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

Nghỉ đã hôm sau làm mệt

Bài 1

a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)

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

b/

\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)

\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

1.

c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)

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

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

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

d/

\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Bài 1:

e/ \(\Leftrightarrow x^4+2x^2-8x+5=0\)

\(\Leftrightarrow x^4-2x^3+x^2+2x^3-4x^2+2x+5x^2-10x+5=0\)

\(\Leftrightarrow x^2\left(x-1\right)^2+2x\left(x-1\right)^2+5\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x^2+2x+5\right)\left(x-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+5=0\left(vn\right)\\x=1\end{matrix}\right.\)

Bài 2:

a/ Đặt \(x^2-5x=t\)

\(t^2+10t+24=0\Rightarrow\left[{}\begin{matrix}t=-4\\t=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=-4\\x^2-5x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2-5x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\\x=2\\x=3\end{matrix}\right.\)

\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)

1.

Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)

Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)

\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=x^2-2x+m\\x=1-x^2+2x-m\end{matrix}\right.\)

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

Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):

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

\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)

Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\): 

Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)

Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)

a,\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\left(đkxđ:x\ne2;4\right)\)

\(< =>\frac{-2}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)

\(< =>-2-\left(x^2-5x+4\right)=x^2+x-5\)

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

\(< =>-2x^2+4x-1=0\)

\(< =>2x^2-4x+1=0\)

đến đây thì pt bậc 2 dể rồi

\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\left(đkxđ:x\ne\pm1\right)\)

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

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

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

\(< =>2+3x-3+x^2-2x+1=0\)

\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=-1\left(loai\right)\\x=0\left(tm\right)\end{cases}}\)

\(\frac{x+2}{x-2}-\frac{2}{x^2-2x}=\frac{1}{x}\left(đkxđ:x\ne0;x\ne2\right)\)

\(< =>\frac{\left(x+2\right)x}{\left(x-2\right)x}-\frac{2}{x\left(x-2\right)}=\frac{x-2}{x\left(x-2\right)}\)

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

\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=0\left(loai\right)\\x=-1\left(tm\right)\end{cases}}\)

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