????  

xin lỗi nha ! 

mình mới học lớp 3 

mà bài này khó nắm 

ko bt thì ko nhắn nha

a.A=\(\sqrt{12}-\sqrt{27}+\sqrt{4+2\sqrt{3}}\)\(=2\sqrt{3}-3\sqrt{3}+\sqrt{\left(\sqrt{3}+1\right)^2}\)               \(=-\sqrt{3}+\sqrt{3}+1\)                                                                                  =1                                                                                                        b. \(x^2-2x-4=0\)                                                                                   Δ= \(\left(-2\right)^2-4\times1\times-4=20>0\)                                             \(\Rightarrow\)   phương trình có 2 nghiệm pb                                                \(x1=\dfrac{2+\sqrt{20}}{2}=1+\sqrt{5}\)     \(x2=\dfrac{2-\sqrt{20}}{2}=1-\sqrt{5}\)                      c. \(\left\{{}\begin{matrix}2x-y=5\\x+3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=5\\2x+6y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-7y=7\\2x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x+1=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)

`a,x^2 +4x-5=0`

`<=> x^2-x+5x-5=0`

`<=> x(x-1)+5(x-1)=0`

`<=>(x-1)(x+5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

`b, x^2 -x-12=0`

`<=> x^2 +3x-4x-12=0`

`<=>(x^2+3x)-(4x+12)=0`

`<=>x(x+3)-4(x+3)=0`

`<=>(x+3)(x-4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

`c, (2x-7)^2 - 6(2x-7)(x-3)=0`

`<=>(2x-7)(2x-7 -6x+18)=0`

`<=>(2x-7) ( -4x+11)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\-4x+11=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\-4x=-11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{11}{4}\end{matrix}\right.\)

a: =>(x+5)(x-1)=0

=>x=1 hoặc x=-5

b: =>(x-4)(x+3)=0

=>x=4 hoặc x=-3

c: =>(2x-7)(2x-7-6x+18)=0

=>(2x-7)(-4x+11)=0

=>x=11/4 hoặc x=7/2

1.

\(\Leftrightarrow6x^2-12x+7-6\sqrt{6x^2-12x+7}-7=0\)

Đặt \(\sqrt{6x^2-12x+7}=t>0\)

\(\Rightarrow t^2-6t-7=0\Rightarrow\left[{}\begin{matrix}t=-1\left(loại\right)\\t=7\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{6x^2-12x+7}=7\)

\(\Leftrightarrow6x^2-12x+7=49\Rightarrow x=1\pm2\sqrt{2}\)

2.

\(\Delta'=\left(m+1\right)^2-m^2-3=2m-2>0\Rightarrow m>1\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+3\end{matrix}\right.\)

\(\left(x_1+x_2\right)^2-2x_1x_2=2x_1x_2+8\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-8=0\)

\(\Leftrightarrow4\left(m+1\right)^2-4\left(m^2+3\right)-8=0\)

\(\Leftrightarrow2m-4=0\Rightarrow m=2\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1

\(PT\Leftrightarrow x^2-2x+\sqrt{6x^2-12x+7}=0\\ \Leftrightarrow x^2-2x+1+\sqrt{6x^2-12x+7}-1=0\\ \Leftrightarrow\left(x-1\right)^2+\dfrac{6\left(x-1\right)^2}{\sqrt{6x^2-12x+7}+1}=0\\ \Leftrightarrow\left(x-1\right)\left(x-1+\dfrac{6}{\sqrt{6x^2-12x+7}+1}\right)=0\\ \Leftrightarrow x=1\left(x-1+\dfrac{6}{\sqrt{6x^2-12x+7}+1}>0\right)\)

\(\sqrt{2x+1}+\sqrt{7-2x}-2\sqrt{\left(2x+1\right)\left(7-2x\right)}+4=0\)

\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{7-2x}\right)^2=-4\)

vô lý

Hoàng Nguyễn Văn bạn có chút nhầm lẫn gì rồi, có dấu căn ở đây nên k đưa vào hằng đẳng thức được

\(DK:x\in\left[-\frac{1}{2};\frac{7}{2}\right]\)

Dat \(\hept{\begin{cases}\sqrt{2x+1}=a\\\sqrt{7-2x}=b\end{cases}}\left(a,b\ge0\right)\)

\(\Rightarrow a^2+b^2=8\)

Ta co HPT 

\(\hept{\begin{cases}a^2+b^2=8\left(1\right)\\a+b-2ab+4=0\left(2\right)\end{cases}}\)

\(\left(1\right)-\left(2\right)\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-12=0\)

\(\Leftrightarrow\left(a+b+3\right)\left(a+b-4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a+b=-3\\a+b=4\end{cases}}\)

Sau do the vo la xong

\(\sqrt{x^2-2x+1}-7=0\left(1\right)\)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=7\Leftrightarrow\left|x-1\right|=7\)

TH1: \(x\ge1\)

\(\left(1\right)\Leftrightarrow x-1=7\Leftrightarrow x=8\)

TH2: \(x< 1\)

\(\left(1\right)\Leftrightarrow1-x=7\Leftrightarrow x=-6\)

Ta có: \(\sqrt{x^2-2x+1}-7=0\)

\(\Leftrightarrow\left|x-1\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)