a, \(16x^2-5=0\)

\(\Rightarrow16x^2=5\)

\(\Rightarrow x^2=\frac{5}{16}\)

\(\Rightarrow x=\sqrt{\frac{5}{16}}\Rightarrow x=\frac{\sqrt{5}}{4}\)

b, \(2\sqrt{x-3}=4\)

\(\Rightarrow\sqrt{x-3}=4:2\)

\(\Rightarrow\sqrt{x-3}=2\)

\(\Rightarrow x-3=4\)

\(\Rightarrow x=4+3\)

\(\Rightarrow x=7\)

c, \(\sqrt{4x^2-4x+1}=3\)

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

\(\Rightarrow2x-1=3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

d, \(\sqrt{x+3}\ge5\)

\(\Rightarrow x+3\ge25\)

\(\Rightarrow x\ge22\)

e, \(\sqrt{3x-1}< 2\)

\(\Rightarrow3x-1< 4\)

\(\Rightarrow3x< 5\)

\(\Rightarrow x< \frac{5}{3}\)

g, \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

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

\(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Rightarrow\sqrt{x-3}=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

a) \(16x^2-5=0\)

\(\Leftrightarrow16x^2=5\)

\(\Leftrightarrow x^2=\frac{5}{16}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{5}{16}}\)

b) \(2\sqrt{x-3}=4\)

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

\(\Leftrightarrow x-3=4\)

\(\Leftrightarrow x=7\)

c) \(\sqrt{4x^2-4x+1}=3\)

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

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

d) \(\sqrt{x+3}\ge5\)

\(\Leftrightarrow x+3\ge25\)

\(\Leftrightarrow x\ge22\)

e) \(\sqrt{3x-1}< 2\)

\(\Leftrightarrow3x-1< 4\)

\(\Leftrightarrow3x< 5\)

\(\Leftrightarrow x< \frac{5}{3}\)

g) \(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

Vì \(\left(\sqrt{x+3}+\sqrt{x-3}\right)>0\)

\(\Leftrightarrow\sqrt{x-3}=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

x\(\varepsilon\)(-\(\frac{1}{2}\);\(\frac{1}{2}\))

a) b) c) bạn bình phương 2 vế

d) pt <=>3-x=x+3+2.căn(x+2)

<=> -2x=2.căn (x+2)

<=>-x=căn (x+2) (x<=0)

<=> x^2=x+2

<=>x=-1 hoặc x=2

Xong bạn xét ĐKXĐ

giải giúp tớ a , b,c luôn đi cậu :<

viết đề khó hiểu quá

a) căn(2x+5) - căn(3-x) = x² -5x + 8 
Điều kiện : \(-\frac{5}{2}\Leftarrow x\Leftarrow3\)
căn(2x+5) - căn(3-x) = x^2-5x+8 
\(\Leftrightarrow\)[căn(2x+5)-3]-[căn(3-x)-1]=x² -5x+6 
nhân liên hợp 
\(\Leftrightarrow\)(2x+5-9) / [căn(2x+5)+3] -(3-x-1) / [căn (3-x)+1]=(x-2)(x-3) 
\(\Leftrightarrow\)(2x-4) / [căn (2x+5)+3] -(2-x) /  [ căn (3-x)+1]-(x-2)(x-3)=0 
\(\Leftrightarrow\)(x-2).M=0 
\(\Leftrightarrow\)x=2 hoặc M=0 
M=2 / [căn(2x+5)+3]+1 / [căn(3-x)+1]-x+3 

2/[can(2x+5)+3]+1/[can(3-x)+1]>0 voi moi x 
voi -5/2<=x<=3 <->3-x thuoc[0;11/2] 
nen M>0 
vay x=2 
b/ 2+ căn(3-8x) = 6x + căn(4x-1) 
dk[1/4;8/3] 
6x-2+căn(4x-1)-căn(3-8x)=0 
<->2(3x-1)+(4x-1-3+8x)/[căn(4x-1)+căn(... 
<->2(3x-1)+(12x-4)/[căn(4x-1)+căn(3-8x... 
<->2(3x-1)+4(3x-1)/[căn(4x-1)+căn(3-8x... 
<->(3x-1){2+4/[căn(4x-1)+căn(3-8x)]}=0 
2+4/[căn(4x-1)+căn(3-8x)>0 
nen 3x-1=0 
x=1/3

 a)  căn(2x+5) - căn(3-x) = x^2-5x+8 
dkxd -5/2<=x<=3 
căn(2x+5) - căn(3-x) = x^2-5x+8 
<->[can(2x+5)-3]-[can(3-x)-1]=x^2-5x+6 
nhan lien hop 
<->(2x+5-9)/[can(2x+5)+3] -(3-x-1)/[can(3-x)+1]=(x-2)(x-3) 
<->(2x-4)/[can(2x+5)+3] -(2-x)/[can(3-x)+1]-(x-2)(x-3)=0 
<->(x-2).M=0 
<->x=2 hoac M=0 
M=2/[can(2x+5)+3]+1/[can(3-x)+1]-x+3 

2/[can(2x+5)+3]+1/[can(3-x)+1]>0 voi moi x 
voi -5/2<=x<=3 <->3-x thuoc[0;11/2] 
nen M>0 
vay x=2 
b/ 2+ căn(3-8x) = 6x + căn(4x-1) 
dk[1/4;8/3] 
6x-2+căn(4x-1)-căn(3-8x)=0 
<->2(3x-1)+(4x-1-3+8x)/[căn(4x-1)+căn(... 
<->2(3x-1)+(12x-4)/[căn(4x-1)+căn(3-8x... 
<->2(3x-1)+4(3x-1)/[căn(4x-1)+căn(3-8x... 
<->(3x-1){2+4/[căn(4x-1)+căn(3-8x)]}=0 
2+4/[căn(4x-1)+căn(3-8x)>0 
nen 3x-1=0 
x=1/3

Đk: `x >=-1`.

`5sqrt(x+1) + sqrt(4x+4) - sqrt(9x+9) = 2`.

`<=> 5sqrt(x+1) + 2 sqrt(x+1) - 3sqrt(x+1) = 2`.

`<=> 4 sqrt(x+1) =2.`

`<=> sqrt(x+1) = 1/2`

`<=> x + 1 = 1/4`

`<=> x = 3/4 (tm)`.

Vậy `x = 3/4`.