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a) I x-7 I= 5 - x <=> x-7 = 5 - x hoặc x-7 = x-5 (x nhỏ hơn hoặc bằng 5)<=>x= 6(loại) hoặc x thuộc rỗng. Vậy phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1:
a. \((x+1)(x+3) - x(x+2)=7 \)
\(x^2+ 3x +x +3 - x^2 -2x =7\)
\(x^2+4x+3-x^2-2x=7\)
\(=> 2x+3=7\)
\(2x=4\)
\(x = 2\)
Bài 2:
a)
\((3x-5)(2x+11) -(2x+3)(3x+7) \)
\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)
\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)
\(=-65\)
\(\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^8+x^7+1=x^8+x^7-x^2-x+x^2+x+1\)
\(=x^7\left(x+1\right)-x\left(x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x+1\right)\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=x.\left(x+1\right)\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\cdot\left(x+1\right)\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=x\cdot\left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x.\left(x^2-1\right)\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
1)
Biểu thức không phân tích được thành nhân tử. Sửa thành:
\(x^2-5x-14=x^2+2x-7x-14\)
\(=x(x+2)-7(x+2)=(x-7)(x+2)\)
2)
\(2x^2+x-6=2x^2+4x-3x-6\)
\(=2x(x+2)-3(x+2)=(2x-3)(x+2)\)
3)
\(15x^2+7x-12\) (biểu thức không phân tích đc thành nhân tử)
4)
\(x^2+11x+30=x^2+5x+6x+30\)
\(=x(x+5)+6(x+5)=(x+6)(x+5)\)
5) \(81x^4+1\) (biểu thức không phân tích được thành nhân tử)
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a ) \(VT=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1=VP\)
Đẳng thức xảy ra khi \(\left(x-1\right)\left(2-x\right)\ge0\Rightarrow1\le x\le2\)
c ) \(VT=\left|x+1\right|+\left|2x+4\right|\ge\left|x+1+2x+4\right|=\left|3x+5\right|\ge3x+5=VP\)
Đẳng thức xảy ra khi \(\begin{cases}\left(x+1\right)\left(2x+4\right)\ge0\\3x+5\ge0\end{cases}\Rightarrow x\ge1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........
![](https://rs.olm.vn/images/avt/0.png?1311)
a) đặt A=(x+1)(x+3)(x+5)(x+7)+ 15
=>A=(x+ 1)(x+7)(x+3)(x+5) + 15
= (x2 +8x+7)(x2 +8x+15) + 15
đặt y= x2+8x+11
=> A= (y-4)(y+4)+15 = y2-16+15 = y2-1
=(y-1)(y+1) = (x2+8x +10)(x2+8x+12)
b) A= (x2+8x +10)(x2+8x+12) \(⋮\) (x2+8x +10)
=> A=A=(x+1)(x+3)(x+5)(x+7)+ 15 \(⋮\) (x2+8x +10)
a) dk x<=5 ta có: x-7=5-x hoặc x-7=x-5 =>x=6 (loại) hoặc 0x=2(vô lý) vậy pt đã cho vô nghiệm b)với 1<x<3 thì x-1>0 ;3-x>0 => (x-1)-(3-x)=6 =>x=5(loại) với x<1 thì x-1<0 ;3-x>0 =>(1-x)-(3-x)=6 =>0x=8 (vô lý) với x>3 thì x-1>0 ;3-x<0 => (x-1)-(x-3)=6=>0x=4 (vô lý) vậy pt đã cho vô nghiệm
tốn diện tích quá mày songoku