3)
√(15-x)+√(x+3)=6
x=[-3;15]
√(15-x)=a
√(x+3)=b=> a^2 +b^2 =18 (2)
a+b=6=> a^2 +b^2 +2ab=36 (3)
(3)-(2)=> 2ab=18
ab=9
a+b=6 => a=3 =>√(15-x)=3
15-x=9 => x=6
4)
√(4x+1)-√(3x+4)=1
x>=-1/4
<=>√(4x+1)=1+√(3x+4)
<=> (4x+1)=1+3x+4+2√(3x+4)
<=> x-4-2√(3x+4)=0
√(3x+4)=t => t>=√13
3x+4 =t^2
x=(t^2-4) /3
<=>(t^2-4) /3 -2t -4 = 0
<=> t^2 -6t -16=0
▲t=9+16 = 25
t=(3-5 ) =-2 (loai)
t=(3+5 )=8 (nhan)
x=(64-4) /3
x=20

5/
√[x-2√(x-1)]-√(x-1) =1
dk x>=1
dat
√(x-1) =t; t>=0 => x=t^2 +1
<=>√(t^2-2t+1)-t=1
<=>√(t-1)^2 = t+1
<=>|t-1| = t+1
0<=t < 1 <=> -t+1 = t+1 =>t= 0 =>x= 1
t>=1 <=> t+1 = t+1 =>(vn)
x={1}
6)
√[x+ √(2x-1)]+√[ x-√(2x-1) ]=√2
x>=1/2
dat
√(2x-1) =t; t>=0 => x=(t^2 +1)/2
<=>√[(t^2 +1)/2 +t]+√[ (t^2 +1)/2-t ]=√2
<=>√[(t^2 +1+2t)/2]+√[ (t^2 +1-2t)/2 ]=√2
<=>√[(t^2 +1+2t) ]+√[ (t^2 +1-2t) ]= 2

<=>|t+1|+|t-1|= 2
0<=t < 1 <=> t+1 -t+1 =2 <=> 2=2
=> 0<=t < 1 => 1/2<=x=1
t>=1 <=> t+1 + t-1 =2 =>t =1 =>x=1
nghiem
1/2≤x ≤1

7/
dk x>3/2
√(6x-9)=t; t>=0
x=(t^2+9)/6
<=>√[(t^2+9)/6+t]+√[(t^2+9)/6-t]=√6
<=>√[(t^2+9+6t]+√[(t^2+9 -6t]= 6
<=>√[(t+3)^2 √[(t-3)^2 = 6
|t+3|+|t-3| =6
BDT trị tuyrtj đôied
|t+3|+|t-3|≥|t+3-t+3=|6| =6
=> nghiẹm
0≤t≤3
<=> 3/2 ≤x≤3

8/
√(x^2-4x+4)+√(x^2-6x+9) =1
√(x-2)^2 +√(x-3)^2 =1
|x-2|+|x-3| =1
|x-2|+|x-3|≥|x -2-x+3|=|1| =1
=> nghiệm
2≤x≤3

9/
√x +√[x+√(1-x)] =1 (1)
0≤x≤1
lhop
<=> x+√(1-x)-x =√[x+√(1-x)]-√x
<=> √(1-x) =√[x+√(1-x)]-√x
<=> √(1-x) -√[x+√(1-x)] =-√x (2)
(1)+(2)
√x+√(1-x)=1-√x
√(1-x)=1-2√x
x<√2/2
(1-x)=1+4x -4√x
5x -4√x = 0
√x(5√x -4) = 0
x=0 (n)
x=16/25(l)
x={0}

10/
√(2x-1) +√(x-2)=√(x+1)
dk x>=2
bp
(2x-1) +x-2+2√[(2x-1)(x-2)]= (x+1)
<=> 2√[(2x-1)(x-2)]=-2x+4
<=> √[(2x-1)(x-2)]= - x+2
-x+2 >=0 <=> x<=2
=>x= 2 là duy nhất
x={2}

11./
√( x+1) +√(x+10)=√(x+2)+√(x+5)
đk x>= -1
BP
( x+1) +x+10 +2√[(x+1)(x+10)]=x+2 +x+5 +2√[(x+2)(x+5)]
<=>4 +2√[(x+1)(x+10)]= 2√[(x+2)(x+5)]
<=> √[(x+2)(x+5)]-√[(x+1)(x+10)] = 2 (a)
LH
<=>(√[(x+2)(x+5)]-√[((x+1)(x+10)])(√[(x+2)(x+5)]+√[(x+1)(x+10)])=2(√[(x+2)(x+5)]+√[(x^2 +1 1x+10)])
<=> x^2 +7x+10-(x^2 +1 1x+10) =2(√[(x+2)(x+5)]+√[(x^2 +1 1x+10)])
<=> 2(√[(x+2)(x+5)]+√[(x^2 +1 1x+10)]) =-4x
<=> √[(x^2 +7x+10]+√[(x^2 +1 1x+10)] =- 2x(b)
x<=0 (a)+(b) <=>2√[(x^2 +7x+10] =2 -2x=-2(x-1)
<=> (x^2 +7x+10) = x^2 -2x +1
9x=-9
x=-1 thỏa mãn dk
x={-1}

√x+3 =5-√x+2

Giúp mk với ak mk cần gấp;((
help mee