a) \(\left(x^2-7x+12\right).\left(x^2-15x+56\right)-60\)

\(=\left(x-3\right)\left(x-4\right)\left(x-7\right)\left(x-8\right)-60\)

b) \(x^4+2000x^2+1999x+2000\)

\(=\left(x^2-x+2000\right)\left(x^2+x+1\right)\)

\(=\left(x^2+x-1\right)^2+1999\left(x^2+x+1\right)+1\)

hình như Thành lộn đề rồi đấy ạ, ý mình là phân tích thành nhân tử á

b) \(x^4+2000x^2+1999x+2000\)

\(=x^4-x+2000x^2+2000x+2000\)

\(=x\left(x^3-1\right)+2000\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2000\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x-2000\right)\)

\(1,\\ a,ĐK:m\ne1\\ \Delta=49+48\left(m-1\right)=48m+1\\ \text{PT vô nghiệm }\Leftrightarrow48m+1< 0\Leftrightarrow m< -\dfrac{1}{48}\\ \text{PT có nghiệm kép }\Leftrightarrow48m+1=0\Leftrightarrow m=-\dfrac{1}{48}\\ \text{PT có 2 nghiệm phân biệt }\Leftrightarrow48m+1>0\Leftrightarrow m>-\dfrac{1}{48};m\ne1\)

\(b,\Delta=4\left(m-1\right)^2+4\left(2m+1\right)=4m^2+8>0,\forall m\\ \text{Vậy PT có 2 nghiệm phân biệt với mọi m}\\ 2,\\ \text{PT có 2 nghiệm phân biệt }\)

\(\Leftrightarrow\Delta=4\left(m+1\right)^2-4\left(m^2-1\right)>0\\ \Leftrightarrow4m^2+8m+4-4m^2+4>0\\ \Leftrightarrow8m+8>0\\ \Leftrightarrow m>-1\)

Pls help nhanh , tui cần gấp lắm ;-;

ĐKXĐ: \(x\ge-5\)

\(\Leftrightarrow\left(x+7\right)^2-2\left(x+7\right)\sqrt{x+5}+x+5-16=0\)

\(\Leftrightarrow\left(x+7-\sqrt{x+5}\right)^2-16=0\)

\(\Leftrightarrow\left(x+7-\sqrt{x+5}-4\right)\left(x+7-\sqrt{x+5}+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=x+3\left(x\ge-3\right)\\\sqrt{x+5}=x+11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2+6x+9\\x+5=x^2+22x+121\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=0\\x^2+21x+116=0\left(vn\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4< -3\left(l\right)\end{matrix}\right.\)

`(3x-1)(x^2 +2)=(3x-1)(7x-10)`

`<=> (3x-1)(x^2 +2)-(3x-1)(7x-10)=0`

`<=> (3x-1)(x^2 +2-7x+10)=0`

`<=> (3x-1)(x^2 -7x+12)=0`

`<=> (3x-1)(x^2 -3x-4x+12)=0`

`<=> (3x-1)[x(x-3)-4(x-3)]=0`

`<=> (3x-1)(x-4)(x-3)=0`

\(< =>\left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\\x=3\end{matrix}\right.\)

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left[\left(x^2-3x\right)-\left(4x-12\right)\right]=0\)

\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(3x-1\right)\left[\left(x-3\right)\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{1}{3};3;4\right\}\)

1/ (x-63)(x+10)(4x² -188x-2520)

2/ 9(x-1)(2x-1)(64x² + 208x+32)/8

3/ (x+2)(x-6)(x² -4x -41)

Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0

=x(x+3)+2(x+3)=(x+2)(x+3)=0

Dễ rồi

2)\(x^2-x-6=0=x^2-3x+2x-6=0\)

=x(x-3)+2(x-3)=0

=(x+2)(x-3)=0

Dễ rồi

3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)

Vì \(x^2+1>0\)

=>\(\left(x+2\right)^2=0\)

Dễ rồi

4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0

=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)

=>x+1=0

=>..................

5)\(x^2-7x+6=x^2-6x-x+6\) =0

=x(x-6)-(x-6)=0

=(x-1)(x-6)=0

=>.....

6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0

=2x(x+1)-5(x+1)=0

=(2x-5)(x+1)=0

7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0

Dễ rồi

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