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

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

Với \(x=28-6\sqrt{3}tmđk\)thay vào P ta có :

\(P=\dfrac{\sqrt{28-6\sqrt{3}}}{28-6\sqrt{3}+\sqrt{28-6\sqrt{3}}+1}=\dfrac{\sqrt{\left(3\sqrt{3}-1\right)^2}}{29-6\sqrt{3}+\sqrt{\left(3\sqrt{3}-1\right)^2}}=\dfrac{3\sqrt{3}-1}{29-6\sqrt{3}+3\sqrt{3}-1}=\dfrac{3\sqrt{3}-1}{28-3\sqrt{3}}=\dfrac{\left(3\sqrt{3}-1\right)\left(28+3\sqrt{3}\right)}{784-27}=\dfrac{81\sqrt{3}-1}{757}\)

\(=\left[\dfrac{1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}-\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]\cdot\dfrac{1-\sqrt{a}}{\sqrt{a}}\)

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

Với \(a=3-2\sqrt{2}tmđk\)thay vào M ta được :

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

Ta có : \(18M=18\cdot\dfrac{2}{\sqrt{a}+1}=\dfrac{36}{\sqrt{a}+1}\)

Đặt \(\dfrac{36}{\sqrt{a}+1}=x^2\left(x\in N\cdot\right)\Rightarrow x^2\left(\sqrt{a}+1\right)=36\)

Ta lại có a².b² = (a.b)² => \(\left\{{}\begin{matrix}x^2\\\sqrt{a}+1\end{matrix}\right.\)phải là bình phương của các số tự nhiên

mà \(x^2\left(\sqrt{a}+1\right)=36\)=> Ta có các trường hợp sau :

\(\left\{{}\begin{matrix}x^2=1\\\sqrt{a}+1=36\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\a=1225\end{matrix}\right.\)(tm)

\(\left\{{}\begin{matrix}x^2=36\\\sqrt{a}+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\a=0\end{matrix}\right.\)(ktm)

\(\left\{{}\begin{matrix}x^2=4\\\sqrt{a}+1=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\a=64\end{matrix}\right.\)(tm)

\(\left\{{}\begin{matrix}x^2=9\\\sqrt{a}+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\a=9\end{matrix}\right.\)(tm)

Bài 12:

a: Ta có: \(P=\dfrac{x+2}{x\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)

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

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

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

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

b: Thay \(x=28-6\sqrt{3}\) vào P, ta được:

\(P=\dfrac{3\sqrt{3}-1}{28-6\sqrt{3}+3\sqrt{3}-1+1}=\dfrac{3\sqrt{3}-1}{28-3\sqrt{3}}=\dfrac{81\sqrt{3}-1}{757}\)

\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

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

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

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

Nãy ghi nhầm =="

a)Hđ gđ là nghiệm pt

`x^2=2x+2m+1`

`<=>x^2-2x-2m-1=0`

Thay `m=1` vào pt ta có:

`x^2-2x-2-1=0`

`<=>x^2-2x-3=0`

`a-b+c=0`

`=>x_1=-1,x_2=3`

`=>y_1=1,y_2=9`

`=>(-1,1),(3,9)`

Vậy tọa độ gđ (d) và (P) là `(-1,1)` và `(3,9)`

b)

Hđ gđ là nghiệm pt

`x^2=2x+2m+1`

`<=>x^2-2x-2m-1=0`

PT có 2 nghiệm pb

`<=>Delta'>0`

`<=>1+2m+1>0`

`<=>2m> -2`

`<=>m> 01`

Áp dụng hệ thức vi-ét:`x_1+x_2=2,x_1.x_2=-2m-1`

Theo `(P):y=x^2=>y_1=x_1^2,y_2=x_2^2`

`=>x_1^2+x_2^2=14`

`<=>(x_1+x_2)^2-2x_1.x_2=14`

`<=>4-2(-2m-1)=14`

`<=>4+2(2m+1)=14`

`<=>2(2m+1)=10`

`<=>2m+1=5`

`<=>2m=4`

`<=>m=2(tm)`

Vậy `m=2` thì ....

Bài 2

5 C

Bài 3

1 D

6 C

Còn lại ol r nhé

2) 5. C

3) 2. D

6. C

Còn lại ok nha

bài đây ạ  

3d:

20<x<45

x chia 4 dư 1 nên x-1 thuộc B(4)

=>\(x-1\in\left\{0;4;...;44;48\right\}\)

=>\(x\in\left\{1;5;...;45;49\right\}\)

mà 20<x<45

nên x thuộc {21;26;31;35;41}

4:

a: A={x∈N|51<=x<=127}

b: B={x∈N|100<=x<=999}

c: C={x∈N|x=7k+5; 0<=k<=8}