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Bài 1.
a) ( x - 3 )( x + 7 ) = 0
<=> x - 3 = 0 hoặc x + 7 = 0
<=> x = 3 hoặc x = -7
Vậy S = { 3 ; -7 }
b) ( x - 2 )2 + ( x - 2 )( x - 3 ) = 0
<=> ( x - 2 )( x - 2 + x - 3 ) = 0
<=> ( x - 2 )( 2x - 5 ) = 0
<=> x - 2 = 0 hoặc 2x - 5 = 0
<=> x = 2 hoặc x = 5/2
Vậy S = { 2 ; 5/2 }
c) x2 - 5x + 6 = 0
<=> x2 - 2x - 3x + 6 = 0
<=> x( x - 2 ) - 3( x - 2 ) = 0
<=> ( x - 2 )( x - 3 ) = 0
<=> x - 2 = 0 hoặc x - 3 = 0
<=> x = 2 hoặc x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(x^2+2x+2x\sqrt{x+3}=9-\sqrt{x+3}\) \(\left(ĐK:x\ge-3\right)\)
\(\Leftrightarrow\left(x^2+2x\sqrt{x+3}+x+3\right)+x+\sqrt{x+3}=12\)
\(\Leftrightarrow\left(x+\sqrt{x+3}\right)^2+\left(x+\sqrt{x+3}\right)-12=0\)
\(\Leftrightarrow\left(x+\sqrt{x+3}\right)\left(x+\sqrt{x+3}+1\right)-12=0\)
Đặt \(a=x+\sqrt{x+3}\)\(\Leftrightarrow\)\(a+1=x+\sqrt{x+3}+1\)
Ta lại có: \(a.\left(a+1\right)-12=0\)
\(\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\)
\(\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+4=0\\a-3=0\end{cases}}\)
+ \(a+4=0\)\(\Leftrightarrow\)\(x+\sqrt{x+3}+4=0\)
\(\Leftrightarrow\)\(x+4=-\sqrt{x+3}\)
\(\Leftrightarrow\)\(\left(x+4\right)^2=\left(-\sqrt{x+3}\right)^2\)
\(\Leftrightarrow\)\(x^2+8x+16=x+3\)
\(\Leftrightarrow\)\(x^2+7x+13=0\)
\(\Leftrightarrow\)\(\left(x^2+7x+\frac{49}{4}\right)+\frac{3}{4}=0\)
\(\Leftrightarrow\)\(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)
Vì \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}>0\forall x\)mà \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)
\(\Rightarrow\)Phương trình \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)vô nghiệm
+ \(a-3=0\)\(\Leftrightarrow\)\(x+\sqrt{x+3}-4=0\)
\(\Leftrightarrow\)\(x-3=-\sqrt{x+3}\)
\(\Leftrightarrow\)\(\left(x-3\right)^2=\left(-\sqrt{x+3}\right)^2\)
\(\Leftrightarrow\)\(x^2-6x+9=x+3\)
\(\Leftrightarrow\)\(x^2-7x+6=0\)
\(\Leftrightarrow\)\(\left(x^2-x\right)-\left(6x-6\right)=0\)
\(\Leftrightarrow\)\(x.\left(x-1\right)-6.\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-6\right).\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-6=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{1;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thực ra 2 câu đầu rất dễ nha bạn ^^!
1) x4 + 2x3 + x2 + 2x + 1 =0 <=> x3(x+2)+x(x+2)+1 = 0
<=> (x3+x)(x+2) + 1=0
1>0
=> (x3+x)(x+2) + 1=0 <=> (x3+x)(x+2) = 0
<=>\(\orbr{\begin{cases}^{x^3+x=0}\\x+2=0\end{cases}}\)<=>\(\orbr{\begin{cases}^{x\left(x^2+1\right)=0}\\x=-2\end{cases}}\) <=>\(\orbr{\begin{cases}^{x=0}\\x=-2\end{cases}}\)
b)
x3+1=\(2\sqrt[3]{2x-1}\)
<=> x^3 - 1 = 2(\(\sqrt[3]{2x-1}\) -1)
<=> (x-1)(x2+x+1) = \(\frac{4\left(x-1\right)}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}\)
<=> (x-1)[(x2+x+1) - \(\frac{1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{2x-1}+1}\) ] =0
<=> x=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x^2+2x\right)^2+5\left(x^2+2x\right)+6-2=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2+5\left(x^2+2x\right)+4=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(x^2+2x+4\right)=0\)
=>x+1=0
hay x=-1
Đặt \(x^2+2x+2=t\)đk t > 0
\(t\left(t+1\right)-2=0\Leftrightarrow t^2+t-2=0\Leftrightarrow t=1;t=2\left(ktm\right)\)
Với t = 1 \(x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện : \(x\ge0\)
Ta có : \(\sqrt{3x+1}-\sqrt{2x+2}=2\sqrt{x}-\sqrt{x+3}\)
\(\Leftrightarrow3x+1+2x+2-2\sqrt{6x^2-8x+2}=4x+x+3-4\sqrt{x^2+3x}\)
\(\Leftrightarrow\sqrt{6x^2+8x+2}=2\sqrt{x^2+3x}\)
\(\Leftrightarrow6x^2+8x+2=4\left(x^2+3x\right)\)
\(\Leftrightarrow2x^2-4x+2=0\Leftrightarrow x=1\)
Vậy nghiệm phương trình đã cho là : \(x=1\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,(x+3)(x^2+2021)=0`
`x^2+2021>=2021>0`
`=>x+3=0`
`=>x=-3`
`2,x(x-3)+3(x-3)=0`
`=>(x-3)(x+3)=0`
`=>x=+-3`
`b,x^2-9+(x+3)(3-2x)=0`
`=>(x-3)(x+3)+(x+3)(3-2x)=0`
`=>(x+3)(-x)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$
`d,3x^2+3x=0`
`=>3x(x+1)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$
`e,x^2-4x+4=4`
`=>x^2-4x=0`
`=>x(x-4)=0`
`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$
1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)
=> S={-3}
\(x^3+2\sqrt{2}x^2+2x=0\)
\(x\left(x^2+2\sqrt{2}x+2\right)=0\)
\(x\left[x^2+2\sqrt{2}x+\left(\sqrt{2}\right)^2\right]=0\)
\(x\left(x+\sqrt{2}\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\\left(x+\sqrt{2}\right)^2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\sqrt{2}\end{cases}}\)
Vậy ....
\(x^3+2\sqrt{2}x^2+2x=0\)
\(x\left(x^2+2\sqrt{2}x+2\right)=0\)
\(x\left(x+\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+\sqrt{2}\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\sqrt{2}\end{cases}}\)
Vậy \(s=\left\{0;-\sqrt{2}\right\}\)