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b: B=căn 49a^2+3a

=|7a|+3a

=7a+3a(a>=0)

=10a

c: C=căn16a^4+6a^2

=4a^2+6a^2

=10a^2

d: \(D=3\cdot3\cdot\sqrt{a^6}-6a^3=6\cdot\left|a^3\right|-6a^3\)

TH1: a>=0

D=6a^3-6a^3=0

TH2: a<0

D=-6a^3-6a^3=-12a^3

e: \(E=3\sqrt{9a^6}-6a^3\)

\(=3\cdot\sqrt{\left(3a^3\right)^2}-6a^3\)

=3*3a^3-6a^3(a>=0)

=3a^3

f: \(F=\sqrt{16a^{10}}+6a^5\)

\(=\sqrt{\left(4a^5\right)^2}+6a^5\)

=-4a^5+6a^5(a<=0)

=2a^5

28 tháng 7 2023

A) \(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\dfrac{\sqrt{x-1}}{3}-\sqrt{x-1}=6\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\)

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

\(\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\)

\(\Leftrightarrow x=17\)

Vậy, x=17

 

A: \(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

=>5/2*căn x-1-căn x-1=6

=>3/2*căn x-1=6

=>căn x-1=4

=>x-1=16

=>x=17

B:

a: ĐKXĐ: x>=0; x<>1

b: Sửa đề: \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}\)

=căn x-1+x-căn x+1

=x

27 tháng 9 2023

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

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 \(A=\dfrac{2\sqrt{x}}{3}\)

#\(Toru\)

27 tháng 9 2023

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

18 tháng 8 2020

/x-25 và /x-2 đấy ạ,máy em bị đánh lỗi. :((

18 tháng 8 2020

\(5\sqrt{x}-\frac{\left(x+10\sqrt{x}+25\right)\left(\sqrt{x}-5\right)}{x-25}=5\sqrt{x}-\frac{\left(\sqrt{x}+5\right)^2\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

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

\(\frac{\sqrt{x^2-4x+4}}{x-2}=\frac{\sqrt{\left(x-2\right)^2}}{x-2}=\frac{\left|x-2\right|}{x-2}=\orbr{\begin{cases}\frac{x-2}{x-2}\left(x>2\right)\\\frac{2-x}{x-2}\left(x< 2\right)\end{cases}=\orbr{\begin{cases}1\left(x>2\right)\\-1\left(x< 2\right)\end{cases}}}\)

19 tháng 11 2023

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

11 tháng 11 2021

Bài 1:

\(a,ĐK:2+8x\ge0\Leftrightarrow x\ge-\dfrac{1}{4}\\ b,ĐK:-\dfrac{1}{5}x+9\ge0\Leftrightarrow-\dfrac{1}{5}x\ge-9\Leftrightarrow x\le45\\ c,ĐK:11-7x\ge0\Leftrightarrow x\le\dfrac{11}{7}\)

Bài 2:

\(a,=\sqrt{144a^2}-2a=12\left|a\right|-2a=12a-2a=10\\ b,=\sqrt{6}-6\sqrt{6}-\sqrt{6}=-6\sqrt{6}\)

Bài 3:

\(a,\Leftrightarrow\left|2x+3\right|=3\Leftrightarrow\left[{}\begin{matrix}2x+3=3\\2x+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ b,ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=4\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

28 tháng 8 2021

\(1,ĐKx\ge5\)

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

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

 

 

 

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

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

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

a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)

\(=14\sqrt{2x}+30\)

b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)

\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)

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

\(=\dfrac{11}{2}\sqrt{x}\)

c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)

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

\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)

\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)