\(\Leftrightarrow2m.2^x+\left(2m+1\right)\left(3-\sqrt{5}\right)^x+\left(3+\sqrt{5}\right)^x=0\)

\(\Leftrightarrow\left(\frac{3+\sqrt{5}}{2}\right)^x+\left(2m+1\right)\left(\frac{3-\sqrt{5}}{2}\right)^x+2m< 0\)

Đặt \(t=\left(\frac{3+\sqrt{5}}{2}\right)^x,0< t\le1\Rightarrow\frac{1}{t}=\left(\frac{3-\sqrt{5}}{2}\right)^x\)

Phương trình trở thành :

\(t+\left(2m+1\right)\frac{1}{t}+2m=0\) (*)

a. Khi \(m=-\frac{1}{2}\) ta có \(t=1\) suy ra \(\left(\frac{3+\sqrt{5}}{2}\right)^x=1\Leftrightarrow x=0\)

Vậy phương trình có nghiệm là \(x=0\)

b. Phương trình (*) \(\Leftrightarrow t^2+1=-2m\left(t+1\right)\Leftrightarrow\frac{t^2+1}{t+1}=-2m\)

Xét hàm số \(f\left(t\right)=\frac{t^2+1}{t+1};t\in\)(0;1]

Ta có : \(f'\left(t\right)=\frac{t^2+2t+1}{\left(t+1\right)^2}\Rightarrow f'\left(t\right)=0\Leftrightarrow=-1+\sqrt{2}\)

Suy ra phương trình đã cho có nghiệm đúng

\(\Leftrightarrow2\sqrt{2}-2\le-2m\le1\Leftrightarrow\sqrt{2}-1\ge m\ge-\frac{1}{2}\)

Vậy \(m\in\left[-\frac{1}{2};\sqrt{2}-1\right]\) là giá trị cần tìm

Ta co:\(\Sigma\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}=\Sigma\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)Ta lai co:

\(\Sigma x+\Sigma\frac{1}{x}=\Sigma\left(x+\frac{1}{4x}\right)+\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+\frac{3}{4}.\frac{9}{x+y+z}\ge3+\frac{3}{4}.\frac{9}{\frac{3}{2}}=\frac{15}{2}\)

Dau '=' xay ra khi \(x=y=z=\frac{1}{2}\)

Vay \(P_{min}=\frac{15}{2}\)khi \(x=y=z=\frac{1}{2}\)

1.

a.

\(A=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{3x-9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(=\frac{3\sqrt{x}}{\sqrt{x}+3}\)

b.

Theo de bai ta co:

\(A\ge0\left(DK:\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\right)\)

\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+3}\ge0\)

\(\Leftrightarrow3\sqrt{x}\ge0\)

\(\Leftrightarrow x\ge0\)

Vay de \(A\ge0\)thi \(\left\{{}\begin{matrix}x\in\left[0;+\infty\right]\\x\ne9\end{matrix}\right.\)

2.

a.

De \(\left(d_1\right)//\left(d_2\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m^2-1=3\\2m\ne4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m=\pm2\\m\ne2\end{matrix}\right.\)

\(\Leftrightarrow m=-2\)

1:

\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)

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

a: \(\text{Δ}=\left(-5\right)^2-4\left(-2m+5\right)\)

=25+8m-20=8m+5

Để phương trình có nghiệm kép thì 8m+5=0

=>m=-5/8

=>x^2-5x+25/4=0

=>x=5/2

b: \(\text{Δ}=\left(2m-1\right)^2-4\left(m^2-2m+3\right)\)

\(=4m^2-4m+1-4m^2+8m-12=4m-11\)

Để phương trình có nghiệm kép thì 4m-11=0

=>m=11/4

=>x^2-9/2x+81/16=0

=>x=9/4

c: TH1: m=-3

=>-(2*(-3)+1)x+(-3-1)=0

=>-(-5x)-4=0

=>5x-4=0

=>x=4/5(nhận)

TH2: m<>-3

\(\text{Δ}=\left(2m+1\right)^2-4\left(m+3\right)\left(m-1\right)\)

\(=4m^2+4m+1-4\left(m^2+2m-3\right)\)

\(=4m^2+4m+1-4m^2-8m+12=-4m+13\)

Để phương trình có nghiệm kép thì -4m+13=0

=>m=13/4

=>25/4x^2-15/2x+9/4=0

=>(5/2x-3/2)^2=0

=>x=3/2:5/2=3/2*2/5=3/5

a: \(\Leftrightarrow\left(2m-4\right)^2-4\left(m^2-3\right)>=0\)

\(\Leftrightarrow4m^2-16m+16-4m^2+12>=0\)

=>-16m>=-28

hay m<=7/4

b: \(\Leftrightarrow16m^2-4\left(2m-1\right)\left(2m+3\right)=0\)

\(\Leftrightarrow16m^2-4\left(4m^2+4m-3\right)=0\)

=>4m-3=0

hay m=3/4

c: \(\Leftrightarrow\left(4m-2\right)^2-4\cdot4\cdot m^2< 0\)

=>-16m+4<0

hay m>1/4

@Akai Haruma @Nguyễn Việt Lâm
cíu giúp em với ạaaa

1.

Đặt \(x+\frac{1}{x}=t\Rightarrow\left|t\right|\ge2\)

Pt trở thành: \(t^2-2+4t-3-2m=0\)

\(\Leftrightarrow t^2+4t-5=2m\)

Xét \(f\left(t\right)=t^2+4t-5\) trên \((-\infty;-2]\cup[2;+\infty)\)

\(-\frac{b}{2a}=-2\) ; \(f\left(-2\right)=-9\) ; \(f\left(2\right)=7\)

\(\Rightarrow f\left(x\right)\ge-9\Rightarrow\) pt có nghiệm khi và chỉ khi \(2m\ge-9\Leftrightarrow m\ge-\frac{9}{2}\)

2.

Đặt \(\sqrt{x^2-2x+5}=\sqrt{\left(x-1\right)^2+4}=t\Rightarrow t\ge2\)

\(t^2-2-\left(m+1\right)t-m=0\)

\(\Leftrightarrow t^2-t-2-m\left(t+1\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(t-2\right)-m\left(t+1\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(t-m-2\right)=0\)

\(\Leftrightarrow m=t+2\ge4\)

Vậy \(m\ge4\) thì pt có nghiệm

Bài làm

a) 3x - 1 = 2x + 4

<=> x = 5

Vậy x = 5 là nghiệm phương trình.

b) x( x + 3 ) = ( 2x + 1 )( x + 3 )

<=> x( x + 3 ) - ( 2x + 1 )( x + 3 ) = 0

<=> ( x + 3 )( x - 2x - 1 ) = 0

<=> ( x + 3 )( -x - 1 ) = 0

<=> x + 3 = 0 hoặc -x - 1 = 0

<=> x = -3 hoặc x = -1

Vậy x = -3 hoặc x = -1 là tập nghiệm phương trình

c) quy đồng mẫu ra r lm, bh ngủ.

Bài 1

a)\(3x-1=2x+4\)

\(\Leftrightarrow x=5\)

b) \(x\left(x+3\right)=\left(2x+1\right)\left(x+3\right)\)

\(x\left(x+3\right)-\left(2x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\\left(x-1\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-3\\x=1\end{matrix}\right.\)

c) \(\frac{1}{x+1}+\frac{5}{x-2}=\frac{3x}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\frac{1\left(x-2\right)+5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3x}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow x-2+5x+5=3x\)

\(\Leftrightarrow3x=3\)

\(\Rightarrow x=1\)

a) Phương trình 1,5x² – 1,6x + 0,1 = 0

Có a + b + c = 1,5 – 1,6 + 0,1 = 0 nên x₁ = 1; x₂ = \(\dfrac{0,1}{15}\)

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

Có \(a+b+c=2-\sqrt{3}+2\sqrt{3}-\left(2+\sqrt{3}\right)=0\)

Nên x₁ = 1, x₂ = \(\dfrac{-\left(2+\sqrt{3}\right)}{2-\sqrt{3}}\) = -(2 + \(\sqrt{3}\))² = -7 - 4\(\sqrt{3}\)

d) (m – 1)x² – (2m + 3)x + m + 4 = 0

Có a + b + c = m – 1 – (2m + 3) + m + 4 = 0

Nên x₁ = 1, x₂ = \(\dfrac{m+4}{m-1}\)

a) Phương trình 1,5x2 – 1,6x + 0,1 = 0

Có a + b + c = 1,5 – 1,6 + 0,1 = 0 nên x1 = 1; x2 =

b) Phương trình √3x2 – (1 - √3)x – 1 = 0

Có a – b + c = √3 + (1 - √3) + (-1) = 0 nên x1 = -1, x2 = =

c) (2 - √3)x2 + 2√3x – (2 + √3) = 0

Có a + b + c = 2 - √3 + 2√3 – (2 + √3) = 0

Nên x1 = 1, x2 = = -(2 + √3)2 = -7 - 4√3

d) (m – 1)x2 – (2m + 3)x + m + 4 = 0

Có a + b + c = m – 1 – (2m + 3) + m + 4 = 0

Nên x1 = 1, x2 =