3:

b: x1^2+x2^2=12

=>(x1+x2)^2-2x1x2=12

=>(2m+2)^2-4m=12

=>4m^2+4m+4=12

=>m^2+m+1=3

=>(m+2)(m-1)=0

=>m=1;m=-2

2:

b: =>|x1|-|x2|=m+3-|-1|=m+2

=>x1^2+x2^2-2|x1x2|=m+2

=>(x1+x2)^2-2x1x2-2|x1x2|=m+2

=>(2m)^2-2(-1)-2|-1|=m+2

=>4m^2-m-2=0

=>m=(1+căn 33)/8; m=(1-căn 33)/8

a) ĐKXĐ: x >= 1/2

Pt <=> 2x - 1 = 9

<=> x = 5 (thỏa ĐKXĐ)

b) ĐKXĐ: x>=4/3

Pt <=> 6x - 8 = 4

<=> 6x = 12 <=> x = 2 (thỏa ĐKXĐ)

c) ĐKXĐ: x >= 1

Pt <=> sqrt(x-1)=4

<=> x - 1 = 16 <=> x = 17 (thỏa ĐKXĐ)

d) ĐKXĐ: x >= -5

Pt <=> sqrt(x+5)=9

<=> x + 5 = 81

<=> x = 76 (thỏa ĐKXĐ)

e) ĐKXĐ: x >= 0

Pt <=> -2sqrt(x) = -6

<=> sqrt(x)=3

<=> x = 9 (thỏa ĐKXĐ)

f) ĐKXĐ: x >= 1

Pt <=> -sqrt(x-1)=-1

<=> sqrt(x-1)=1

<=> x - 1 = 1 <=> x = 2(thỏa ĐKXĐ)

a) \(đk:x\ge\dfrac{1}{2}\)

\(\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\left(tm\right)\)

b) \(đk:x\ge\dfrac{4}{3}\)

\(pt\Leftrightarrow6x-8=4\Leftrightarrow6x=12\Leftrightarrow x=2\left(tm\right)\)

c) \(đk:x\ge1\)

\(pt\Leftrightarrow\sqrt{x-1}=4\Leftrightarrow x-1=16\Leftrightarrow x=17\left(tm\right)\)

d) \(đk:x\ge-5\)

\(pt\Leftrightarrow\sqrt{x+5}=9\Leftrightarrow x+5=81\Leftrightarrow x=76\left(tm\right)\)

e) \(đk:x\ge0\)

\(pt\Leftrightarrow2\sqrt{x}=6\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\left(tm\right)\)

f) \(đk:x\ge1\)

\(pt\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=0\)

\(\Leftrightarrow\sqrt{x-1}-1=0\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

lỗi ảnh rùi e

lx

a. \(\widehat{DAB}=\widehat{ABC}=\widehat{BCE}=90^0\)

\(\widehat{ABD}=180^0-\widehat{ABC}-\widehat{EBC}=180^0-60^0-\left(180^0-\widehat{BCE}-\widehat{CEB}\right)=180^0-60^0-\left(180^0-60-\widehat{CEB}\right)=\widehat{CEB}\)\(\Rightarrow\)△ABD∼△CEB (g-g).

\(\Rightarrow\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow AD.CE=CB.AB\Rightarrow AD.CE=a^2\) không đổi

b. \(\widehat{CAD}=\widehat{BAD}+\widehat{BAC}=60^0+60^0=\widehat{BCE}+\widehat{ACB}=\widehat{ACE}\)

 \(\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow\dfrac{AD}{AC}=\dfrac{AC}{CE}\)

\(\Rightarrow\)△ACD∼△CEA (c-g-c) 

\(\Rightarrow\left\{{}\begin{matrix}\widehat{ACD}=\widehat{CEA}\\\dfrac{CE}{AC}=\dfrac{EA}{CD}\end{matrix}\right.\)

\(\Rightarrow\)△ACK∼△AEC (g-g).

\(\Rightarrow\dfrac{CK}{EC}=\dfrac{AK}{AC}\Rightarrow\dfrac{CE}{AC}=\dfrac{CK}{AK}\)

\(\Rightarrow\dfrac{AE}{CD}=\dfrac{CK}{AK}\Rightarrow AE.AK=CD.CK\)

TH1 x>0

<=>3x-9=5x-17

    -2x=-6

x=3

\(TH2 x<0 => -3x+9=5x-17 =>-8x=-16 =>x=2\)

a: Xét ΔSBM và ΔSNB có 

\(\widehat{SBM}=\widehat{SNB}\)

\(\widehat{BSM}\) chung

Do đó: ΔSBM\(\sim\)ΔSNB

Suy ra: SB/SN=SM/SB

hay \(SB^2=SM\cdot SN\)

b: Xét (O) có

SA là tiếp tuyến

SB là tiếp tuyến

Do đó: SA=SB

mà OA=OB

nên SO là đường trung trực của AB

=>SO⊥AB

Xét ΔOBS vuông tại B có BH là đường cao

nên \(SH\cdot SO=SB^2=SM\cdot SN\)

\(a,\left(-8,5\right)+16,35+\left(-4,5\right)-\left(-2,25\right)\\ =\left[\left(-8,5\right)+\left(-4,5\right)\right]+\left[16,35-\left(-2,25\right)\right]\\ =-13+18,6=5,6\\ b,5,63+\left(-2,75\right)-\left(-8,94\right)+9,06-15,25\\ =5,63-2,75+8,94+9,06-15,25\)

\(=5,63-\left(2,75+15,25\right)+\left(8,94+9,06\right)\\ =5,63-18+18\\ =5,63\)

a) (-8,5) + 16,35 + (-4,5) - (-2,25) = 5,6       

b)  5,63 + (-2,75) - (-8,94) + 9,06 - 15,25 = 5,63