a: =>(x+10)(x-1)=0

=>x=-10 hoặc x=1

b: \(A=x^3-1-\left(x+5\right)\left(x^2-3\right)-5x^2-10x-5\)

\(=x^3-5x^2-10x-6-x^3+3x-5x^2+15\)

=-7x+9

=110/13

Chọn A

A

chon A bạn nhé

Câu 1:\(\frac{6}{5}\)
Câu 2:  X =  \(\frac{13}{16}\)

\(\frac{3}{2}:\frac{5}{4}=\frac{3}{2}\times\frac{4}{5}=\frac{12}{20}=\frac{3}{5}\)

X x 18/5 = 7/10 + 3/5 

X x 18/5 =  13/10

X            = 13/10 : 18/5

 X            = 13/36 

k cho mik với 

NGU DỐT

Với `x \ne -5,x \ne -1` có:

`A=[x+2]/[x+5]+[-5x-1]/[x^2+6x+5]-1/[1+x]`

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+5)(x+1)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+5)(x+1)]`

`A=[x^2-3x-4]/[(x+5)(x+1)]`

`A=[(x-4)(x+1)]/[(x+5)(x+1)]`

`A=[x-4]/[x+5]`

\(=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{x^2+x+5x+5}-\dfrac{1}{x+1}\\ =\dfrac{x+2}{x+5}+\dfrac{-5x-1}{\left(x^2+x\right)+\left(5x+5\right)}-\dfrac{1}{x+1}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{x\left(x+1\right)+5\left(x+1\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{\left(x+1\right)\left(x+5\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+2x+x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+x-4x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x^2+x\right)-\left(4x+4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x\left(x+1\right)-4\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x-4}{x+5}\)

a: ĐKXĐ: \(x\notin\left\{5;-5\right\}\)

b: \(P=\dfrac{x-5+2x+10-2x-10}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)​

\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{x\sqrt{x}+2x+2x-50+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{x\sqrt{x}-5\sqrt{x}+4x}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}\left(x+4\sqrt{x}-5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-1}{2}\)

\(\dfrac{x+2\sqrt{x}}{2\sqrt{x}+10}+\dfrac{\sqrt{x}-5}{\sqrt{x}}+\dfrac{50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\left(đk:x>0\right)\)

\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)+2\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}\)

\(=\dfrac{x\sqrt{x}+2x+2x-50+50-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{x\sqrt{x}+4x-5\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+5\right)}{2\sqrt{x}\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-1}{2}\)

1:

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

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)

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

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

\(=\dfrac{x-4\sqrt{x}+3-x+4-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

ĐKXĐ: \(x\ne4;x\ne9;x\ge0\)

A \(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}-\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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