1. \(\sqrt{\left(x+3\right)\left(x+7\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)

\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+7\right)}-3\sqrt{x+3}-2\sqrt{x+7}+6=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+7}-3\right)-2\left(\sqrt{x+7}-3\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+7}-3\right)\left(\sqrt{x+3}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+7}-3=0\\\sqrt{x+3}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+7}=3\\\sqrt{x+3}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

Vậy...

2. \(2x^2+2x+1=\sqrt{4x+1}\)

\(\Leftrightarrow2x^2+2x+1-\sqrt{4x+1}=0\)

\(\Leftrightarrow4x^2+4x+2-2\sqrt{4x+1}=0\)

\(\Leftrightarrow4x+1-2\sqrt{4x+1}+1+4x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+1}=1\\2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4x+1=1\\x=0\end{matrix}\right.\)\(\Leftrightarrow x=0\)

Vậy...

3. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}=\frac{x+3}{2}\)

\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}=\frac{x+3}{2}\)

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

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

Đặt \(\sqrt{x-1}=a\)

\(\Leftrightarrow x-1=a^2\Leftrightarrow x+3=a^2+4\)

\(pt\Leftrightarrow\left|a-1\right|+a+1=\frac{a^2+4}{2}\)

+) Xét \(a\le1\Leftrightarrow a-1\le0\Leftrightarrow1\le x\le2\)

\(pt\Leftrightarrow1-a+a+1=\frac{a^2+4}{2}\)

\(\Leftrightarrow2=\frac{a^2+4}{2}\)

\(\Leftrightarrow a^2+4=4\)

\(\Leftrightarrow a=0\)

\(\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\) ( thỏa )

+) Xét \(a\ge1\Leftrightarrow a-1\ge0\Leftrightarrow x>2\)

\(pt\Leftrightarrow a-1+a+1=\frac{a^2+3}{2}\)

\(\Leftrightarrow2a=\frac{a^2+3}{2}\)

\(\Leftrightarrow a^2+3=4a\)

\(\Leftrightarrow a^2-4a+3=0\)

\(\Leftrightarrow\left(a-1\right)\left(a-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loai\right)\\x=10\left(thoa\right)\end{matrix}\right.\)

Vậy...

a: =4x^2+8x-3x-6

=4x(x+2)-3(x+2)

=(x+2)(4x-3)

b: =3(3x^2-2x-1)

=3(3x^2-3x+x-1)

=3(x-1)(3x+1)

c: =2x^2-4x+x-2

=2x(x-2)+(x-2)

=(x-2)(2x+1)

d: =3x^2+3x-2x-2

=3x(x+1)-2(x+1)

=(x+1)(3x-2)

e: =3x^2+9x+x+3

=3x(x+3)+(x+3)

=(x+3)(3x+1)

a) \(4x^2+5x-6\)

\(=4x^2+8x-3x-6\)

\(=\left(4x^2+8x\right)-\left(3x+6\right)\)

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

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

b) \(9x^2-6x-3\)

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

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

\(=3\left[3x\left(x-1\right)+\left(x-1\right)\right]\)

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

c) \(2x^2-3x-2\)

\(=2x^2-4x+x-2\)

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

\(=2x\left(x-2\right)+\left(x-2\right)\)

\(=\left(2x+1\right)\left(x-2\right)\)

d) \(3x^2+x-2\)

\(=3x^2+3x-2x-2\)

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

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

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

e) \(3x^2+10x+3\)

\(=3x^2+9x+x+3\)

\(=3x\left(x+3\right)+\left(x+3\right)\)

\(=\left(x+3\right)\left(3x+1\right)\)

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\dfrac{1}{4}x^2=\left(x^2+\dfrac{11}{2}x+8\right)^2-\left(\dfrac{1}{2}x\right)^2=\left(x^2+\dfrac{11}{2}x+8-\dfrac{1}{2}x\right)\left(x^2+\dfrac{11}{2}x+8+\dfrac{1}{2}x\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)=\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)

\(=\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)\)

\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)

\(x^2+6x-3=4x\sqrt{2x-1}\left(1\right)\)      ĐK: \(x\ge\frac{1}{2}\)

Đặt \(\sqrt{2x-1}=a\ge0\)

\(\Rightarrow6x-3=3a^2\)

=> (1) <=> x^2 +3a^2 = 4ax

<=> x^2 -4ax +3a^2 =0

<=> x^2 -ax - 3ax +  3a^2 =0

<=> x(x-a) -3a(x-a) =0

<=> (x-a) ( x-3a ) =0

\(\Leftrightarrow\orbr{\begin{cases}x=a\\x=3a\end{cases}}\)

TH1: x=a

\(\Rightarrow x=\sqrt{2x-1}\)\(\left(x\ge0\right)\)

\(\Leftrightarrow x^2=2x-1\)

\(\Leftrightarrow\left(x-1\right)^2=0\)

<=> x=1 (tm)

TH2: x= 3a

\(\Rightarrow x=3\sqrt{2x-1}\left(x\ge0\right)\)

\(\Leftrightarrow x^2=18x-9\)

\(\Leftrightarrow x^2-18x+9=0\)

\(\Delta=288\)

=> pt có 2 nghiệm pb \(\orbr{\begin{cases}x=\frac{18+12\sqrt{2}}{2}=9+6\sqrt{2}\left(tm\right)\\x=\frac{18-12\sqrt{2}}{2}=9-6\sqrt{2}\left(tm\right)\end{cases}}\)

Vậy ...

a) Đề đúng: \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=\left(x^2+x\right)+4\left(x^2+x\right)-12\)

Đặt \(x^2+x=y\)

BT = \(y^2+4y-12\)

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

\(=\left(y-2\right)\left(y+6\right)\)

\(=\left(x^2+x-2\right)\left(x^2+x-6\right)\)

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

b) Đặt \(x^2+x+1=y\)

=> BT = \(y\left(y+1\right)-12\)

\(=y^2+y-12\)

\(=\left(y-3\right)\left(y+4\right)\)

\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

cảm ơn các cậu nhiều

a: \(x^3-2x+4\)

\(=x^3+2x^2-2x^2-4x+2x+4\)

\(=\left(x+2\right)\left(x^2-2x+2\right)\)

b: \(x^3-4x^2+12x-27\)

\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)

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

c: \(x^3+2x^2+2x+1\)

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

\(=\left(x+1\right)\left(x^2+x+1\right)\)

Áp dụng hàm đẳng thức của lớp 8 là ra.

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)