K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

26 tháng 6 2021

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

\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}=\dfrac{\left(\sqrt{5}-1\right).\left(1-\sqrt{5}\right)+6.\left(\sqrt{5}+1\right)}{-4}=\dfrac{6-2\sqrt{5}-6\sqrt{5}-6}{4}=\dfrac{-8\sqrt{5}}{4}=-2\sqrt{5}\)

\(\dfrac{\sqrt{2}-\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}=\dfrac{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{6}+2\right)+\left(\sqrt{3}-\sqrt{2}\right).\left(2-\sqrt{6}\right)}{-2}=\dfrac{2\left(\sqrt{12}-\sqrt{18}\right)}{-2}=\sqrt{18}-\sqrt{12}\)

\(\dfrac{-31+8\sqrt{x}-x}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)

\(=\dfrac{-31+8\sqrt{x}-x}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}+\dfrac{3\sqrt{x}-1}{\sqrt{x}-5}\)

\(=\dfrac{-31+8\sqrt{x}-x-x+25+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

 

 

26 tháng 6 2021

bài 14 mà bạn

 

26 tháng 6 2021

14, \(\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{5x+9\sqrt{x}-2}\)

\(=\frac{-7\sqrt{x}+7}{5\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}+2}+\frac{39\sqrt{x}+12}{\left(\sqrt{x}+2\right)\left(5\sqrt{x}-1\right)}\)

\(=\frac{\left(-7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)+\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{-7x-14\sqrt{x}+7\sqrt{x}+14+10x-2\sqrt{x}-10\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3x+20\sqrt{x}+28}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(3\sqrt{x}+14\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}+14}{5\sqrt{x}-1}\)

27 tháng 6 2021

thank

27 tháng 6 2021

16. \(\dfrac{2\sqrt{x}-4}{3\sqrt{x}-4}-\dfrac{4+2\sqrt{x}}{\sqrt{x}-2}+\dfrac{x+13\sqrt{x}-20}{3x-10\sqrt{x}+8}\)

=\(\dfrac{\left(2\sqrt{x}-4\right)\left(\sqrt{x}-2\right)-\left(4+2\sqrt{x}\right)\left(3\sqrt{x}-4\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

=\(\dfrac{2x-8\sqrt{x}+8-\left(4\sqrt{x}+6x-16\right)+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

=\(\dfrac{2x-8\sqrt{x}+8-4\sqrt{x}-6x+16+x+13\sqrt{x}-20}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)

=\(\dfrac{-3x+\sqrt{x}+4}{\left(3\sqrt{x}-4\right)\left(x+2\right)}\)

=\(\dfrac{-\left(3x+3\sqrt{x}-4\sqrt{x}-4\right)}{\left(3\sqrt{x}-4\right)\left(x+2\right)}\)

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

27 tháng 6 2021

14.

=\(\dfrac{-\left(7\sqrt{x}+7\right)\left(\sqrt{x}+2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{\left(2\sqrt{x}-2\right)\left(5\sqrt{x}-1\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

=\(\dfrac{-7x-21\sqrt{x}-14}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{10x-12\sqrt{x}+2}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)+\(\dfrac{39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

=\(\dfrac{-7x-21\sqrt{x}-14+10x-12\sqrt{x}+2+39\sqrt{x}+12}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

=\(\dfrac{3x-6\sqrt{x}}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)=\(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(5\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

17.

17 tháng 7 2021

undefined

17 tháng 7 2021

14a) \(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

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

\(=\sqrt{5}+2-\sqrt{5}+2=4\)

b) \(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}-\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)

\(=\sqrt{7}-1-\sqrt{7}-1=-2\)

15a) \(P=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{3^2+2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

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

b) \(Q=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}+\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

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

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

 

27 tháng 6 2021

`16)(2sqrtx-4)/(3sqrtx-4)-(4+2sqrtx)/(sqrtx-2)+(x+13sqrtx-20)/(3x-10sqrtx+8)`
`=((2sqrtx-4)(sqrtx-2)-(4+2sqrtx)(3sqrtx-4)+x+13sqrtx-20)/((3sqrtx-4)(sqrtx-2)`
`=(2x-8sqrtx+8-6x-4sqrtx+16+x+13sqrtx-20)/((3sqrtx-4)(sqrtx-2)`
`=(-3x+sqrtx+4)/((3sqrtx-4)(sqrtx-2)`
`=(-(3sqrtx-4)(sqrtx+1))/((3sqrtx-4)(sqrtx-2)`
`=(-(sqrtx+1))/(sqrtx-2)`

27 tháng 6 2021

thank

27 tháng 6 2021

`15)(-5sqrtx+4)/(3sqrtx-2)+(6sqrtx+4)/(2sqrtx+3)+(29sqrtx-28)/(3(6x+5sqrtx-6))`
`=(3(-5sqrtx+4)(2sqrtx+3)+3(6sqrtx+4)(3sqrtx-2)+29sqrtx-28)/(3(3sqrtx-2)(2sqrtx+3))`
`=(-30x-21sqrtx+36+54x-24+29sqrtx-28)/(3(3sqrtx-2)(2sqrtx+3))`
`=(24x+8sqrtx-16)/(3(3sqrtx-2)(2sqrtx+3))`

`=(8(3sqrtx-2)(sqrtx+1))/(3(3sqrtx-2)(2sqrtx+3))`

`=(8(sqrtx+1))/(3(2sqrtx+3))`

27 tháng 6 2021

thank

AH
Akai Haruma
Giáo viên
26 tháng 6 2021

11.

\(=\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}+\frac{9-x}{(2-\sqrt{x})(\sqrt{x}+3)}\)

\(=\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(2-\sqrt{x})(\sqrt{x}+3)}+\frac{\sqrt{x}-2}{3+\sqrt{x}}+\frac{9-x}{(2-\sqrt{x})(\sqrt{x}+3)}\)

\(=\frac{x-9}{(2-\sqrt{x})(\sqrt{x}+3)}+\frac{\sqrt{x}-2}{3+\sqrt{x}}+\frac{9-x}{(2-\sqrt{x})(\sqrt{x}+3)}\)

\(=\frac{\sqrt{x}-2}{3+\sqrt{x}}\)

 

 

 

AH
Akai Haruma
Giáo viên
26 tháng 6 2021

12.

\(=\frac{(3-\sqrt{x})(3\sqrt{x}-2)+(5\sqrt{x}+7)(3\sqrt{x}+4)}{(5\sqrt{x}+7)(3\sqrt{x}-2)}-\frac{42\sqrt{x}+34}{(5\sqrt{x}+7)(3\sqrt{x}-2)}\) 

\(=\frac{12x+52\sqrt{x}+22}{(5\sqrt{x}+7)(3\sqrt{x}-2)}-\frac{42\sqrt{x}+34}{(5\sqrt{x}+7)(3\sqrt{x}-2)}\)

\(=\frac{12x+10\sqrt{x}-12}{(5\sqrt{x}+7)(3\sqrt{x}-2)}=\frac{2(3\sqrt{x}-2)(2\sqrt{x}+3)}{(5\sqrt{x}+7)(3\sqrt{x}-2)}=\frac{2(2\sqrt{x}+3)}{5\sqrt{x}+7}\)

 

 

27 tháng 6 2021

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne49\end{matrix}\right.\)

Ta có : \(\dfrac{7\sqrt{x}-1}{\sqrt{x}-7}-\dfrac{6\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(7\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)\left(6\sqrt{x}+1\right)+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{7x+7\sqrt{x}-\sqrt{x}-1-\left(6x-42\sqrt{x}+\sqrt{x}-7\right)+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{7x+7\sqrt{x}-\sqrt{x}-1-6x+42\sqrt{x}-\sqrt{x}+7+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-8\sqrt{x}+7}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-7\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

19) Ta có: \(\dfrac{7\sqrt{x}-1}{\sqrt{x}-7}-\dfrac{6\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{1-55\sqrt{x}}{x-6\sqrt{x}-7}\)

\(=\dfrac{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(6\sqrt{x}+1\right)\left(\sqrt{x}-7\right)}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}+\dfrac{1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{7x+7\sqrt{x}-\sqrt{x}-1-6x+42\sqrt{x}-\sqrt{x}+7+1-55\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-8\sqrt{x}+7}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+1\right)}\)

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

Bài 14:

a)

Sửa đề: \(AE\cdot AB=AD\cdot AC\)

Xét ΔADB vuông tại D và ΔAEC vuông tại E có 

\(\widehat{BAD}\) chung

Do đó: ΔADB\(\sim\)ΔAEC(g-g)

Suy ra: \(\dfrac{AD}{AE}=\dfrac{AB}{AC}\)

hay \(AE\cdot AB=AD\cdot AC\)(đpcm)

b) Ta có: \(\dfrac{AD}{AE}=\dfrac{AB}{AC}\)(cmt)

nên \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)

Xét ΔADB vuông tại D có 

\(\cos\widehat{A}=\dfrac{AD}{AB}\)

Xét ΔAED và ΔACB có 

\(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)(cmt)

\(\widehat{A}\) chung

Do đó: ΔAED∼ΔACB(c-g-c)

Suy ra: \(\dfrac{AD}{AB}=\dfrac{ED}{CB}\)(Các cặp cạnh tương ứng tỉ lệ)

hay \(\dfrac{AD}{AB}\cdot BC=DE\)

\(\Leftrightarrow DE=BC\cdot\cos\widehat{A}\)(đpcm)

c) Ta có: \(DE=BC\cdot\cos\widehat{A}\)(cmt)

nên \(DE=BC\cdot\cos60^0=\dfrac{1}{2}BC\)(1)

Ta có: ΔEBC vuông tại E(gt)

mà EM là đường trung tuyến ứng với cạnh huyền BC(M là trung điểm của BC)

nên \(EM=\dfrac{1}{2}BC\)(2)

Ta có: ΔDBC vuông tại D(gt)

mà DM là đường trung tuyến ứng với cạnh huyền BC(M là trung điểm của BC)

nên \(DM=\dfrac{1}{2}BC\)(3)

Từ (1), (2) và (3) suy ra ME=MD=DE

hay ΔMDE đều(đpcm)

1 tháng 7 2021

Dạ em cảm ơn ạ!