Xét \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
Phương trình trên tương đương:
x³+4x+5=0
<=>x(x²-1)+5(x+1)=0
<=>x(x-1)(x+1)+5(x+1)=0
<=>(x+1)(x²-x+5)=0
<=>x+1=0 hoặc x²-x+5=0(vô nghiệm)
<=>x=-1
Vậy pt trên có nghiệm là x=-1

x=-1 nha 

Bài này đi thi vio mk cũng gặp ..

    bằng 1 ak
 

Câu 1: \(\sqrt{8}\) − \(\sqrt{18}\) + \(2\sqrt{32}\) = \(\sqrt{4\text{×}2}\) −  \(\sqrt{\text{9×2}}\) + 2\(\sqrt{\text{16×2}}\)

                                           =2\(\sqrt{2}\) − 3\(\sqrt{2}\) + 2×4\(\sqrt{2}\) 

                                           =(2− 3+ 8)\(\sqrt{2}\)

                                           =7\(\sqrt{2}\)

Câu 2: Mik ko chắc làm đúng hay ko  nên ko làm

b: \(=\dfrac{\sqrt{x}-4+\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}+4}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-4}\)

`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`

`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`

`=((x+3\sqrtx-4)/(x-4)).((x-4)/\sqrtx))`

`=(x+3\sqrtx)/\sqrtx`

`=(\sqrtx(\sqrtx+3))/\sqrtx`

`=\sqrtx+3(đpcm)`

`2)P=x+3

`<=>\sqrtx+3=x+3`

`<=>x-\sqrtx=0`

`<=>\sqrtx(\sqrtx-1)=0`

Vì `x>0=>\sqrtx>0`

`=>\sqrtx-1=0<=>x=1(tm)`

Vậy `x=1=>\sqrtx+3=x+3`

`1)P((\sqrtx+1)/(\sqrtx-2)-2/(x-4)).(\sqrtx-1+(\sqrtx-4)/\sqrtx)(x>0,x ne 4)`

`=((x+3\sqrtx+2-2)/(x-4)).((x-\sqrtx+\sqrtx-4)/\sqrtx)`

`=((x+3\sqrtx)/(x-4)).((x-4)/\sqrtx))`

`=(x+3\sqrtx)/\sqrtx`

`=(\sqrtx(\sqrtx+3))/\sqrtx`

`=\sqrtx+3(đpcm)`

`2)P=x+3

`<=>\sqrtx+3=x+3`

`<=>x-\sqrtx=0`

`<=>\sqrtx(\sqrtx-1)=0`

Vì `x>0=>\sqrtx>0`

`=>\sqrtx-1=0<=>x=1(tm)`

Vậy `x=1=>\sqrtx+3=x+3`

A = \(\frac{8\sqrt{41}}{2\sqrt{2^2+2.2.\sqrt{41}+\sqrt{41}^2}}\)

A = \(\frac{8\sqrt{41}}{2\sqrt{\left(2+\sqrt{41}\right)^2}}\)

A = \(\frac{8\sqrt{41}}{2\left|2+\sqrt{41}\right|}\)

A = \(\frac{8\sqrt{41}}{4+2\sqrt{41}}\)

B = \(\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1+x+4}{x+\sqrt{x}+1}\)

B = \(\left(\frac{2x+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right).\frac{x+\sqrt{x}+1}{2x+\sqrt{x}+5}\)

Bạn tự làm tiếp nhé, mỏi tay quá!!

\(A=\frac{8\sqrt{41}}{2\sqrt{45+4\sqrt{41}}}=\frac{8\sqrt{41}}{2\sqrt{41+4\sqrt{41}+4}}=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}\right)^2+2\cdot\sqrt{41}\cdot2+2^2}}\)

\(=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}+2\right)^2}}=\frac{8\sqrt{41}}{2\left(\sqrt{41}+2\right)}=\frac{8\sqrt{41}\left(\sqrt{41}-2\right)}{2\left(41-4\right)}=\frac{328-16\sqrt{41}}{74}=\frac{164-8\sqrt{41}}{37}\)

\(B=\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)

\(=\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\right)\)

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

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

\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\frac{x+3\sqrt{x}}{x-9}\)

ĐkXĐ: x \(\ne\)32

ta có:\(\frac{\sqrt{x}+2}{x+32}=\frac{1}{\frac{x-4+36}{\sqrt{x}+2}}=\frac{1}{\sqrt{x}-2+\frac{36}{\sqrt{x}+2}}=\frac{1}{\sqrt{x}+2+\frac{36}{\sqrt{x}+2}-4}\)

vì \(\sqrt{x}+2>0\)Áp dụng BĐT cauchy:

\(\sqrt{x}+2+\frac{36}{\sqrt{x}+2}\ge2\sqrt{\left(\sqrt{x}+2\right)\left(\frac{36}{\sqrt{x}+2}\right)}=12\)

\(\Rightarrow\frac{\sqrt{x}+2}{x+32}\le\frac{1}{12-4}=\frac{1}{8}\)

dấu = xảy ra khi \(\sqrt{x}+2=6\Leftrightarrow x=16\)(t/m đk)

đặt y=\(y=\sqrt{x}\Rightarrow y\ge0\)

\(A=\dfrac{y+2}{y^2+32}\)\(\Rightarrow A+\dfrac{1}{16}=\dfrac{1}{16}+\dfrac{y+2}{y^2+32}=\dfrac{y^2+16y+64}{16.\left(y^2+32\right)}\)

\(A+\dfrac{1}{16}=\dfrac{\left(y+4\right)^2}{16\left(y^2+32\right)}\ge0\) đẳng thức khi y=-4 nằm ngoài khoảng đang xét.

A(0)=1/16 xét y>0 chia cả tử mấu A cho y khác 0

\(A=\dfrac{1+\dfrac{2}{y}}{y+\dfrac{32}{y}}\) hiển nhiên khi y đạt giá trị nào đó thì y càng lớn --> A càng nhỏ:

kết luận không có giá trị nhỏ nhất

Bấm vào chữ "Đúng 0" là hiện ra câu trả lời của mik ák

a, x + 32...  .<......41+  x
b,56- y .. >  ..... 45-y
c, x-26 ..<.  ......x-18
d,42 + 21 + x...> .... 42+ 21
g, 52- 47..> ..... 52-47-x
h, 29+ 42- x..<. 42+29+ x

Học tốt!!!

#Bo

Giải

a, x + 32         <  41+  x
b,56- y            >  45-y
c, x-26            <  x-18
d,42 + 21 + x  >  42+ 21
g, 52- 47        >   52-47-x
h, 29+ 42- x   <  42+29+ x

-Chúc họk tốt-

Q