Toán

2: \(\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{2}-1}-\dfrac{1}{\sqrt{5}-2}\)

\(=\left(2\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}+1\right)-\sqrt{5}-2\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}-\sqrt{5}-2\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-2\sqrt{5}-2\)

3: \(=2\cdot3\sqrt{3}-6\cdot\dfrac{1}{\sqrt{3}}+2-\sqrt{3}-3\sqrt{3}\)

\(=6\sqrt{3}-2\sqrt{3}+2-4\sqrt{3}=2\)

1) \(\dfrac{3+\sqrt{3}}{\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)

\(=\dfrac{\sqrt{5}\cdot\left(3+\sqrt{3}\right)}{\sqrt{5}\cdot\sqrt{5}}-\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\dfrac{3\sqrt{5}+\sqrt{15}}{5}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}\)

\(=\dfrac{3\sqrt{5}+\sqrt{15}}{5}-\left(\sqrt{3}-1\right)\)

\(=\dfrac{3\sqrt{5}+\sqrt{15}-5\sqrt{3}+5}{5}\)

2) \(\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{2}-1}-\dfrac{1}{\sqrt{5}-2}\)

\(=\dfrac{\left(2\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)

\(=\dfrac{2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}}{2-1}-\dfrac{\sqrt{5}+2}{5-4}\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}-\left(\sqrt{5}+2\right)\)

\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-2\sqrt{5}-2\)

3) \(2\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{1}{2}+\dfrac{\sqrt{3}-9}{\sqrt{3}}\)

\(=2\cdot3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{1}{2}+\dfrac{\sqrt{3}\left(1-3\sqrt{3}\right)}{\sqrt{3}}\)

\(=6\sqrt{3}-\dfrac{\sqrt{3}\cdot2\sqrt{3}}{\sqrt{3}}+\dfrac{1}{2}+1-3\sqrt{3}\)

\(=6\sqrt{3}-2\sqrt{3}+\dfrac{1}{2}+1-3\sqrt{3}\)

\(=\dfrac{1}{2}+1+\sqrt{3}\)

\(=\dfrac{3}{2}+\sqrt{3}\)

\(a)O\widehat{A}T=80^o\Rightarrow x\widehat{At}=100^{^{ }o}\)

\(\Rightarrow x\widehat{At}'=50^o\)

Do đó,\(x\widehat{O}y=x\widehat{At}'\Rightarrow OY//AT\)

B)\(x\widehat{Oy}=O\widehat{Bn}=50^o\Rightarrow OX//BN\)

1) \(x^2-25\)

\(=x^2-5^2\)

\(=\left(x-5\right)\left(x+5\right)\)

2) \(9x^2-\dfrac{1}{16}y^2\)

\(=\left(3x\right)^2-\left(\dfrac{1}{4}y\right)^2\)

\(=\left(3x-\dfrac{1}{4}y\right)\left(3x+\dfrac{1}{4}y\right)\)

3) \(x^6-y^4\)

\(=\left(x^3\right)^2-\left(y^2\right)^2\)

\(=\left(x^3-y^2\right)\left(x^3+y^2\right)\)

4) \(\left(2x-5\right)^2-64\)

\(=\left(2x-5\right)^2-8^2\)

\(=\left(2x-5-8\right)\left(2x-5+8\right)\)

\(=\left(2x-13\right)\left(2x+3\right)\)

5) \(81-\left(3x+2\right)^2\)

\(=9^2-\left(3x+2\right)^2\)

\(=\left(9-3x-2\right)\left(9+3x+2\right)\)

\(=\left(7-3x\right)\left(3x+11\right)\)

6) \(9\left(x-5y\right)^2-16\left(x+y\right)^2\)

\(=\left[3\left(x-5y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)

\(=\left(3x-15y-4x-4y\right)\left(3x-15y+4x+4y\right)\)

\(=\left(-x-19y\right)\left(8x-11y\right)\)

7: x^3-8

=x^3-2^3

=(x-2)(x^2+2x+4)

8: 27x^3+125y^3

=(3x)^3+(5y)^3

=(3x+5y)(9x^2-15xy+25y^2)

9: x^6+216

=(x^2)^3+6^3

=(x^2+6)(x^4-6x^2+36)

10: x^2+8x+16=(x+4)^2

11: 9x^2-12xy+4y^2=(3x-2y)^2

12: =-(25x^2y^2-10xy+1)=-(5xy-1)^2

13: =x^3-3*x^2*2+3*x*2^2-2^3

=(x-2)^3

14: =(2x)^3+3*(2x)^2*y+3*2x*y^2+y^3

=(2x+y)^3

15: \(=2xy\left(x^2+y^2+2xy-1\right)\)

\(=2xy\left[\left(x+y\right)^2-1\right]\)

=2xy(x+y+1)(x+y-1)

16: =(x-y)^2+4(x-y)

=(x-y)(x-y+4)

17: =x^3+3x^2y+3xy^2+y^3-x-y

=(x+y)^3-(x+y)

=(x+y)[(x+y)^2-1]

=(x+y)(x+y+1)(x+y-1)

18: =(x-y)^2-4z^2

=(x-y-2z)(x-y+2z)

19: =x^2-y^2-(x+y)

=(x+y)(x-y)-(x+y)

=(x+y)(x-y-1)

20: =(x-y)^2-z^2

=(x-y-z)(x-y+z)

=3(6x^2-5x+1)-(18x^2-2x-27x+3)

=18x^2-15x+3-18x^2+29x-3

=14x

\(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)\)

\(=3\cdot\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)\)

\(=3\cdot\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)\)

\(=18x^2-15x+3-18x^2+29x-3\)

\(=\left(18x^2-18x^2\right)+\left(-15x+29x\right)+\left(3-3\right)\)

\(=0+14x+0\)

\(=14x\)

a) Đường thẳng c cắt 2 đường thẳng a,b tại A và B tạo thành cặp góc trong cùng phía bù nhau

=> a // b

Vì a // b

=> Hai cặp góc so le trong bẳng nhau

b) Vì a //b (câu a)

=> Hai cặp góc đồng vị bằng nhau

Sửa đề: \(B=\sqrt{\left(x-2023\right)^2}+\sqrt{\left(x-1\right)^2}\)

B=|x-2023|+|x-1|

=|x-2023|+|1-x|

=>B>=|x-2023+1-x|=2022

Dấu = xảy ra khi

(x-2023)(x-1)<=0

TH1: x-2023<=0 và x-1>=0

=>x<=2023 và x>=1

=>1<=x<=2023

TH2: x-2023>=0 và x-1<=0

=>x>=2023 hoặc x<=1

=>Loại

=>7x+15=391/23=17

=>7x=2

=>x=2/7

\(\left(x.7+15\right).23=391\) 

 \(x.7+15=\dfrac{391}{23}\) 

\(x.7+15=17\) 

\(x.7=17-15\) 

\(x.7=2\)

   \(x=\dfrac{2}{7}\) 

Gọi ba số lẻ liên tiếp là 2k+1;2k+3;2k+5

Theo đề, ta có: (2k+3)(2k+5)-(2k+1)(2k+3)=212

=>(2k+3)(2k+5-2k-1)=212

=>2k+3=212/4=53

=>2k=50

=>k=25

Vậy: Ba số cần tìm là 51;53;55

Gọi k là số tự nhiên \(k\in N\)

Số lẻ thứ nhất là: \(2k+1\)

Số lẻ thứ hai là: \(2k+3\)

Số tự nhiên thứ ba là: \(2k+5\)

Tích của 2 số lẻ đầu tiên là: \(\left(2k+1\right)\left(2k+3\right)\)

Tích của hai số lẻ sau là: \(\left(2k+3\right)\left(2k+5\right)\)

Mà tích của hai số lẻ sau lớn hơn tích của hai số lẻ đầu 212 nên ta có:

\(\left(2k+3\right)\left(2k+5\right)-\left(2k+1\right)\left(2k+3\right)=212\)

\(\Leftrightarrow4k^2+10k+6k+15-\left(4k^2+6k+2k+3\right)=212\)

\(\Leftrightarrow4k^2+16k+15-4k^2-8k-3=212\)

\(\Leftrightarrow\left(4k^2-4k^2\right)+\left(16k-8k\right)+\left(15-3\right)=212\)

\(\Leftrightarrow8k+12=212\)

\(\Leftrightarrow8k=212-12\)

\(\Leftrightarrow8k=200\)

\(\Leftrightarrow k=\dfrac{200}{8}\)

\(\Leftrightarrow k=25\left(tm\right)\)

Số lẻ thứ nhất là: \(2\cdot25+1=51\)

Số lẻ thứ hai là: \(2\cdot25+3=53\)

Số lẻ thứ ba là: \(2\cdot25+5=55\)

1: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)

\(=3+2\sqrt{2}+\sqrt{5}-2=2\sqrt{2}+\sqrt{5}+1\)

2: \(\dfrac{1}{\sqrt{3}+\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)

\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{2\left(1+\sqrt{7}\right)}{-6}\)

\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}-\dfrac{1+\sqrt{7}}{3}\)

\(=\dfrac{3\left(\sqrt{7}-\sqrt{3}\right)-4\left(\sqrt{7}+1\right)}{12}=\dfrac{-\sqrt{7}-3\sqrt{3}-4}{12}\)

3:

\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=-\sqrt{a}\)

4:

\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{xy}\)

1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

\(=\dfrac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

\(=\dfrac{3+2\sqrt{2}}{3^2-\left(2\sqrt{2}\right)^2}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}\right)^2-2^2}\)

\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)

\(=3+2\sqrt{2}+\sqrt{5}-2\)

\(=2\sqrt{2}+\sqrt{5}+1\)

2) \(\dfrac{1}{\sqrt{3}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)

\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{1^2-\left(\sqrt{7}\right)^2}\)

\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{2\cdot\left(1+\sqrt{7}\right)}{6}\)

\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{1+\sqrt{7}}{3}\)

\(=\dfrac{-3\sqrt{3}-3\sqrt{7}}{12}-\dfrac{4+4\sqrt{7}}{12}\)

\(=\dfrac{-3\sqrt{3}-3\sqrt{7}-4-4\sqrt{7}}{12}\)

\(=\dfrac{-3\sqrt{3}-7\sqrt{7}-4}{12}\)

3) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}\)

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

\(=-\dfrac{\sqrt{a}\cdot\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)

\(=-\sqrt{a}\)

4) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{xy}\)

Để hàm số đồng biến thì 6-2m>0

=>2m<6

=>m<3

Ta có hàm số :

\(y=\left(6-2m\right)x+m-2\)

Hàm số y đồng biến khi:

\(6-2m>0\)

\(\Leftrightarrow6>2m\)

\(\Leftrightarrow2m< 6\)

\(\Leftrightarrow m< \dfrac{6}{2}\)

\(\Leftrightarrow m< 3\) 

Hàm số y đồng biến khi \(m< 3\)