Điều kiện : \(x\le3\)

Đặt \(3-x=t\left(t\ge0\right)\Rightarrow5-x=2+t\)

Khi này ta có phương trình :

\(\sqrt{t}+\sqrt{t+2}< 2\)

\(\Leftrightarrow t+2\sqrt{t\left(t+2\right)}+t+2< 2\)

\(\Leftrightarrow2t+2\sqrt{t\left(t+2\right)}< 0\)

\(\Leftrightarrow t+\sqrt{t\left(t+2\right)}< 0\) ( vô lí do \(t\ge0\) )

Vậy bất phương trình đã cho vô nghiệm

Lời giải:

Thay vì dấu < thì dấu $\leq$ đúng hơn 

CMR: $(ax+by)^2\leq (a^2+b^2)(x^2+y^2)$

$\Leftrightarrow (a^2+b^2)(x^2+y^2)-(ax+by)^2\geq 0$

$\Leftrightarrow (a^2x^2+a^2y^2+b^2x^2+b^2y^2)-(a^2x^2+b^2y^2+2axby)\geq 0$

$\Leftrightarrow a^2y^2+b^2x^2-2axby\geq 0$

$\Leftrightarrow (ay-bx)^2\geq 0$ (luôn đúng) 

Vậy ta có đpcm.

Dấu "=" xảy ra khi $ay=bx$

bạn thấy đó:\(A^2-B=0\)

nên : Nên theo cái dấu mà bạn viết thì VP=0

bạn thấy vế trái có thể =0 ko?

\(A=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(=\sqrt{2}\left(\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\right)\)

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

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

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

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

\(=\sqrt{2}\left(2-\dfrac{6}{9-3}\right)\)

\(=\sqrt{2}\)

\(\dfrac{1}{\sqrt{2}}A=\dfrac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2-\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{2}}{3-\sqrt{3}}=\dfrac{6-2\sqrt{3}+3\sqrt{3}-3+6-3\sqrt{2}+2\sqrt{3}-\sqrt{6}}{6}\)

\(=\dfrac{9-3\sqrt{2}-\sqrt{6}+3\sqrt{3}}{6}.\sqrt{2}=\dfrac{9\sqrt{2}-6-2\sqrt{3}+3\sqrt{6}}{6}=\dfrac{3}{2}\sqrt{2}-1-\dfrac{1}{3}\sqrt{3}+\dfrac{1}{2}\sqrt{6}\)

Lời giải:

$4x^2-y^2+4x+1=(4x^2+4x+1)-y^2=(2x+1)^2-y^2$

$=(2x+1-y)(2x+1+y)=(2.10+1-5)(2.10+1+5)$

$=16.26=416$

a, \(=\left(2x+1\right)^2-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)

Thay x = 10 ; y = 5 

\(\left(20+1-5\right)\left(20+1+5\right)=16.26=416\)

b, \(=x^2-\left(y^2+2y+1\right)=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)

Thay x = 93 ; y = 6 

\(\left(93-6-1\right)\left(93+6+1\right)=8600\)

1, \(\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)

2, \(\left(x-y\right)^2-4z^2=\left(x-y-2z\right)\left(x-y+2z\right)\)

3, \(x\left(x-1\right)+y\left(x-1\right)=\left(x+y\right)\left(x-1\right)\)

4, \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)

a, \(x\left(x-y\right)+x-y=\left(x+1\right)\left(x-y\right)\)

b, \(z\left(x+y\right)-5\left(x+y\right)=\left(z-5\right)\left(x+y\right)\)

c, \(3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)

d, \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)

e, \(9\left(5-x\right)+x^2\left(x-5\right)=\left(3-x\right)\left(x+3\right)\left(x-5\right)\)

f, \(x^3\left(x+1\right)+x+1=\left(x+1\right)^2\left(x^2-x+1\right)\)

đk x >= 0 

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