sqrt{\sqrt{81}}\)
= \(\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-3\)
=\(\sqrt{9}+\sqrt{4}-3\)
= 3+2-3
= 2
a) ĐK: a>0,a\(\ne1\)
b) \(M=\left(\dfrac{\sqrt{a}+2}{a+2\sqrt{a}+1}-\dfrac{\sqrt{a}-2}{a-1}\right)\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{a+\sqrt{a}-2}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}-\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\left[\dfrac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\right]\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}=\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{2}{a-1}\)
c) Để M là số nguyên thì \(\dfrac{2}{a-1}\in Z\Rightarrow a-1\inƯ\left(2\right)\in\left(\pm1,\pm2\right)\)\(\Rightarrow\left[{}\begin{matrix}a-1=1\\a-1=-1\\a-1=2\\a-1=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=2\left(tm\right)\\a=0\left(ktm\right)\\a=3\left(tm\right)\\a=-1\left(ktm\right)\end{matrix}\right.\)
Vậy để M là số nguyên thì a=2 hoặc a=3
a) \(9+4\sqrt{5}=\left(\sqrt{5}\right)^2+2.\sqrt{5}.2+2^2=\left(\sqrt{5}+2\right)^2\)
b) \(23-8\sqrt{7}=4^2-2.4.\sqrt{7}+\left(\sqrt{7}\right)^2=\left(4-\sqrt{7}\right)^2\)
c) \(4-2\sqrt{3}=\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2=\left(\sqrt{3}-1\right)^2\)
d) \(11+6\sqrt{2}=3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(3+\sqrt{2}\right)^2\)
a) \(9+4\sqrt{5}=\left(\sqrt{5}+2\right)^2\)
b) \(23-8\sqrt{7}=\left(4-\sqrt{7}\right)^2\)
c) \(4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)
d) \(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
\(a+\sqrt{1-a^2}=b+\sqrt{1-b^2}\)
\(\Rightarrow a\sqrt{1-a^2}=b\sqrt{1-b^2}\)
\(\Rightarrow a^2\left(1-a^2\right)=b^2\left(1-b^2\right)\)
\(\Rightarrow a^2-a^4=b^2-b^4\)
\(\Rightarrow a^2-b^2-\left(a^2-b^2\right)\left(a^2+b^2\right)=0\)
\(\Rightarrow\left(a^2-b^2\right)\left(a^2+b^2-1\right)=0\)
\(\Rightarrow a^2+b^2=1\)
- Với \(y=0\Rightarrow x=1\) là 1 nghiệm của pt
- Với \(y>0\Rightarrow2016^y\) luôn chẵn
\(VT=1+x\left(x^3+x^2+x+1\right)=1+x\left(x+1\right)\left(x^2+1\right)\)
Do \(x\left(x+1\right)\) là tích 2 STN liên tiếp nên luôn chẵn
\(\Rightarrow VT\) luôn lẻ \(\Rightarrow\) pt vô nghiệm
Vậy pt có cặp nghiệm duy nhất \(\left(x;y\right)=\left(0;0\right)\)
a) \(A=sin42^0+tan45^0-cos48^0=sin42^0+1-sin42^0=1\)
Bài b bạn coi thử có nhầm lẫn chỗ nào không
đúng đề r nha bn