nhathongthai123

Ta có : $\frac{b}{2}\geq \sqrt{b-1}$
CM:$\Leftrightarrow b-2\sqrt{b-1}\geq0
\Leftrightarrow b-1-2\sqrt{b-1} +1 \geq 0
\Leftrightarrow (\sqrt{b-1}-1)^{2}\geq 0$(luôn đúng)
CM tương tự
$\frac{a}{2}\geq \sqrt{a-1}
\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\leq a.\frac{b}{2}+b\frac{a}{2}=ab$

$a+b=a.b=\frac{a}{b}
\Rightarrow b(a+b)=ab^2=a(b\neq 0)
\Rightarrow a(b^2-1)=0
a=0\Rightarrow a+b=b=a.b=0\Rightarrow b=0 (loại)
b=1\Rightarrow a+1=a (loại)
\ b=-1\Rightarrow a+b=a-1=a.b=-a\Rightarrow a-1+a=1\Rightarrow 2a=1\Rightarrow a=\frac{1}{2}$
(chọn)

1)
$\frac{4x-11}{3-x}=\frac{-4(3-x)+1}{3-x}=-4+\frac{1}{3-x}$
Để phân số nhận giá trị nguyên thì $\left\{\begin{matrix}
3-x=1 & \\
3-x=-1 &
\end{matrix}\right.$
Vậy x=2;4
2)
$B=\frac{2x+3}{5x+2}$
$\Leftrightarrow 5B=\frac{10x+15}{5x+2}=2+\frac{11}{5x+2}$
Để B nguyên thì
$\left\{\begin{matrix}
5x+2=11 & \\
5x+2=-11 &
\end{matrix}\right.$
Vậy $x=\frac{9}{5};x=\frac{-13}{5}$

Bạn đã thử thay x vào chưa (phần 2)
Nếu 5B là số nguyên chưa chắc B là số nguyên đâu

Bài 7. Cho $\frac{2a + 13b}{3a - 7b}= \frac{2c + 13d}{3c - 7d}$. CMR: $\frac
{a}{b}=\frac{c}{d}$.

7
$\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}
\Rightarrow (2a+13b)(3c-7d)=(2c+13d)(3a-7b)
\Rightarrow 6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd
\Rightarrow 6ac-14ad+39bc-91bd-6ac+14bc-39ad+91bd=0
\Rightarrow 39bc+14bc-14ad-39ad=0
\Rightarrow 53(bc-ad)=0 \Rightarrow bc-ad=0
\Rightarrow bc=ad\Rightarrow \frac{a}{b}=\frac{c}{d}$

Bài 2. Tính:
a) A = $2008^{(1.9.4.6).(1.9.4.7)...(1.9.9.9)}$
b) B = $(1000 - 1^3) (1000 - 2^3) (1000 - 3^3) ...(1000 - 50^3)$

Bài 2.
a) A = $2008^{(1.9.4.6).(1.9.4.7)...(1.9.5.0)....(1.9.9.9)}=2008^0 =1$
b) B = $(1000 - 1^3) (1000 - 2^3) (1000 - 3^3) ...(1000 - 10^3).....(1000 - 50^3)=0$

$f(x)=\frac{3}{x^4-x^3+x-1}-\frac{1}{x^4+x^3-x-1}-\frac{4}{x^5-x^4+x^3-x^2+x-1}
=\frac{3}{(x-1)(x+1)(x^2-x+1)}-\frac{1}{(x+1)(x-1)(x^2+x+1)}-\frac{4}{(x-1)(x^4+x^2+1)}
=\frac{(3x^2+3x+3)-(x^2-x+1)}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}-\frac{4}{(x-1)(x^4+x^2+1)}
=\frac{2(x+1)^2}{(x-1)(x+1)(x^2-x+1)(x^2+x+1)}$-$\frac{4}{(x-1)(x^4+x^2+1)}$
$=\frac{2(x+1)}{(x-1)(x^2-x+1)(x^2+x+1)}-\frac{4}{(x-1)(x^4+x^2+1)}=
\frac{2(x+1)}{(x-1)(x^4+x^2+1)}-\frac{4}{(x-1) (x^4+x^2+1)}
=\frac{2x-2}{(x^4+x^2+1)(x-1)}
=\frac{2}{x^4+x^2+1}$

( chắc sai)

b) (nếu a đúng)
$\frac{2}{x^4+x^2+1}> 0 ()$
$\frac{32}{9}-\frac{2}{x^4+x^2+1}=\frac{32x^4+32x^2+32-18}{9(x^4+x^2+1)}
=\frac{32x^4+32x^2+14}{9(x^4+x^2+1)}=\frac{32(x+\frac{1}{2})^2+6}{9(x^4+x^2+1)}> 0$
=>dpcm

$x^4+9x^3+20x+9x+1=0
\Rightarrow x^2(x^2+9x+20+\frac{9}{x}+\frac{1}{x^2})=0
\Rightarrow x^2+9x+20+\frac{9}{x}+\frac{1}{x^2} =0$
Đặt $x+\frac{1}{x}=y$
$x^2 +9x+20+\frac{9}{x}+\frac{1}{x^2}
=(y^2-2)+9y+20=0
=y^2+9y+18=(y+3)(y+6)=0
\Rightarrow y=-3;-6
\Rightarrow(1)\Rightarrow x+\frac{1}{x}=-3\Rightarrow Vô nghiệm$
$(2)\Rightarrow x+\frac{1}{x}=-6\Rightarrow Vô nghiệm$
$\Rightarrow Vô nghiệm$
Có gì post hộ đáp án nhé

Dùng mọi thủ đoạn

N=$N=\frac{3-2\sqrt{2}}{\sqrt{2}-1}=\frac{2-2\sqrt{2}+1}{\sqrt{2}-1}=\frac{(\sqrt{2}-1)^2}{\sqrt{2}-1}=\sqrt{2}-1$
neu thay can thiet thi thanks minh nha

$M=\frac{\sqrt{12}+3}{\sqrt{3}}=\frac{2\sqrt{3}+\sqrt{3}.\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.(2+\sqrt{3})}{\sqrt{3}}=2+\sqrt{3}$

$CMR : \sqrt{1^3+{2^3+...{n^3{}}}} \epsilon \mathbb{N} (n\epsilon \mathbb{N})$