a) \(6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9x-9}+\dfrac{7}{2}\sqrt{4x-4}=24\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9\left(x-1\right)}+\dfrac{7}{2}\sqrt{4\left(x-1\right)}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot3\sqrt{x-1}+\dfrac{7}{2}\cdot2\sqrt{x-1}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\sqrt{x-1}+7\sqrt{x-1}=24\)

\(\Leftrightarrow12\sqrt{x-1}=24\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{24}{12}\)

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

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=4+1\)

\(\Leftrightarrow x=5\left(tm\right)\)

b) \(\dfrac{1}{2}\sqrt{4x+8}-2\sqrt{x+2}-\dfrac{3}{7}\sqrt{49x+98}=-8\) (ĐK: \(x\ge-2\))

\(\Leftrightarrow\dfrac{1}{2}\cdot2\sqrt{x+2}-2\sqrt{x+2}-\dfrac{3}{7}\cdot7\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}-3\sqrt{x+2}=-8\)

\(\Leftrightarrow-4\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{-8}{-4}\)

\(\Leftrightarrow\sqrt{x+2}=2\)

\(\Leftrightarrow x+2=4\)

\(\Leftrightarrow x=4-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

\(x^2-x+\sqrt{x+1}-8=0\)

như ngày ak

như này ak nhầm

pt <=> x^2+2x+1+căn(x+1)-3(x+1)-6=0

(x+1)^2+căn(x+1)-3(x+1)-6=0

đặt x+1=t

<=>t^2+cănt-3t-6=0

giải tiếp đi

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http://olm.vn/hoi-dap/question/604325.html

x = 3

\(x^2-x+\sqrt{x+1}-8=0.\)(1) ĐK: x >= -1

Đặt: \(t=\sqrt{x+1}\mid t\ge0\)

Thay vào (1):

(1) \(\Leftrightarrow t^4-3t^2+t-6=0\)

\(\Leftrightarrow t^4-4t^2+t^2-2t+3t-6=0\)

\(\Leftrightarrow\left(t-2\right)\left(t^3+2t^2+t+3\right)=0\)(*)

Vì t>=0 nên t³ + 2t² + t + 3 >0 với mọi t

(*) \(\Leftrightarrow t-2=0\Rightarrow t=2\)

\(\sqrt{x+1}=2\Rightarrow x=3\)(TMĐK >= -1)

Vậy, PT có nghiệm duy nhất x = 3.

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

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

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)