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5 tháng 2 2022

\(H=\left(\sqrt{5}-2\right)^2.\left(\sqrt{5}+2\right)^2\)

\(H=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)

\(H=\left[\left(\sqrt{5}\right)^2-2^2\right].\left[\left(\sqrt{5}\right)^2-2^2\right]\)

\(H=1.1=1\)

5 tháng 2 2022

\(I=\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)\)

\(I=11^2-\left(4\sqrt{3}\right)^2\)

\(I=121-48=73\)

24 tháng 9 2021

1) \(\sqrt{\dfrac{1}{200}}\)                            2) \(\dfrac{5}{1-\sqrt{6}}\)

\(=\sqrt{\dfrac{1^2}{10^2.2}}\)                          \(=\dfrac{1-\sqrt{6}+4+\sqrt{6}}{1-\sqrt{6}}\)

\(=\dfrac{1}{10\sqrt{2}}\)                              \(=1+\dfrac{4+\sqrt{6}}{1-\sqrt{6}}\)

24 tháng 9 2021

Bài 2: 

1. \(\sqrt{2x-5}=7\)    ĐKXĐ: \(x\ge\dfrac{5}{2}\)

<=> 2x - 5 = 72

<=> 2x - 5 = 49

<=> 2x = 54

<=> x = 27 (TM)

2. \(3+\sqrt{x-2}=4\)     ĐKXĐ: \(x\ge2\)

<=> \(\sqrt{x-2}=1\)

<=> x - 2 = 1

<=> x = 3 (TM)

3. \(\sqrt{x^2-2x+1}=1\)

<=> \(\sqrt{\left(x-1\right)^2}=1\)

<=> \(|x-1|=1\)

<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

4. \(\sqrt{x^2-4x+4}=1\)

<=> \(\sqrt{\left(x-2\right)^2}=1\)

<=> \(|x-2|=1\)

<=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

5. \(\sqrt{4x^2+1-4x}=\sqrt{x^2+16+8x}\)

<=> \(\left(\sqrt{4x^2+1-4x}\right)^2=\left(\sqrt{x^2+16+8x}\right)^2\)

<=> \(|4x^2+1-4x|=|x^2+16+8x|\)

<=> \(\left[{}\begin{matrix}4x^2+1-4x=x^2+16+8x\\4x^2+1-4x=-\left(x^2+16+8x\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x^2-x^2-4x-8x+1-16=0\\4x^2+1-4x=-x^2-16-8x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}3x^2-12x-15=0\\5x^2+4x+17=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}3x^2+3x-15x-15=0\\VNghiệm\end{matrix}\right.\)

<=> 3x(x + 1) - 15(x + 1) = 0

<=> (3x - 15)(x + 1) = 0

<=> \(\left[{}\begin{matrix}3x-15=0\\x+1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

1: \(\sqrt{\dfrac{1}{200}}=\dfrac{\sqrt{2}}{20}\)

2: \(\dfrac{5}{1-\sqrt{6}}=-1-\sqrt{6}\)

3: \(\dfrac{1}{1-\sqrt{2}}-\dfrac{1}{1+\sqrt{2}}\)

\(=\dfrac{1+\sqrt{2}-1+\sqrt{2}}{-1}\)

\(=-2\sqrt{2}\)

2: Ta có: \(\sqrt{16-6\sqrt{7}}\cdot\left(3+\sqrt{7}\right)\)

\(=\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)\)

=9-7

=2

3: Ta có: \(\left(\sqrt{6}+\sqrt{14}\right)\cdot\sqrt{5-2\sqrt{21}}\)

\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)\)

=7-3

=4

27 tháng 9 2021

\(1,=\sqrt{\left(5+2\sqrt{6}\right)^2}-\sqrt{\left(3-\sqrt{6}\right)^2}=5+2\sqrt{6}-3+\sqrt{6}=2+3\sqrt{6}\\ 2,=\sqrt{\left(3-\sqrt{7}\right)^2}\left(3+\sqrt{7}\right)=\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)=9-7=2\\ 3,=\left(\sqrt{3}+\sqrt{7}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{3}+\sqrt{7}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\\ =\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=7-3=4\\ 4,=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{6+2\sqrt{5}}+\sqrt{4-2\sqrt{3}}\right)\\ =\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+1+\sqrt{3}-1\right)\\ =\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)

\(5,\\ =\sqrt{\left(3\sqrt{3}-5\right)^2}+\sqrt{\left(5-2\sqrt{3}\right)^2}=3\sqrt{3}-5+5-2\sqrt{3}=\sqrt{3}\\ 6,=\sqrt{13-4\sqrt{10}}-\sqrt{53+12\sqrt{10}}\\ =\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}-2\sqrt{2}\right)^2}\\ =2\sqrt{2}-\sqrt{5}-3\sqrt{5}+2\sqrt{2}=4\sqrt{2}-4\sqrt{5}\)

11 tháng 12 2023

Câu II:

a: \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{4\sqrt{x}}{x-1}\right):\left(\dfrac{x+\sqrt{x}+5}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right)\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+5-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+4}\)

\(=\dfrac{8\sqrt{x}}{x+4}\)

Câu IV

3: xy-x-y=2

=>xy-x-y+1=3

=>x(y-1)-(y-1)=3

=>(x-1)(y-1)=3

=>\(\left(x-1\right)\cdot\left(y-1\right)=1\cdot3=3\cdot1=\left(-1\right)\cdot\left(-3\right)=\left(-3\right)\cdot\left(-1\right)\)

=>\(\left(x-1;y-1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;4\right);\left(4;2\right);\left(0;-2\right);\left(-2;0\right)\right\}\)

2:

Khoảng cách từ O(0;0) đến (d) là:

\(d\left(O;\left(d\right)\right)=\dfrac{\left|0\cdot\left(m-4\right)+0\cdot\left(m-3\right)-1\right|}{\sqrt{\left(m-4\right)^2+\left(m-3\right)^2}}\)

\(=\dfrac{1}{\sqrt{2m^2-14m+25}}\)

Để d(O;(d)) lớn nhất thì \(2m^2-14m+25\) nhỏ nhất

\(2m^2-14m+25\)=\(2\left(m^2-7m+12,5\right)\)

\(=2\left(m^2-2\cdot m\cdot\dfrac{7}{2}+\dfrac{49}{4}+\dfrac{1}{4}\right)\)

\(=2\left(m-\dfrac{7}{2}\right)^2+\dfrac{1}{2}>=\dfrac{1}{2}\)

Dấu '=' xảy ra khi m=7/2

Vậy: m=7/2

a: Thay x=0 và y=5 vào (d), ta được:

(m-2)x0+m=5

=>m=5

c: Để hai đườg song song thì m-2=2

hay m=4

Câu b bạn chụp thiếu đề rồi 

21 tháng 12 2021

b: \(=3-\sqrt{5}+4-\sqrt{5}=7-2\sqrt{5}\)