Có \(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=1-\dfrac{10}{\sqrt{x}+5}\)

Dễ thấy \(\dfrac{10}{\sqrt{x}+5}>0\forall x\Rightarrow A=1-\dfrac{10}{\sqrt{x}+5}< 1\)

=> A < 2

Bài 1: Cho A + B = 12. Tính các tổng sau:

a. 13 x A + 5 x B + 13 x B + 5 x A

= (13+5) x (A+B)

= 18 x (A+B)= 18 x 12 = 216

b. 4 x A + 25 x B + 16 x A – 5 x B

= (4+16) x A + (25 - 5) x B

= 20 x A + 20 x B

= 20 x (A+B)= 20 x 12= 240

tính A à bạn

À CÒN BÀI

X^1+X^2+X^3+X^4+...+X^100 X THUỘC N KO CÓ 0

rút gọn biểu thức thôi

a) Ta có: \(\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}\)

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

b) Ta có: \(B=\dfrac{a-2\sqrt{a}-3}{a-9}\)

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

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

c) Ta có: \(C=\sqrt{x-1-2\sqrt{x-2}}\)

\(=\sqrt{x-2-2\cdot\sqrt{x-2}\cdot1+1}\)

\(=\sqrt{\left(\sqrt{x-2}-1\right)^2}\)

\(=\left|\sqrt{x-2}-1\right|\)

`a)A=(x+sqrt5)(x^2+2xsqrt5+5)`

`=(x+sqrt5)/(x+sqrt5)^2=1/(x+sqrt5)`

`b)B=(a-2sqrta-3)/(a-9)(a>=0,a ne 9)`

`=(a+sqrta-3sqrta-3)/(a-9)`

`=((sqrta+1)(sqrta-3))/((sqrta-3)(sqrta+3))`

`=(sqrta+1)/(sqrta+3)`

`c)C=sqrt{x-1-2sqrt{x-2}}(x>=2)`

`=sqrt{x-2-2sqrt{x-2}+1}`

`=sqrt{(sqrt{x-2}-1)^2}`

`=|sqrt(x-2)-1|`

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

2: \(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}=\dfrac{\sqrt{x}+5-10}{\sqrt{x}+5}\)

\(=1-\dfrac{10}{\sqrt{x}+5}\)

\(\sqrt{x}+5>=5\forall x\)

=>\(\dfrac{10}{\sqrt{x}+5}< =\dfrac{10}{5}=2\forall x\)

=>\(-\dfrac{10}{\sqrt{x}+5}>=-2\forall x\)

=>\(-\dfrac{10}{\sqrt{x}+5}+1>=-2+1=-1\forall x\)

Dấu '=' xảy ra khi x=0

Vậy: \(A_{min}=-1\) khi x=0

\(A=\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\left(x\ge0;x\ne25\right)\)

Để \(A=\dfrac{2\sqrt{x}}{3}\) thì:

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

\(\Leftrightarrow3\sqrt{x}-15=2x+10\sqrt{x}\)

\(\Leftrightarrow2x+10\sqrt{x}-3\sqrt{x}+15=0\)

\(\Leftrightarrow2x+7\sqrt{x}+15=0\) 

Mà \(2x+7\sqrt{x}+15>0\) (vì \(x\ge0\))

nên không tìm được giá trị nào của \(x\) thoả mãn \(A=\dfrac{2\sqrt{x}}{3}\)

#\(Toru\)

câu đầu tiên : a = 3,5

câu thứ hai : a = 3,4

ai thích k cho mk thì cứ k thoải mái đi nha, mk muốn có nhiều điểm 

a x 2 + a x 5 + a x 7 = 49

a x ( 2 + 5 + 7 ) = 49

a x 14 = 49

==> a = 3,5

a x 5 + a x 8 + a x 2 = 51

a x (  5 + 8 + 2 ) = 51

a x 15 = 51

==> a  = 3,4                         Tk mk,mk k lại!!!!!!!!!!

ax2+ax5+ax7+=49

ax(2+5+7)=49

ax14=49

a=49:14=3,5

ax5+ax8+ax2=51

ax(5+8+2)=51

ax15=51

a=51:15=3,4

Chúc bạn may mắn!

Sửa đề: căn x-5/căn x-3

a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

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

b: x-5căn x+6=0

=>căn x=2 hoặc căn x=3

=>x=9(loại) hoặc x=4(nhận)

Khi x=4 thì A=5/(2+3)=5/5=1