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\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)+9=0\)
\(\Leftrightarrow\left(x^2-3x+4\right)\left(x^2+3x-10\right)+9=0\)
\(\Leftrightarrow\left(x^2+3x-7+3\right)\left(x^2+3x-7-3\right)+9=0\)
\(x^2+3x-7=0\)
\(x^2+3x=7\)
\(\Rightarrow x^2+2x.\frac{3}{2}+\frac{9}{4}=7+\frac{9}{4}\)
\(\Rightarrow\left(x+\frac{3}{2}\right)^2=\frac{37}{4}\)
\(\Rightarrow x+\frac{3}{2}=\pm\sqrt{\frac{37}{4}}\)
\(\Rightarrow x=\frac{-3}{2}-\sqrt{\frac{37}{4}}\)
\(\Rightarrow x=\frac{-3}{2}+\sqrt{\frac{37}{4}}\)
Vậy \(S=\left\{\frac{-3}{2}-\sqrt{\frac{37}{4}};\frac{-3}{2}+\sqrt{\frac{37}{4}}\right\}\)
<=> [3(x-1)]2- [2(2x+1)]2= 0
<=> (3x-3)2 - (4x+2)2= 0
<=> (3x-3-4x-2)(3x-3+4x+2) = 0
<=> (-x-5)(7x-1) = 0
=> -x-5= 0 hoặc 7x-1= 0
=> x= -5 => x = 1/7
\(9\left(x-1\right)^2-4\left(2x+1\right)^2=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow9x^2-18x+9-16x^2-16x-4=0\)
\(\Leftrightarrow-7x^2-34x+5=0\)
\(\Leftrightarrow-7x^2+35x-x+5=0\)
\(\Leftrightarrow-7x\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(-7x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\-7x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1}{7}\end{matrix}\right.\)
c: =>(x+2)(x+3)(x-5)(x-6)=180
=>(x^2-3x-10)(x^2-3x-18)=180
=>(x^2-3x)^2-28(x^2-3x)=0
=>x(x-3)(x-7)(x+4)=0
=>\(x\in\left\{0;3;7;-4\right\}\)
c: =>(x-3)(x+2)(2x+1)(3x-1)=0
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
bai 1
1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0
<=>(2x)^2-5^2=0
<=>(2x+5)*(2x-5)=0
<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự
\(\left(3x-1\right)^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)\)
\(\Leftrightarrow9x^2-6x+1-9x+6=9\left(x^2-2x-3\right)\)
\(\Leftrightarrow9x^2-15x+7=9x^2-18x-27\)
\(\Leftrightarrow-15x+18x+7+27=0\)
\(\Leftrightarrow3x+34=0\)
\(\Leftrightarrow x=\frac{-34}{3}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{34}{3}\right\}\)
*\(\dfrac{x-1}{x+2}\)-\(\dfrac{x}{x+2}\)=\(\dfrac{5x-2}{4-x^2}\).ĐKXĐ: x\(\ne\pm2\)
<=>\(\dfrac{\left(x-1\right)\left(2-x\right)}{4-x^2}\)-\(\dfrac{x\left(2-x\right)}{4-x^2}\)=\(\dfrac{5x-2}{4-x^2}\)
=>2x-\(x^2\)-2+x-2x+\(x^2\)=5x-2
<=>x-2=5x-2
<=>x-5x=2-2
<=>-4x=0
<=> x = 0(TM)
Vậy phương trình có tập nghiệm là S={0}
\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)
<=> \(\left(x+1\right)^2.\left(x-2\right)^2.\left(x-4\right)^2+\frac{x+1}{x-4}.\left(x-2\right)^2.\left(x-4\right)^2-\frac{3\left(2x-4\right)^2}{\left(x-4\right)^2}.\left(x-2\right)^2.\left(x-4\right)^2\)\(=0.\left(x-2\right)^2.\left(x-4\right)^2\)
<=> \(\left(x+1\right)^2.\left(x-4\right)^2+\left(x+1\right).\left(x-2\right)^2.\left(x-4\right)^2-3\left(2x-4\right)^2.\left(x-2\right)^2=0\)
<=> \(-\left(x-3\right)\left(5x-4\right)\left(2x^2-9x+16\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
Mà vì: \(2x^2-9x+16\ne0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
\(4.\left(x+1\right)^2-9.\left(x-1\right)^2=0\)
\(\Leftrightarrow4.\left(x^2+2x+1\right)-9.\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow4x^2+8x+4-9x^2+18x-9=0\)
\(\Leftrightarrow\left(4x^2+8x+4\right)-\left(9x^2-18x+9\right)=0\)
\(\Leftrightarrow\left(2x+2\right)^2-\left(3x-3\right)^2=0\)
\(\Leftrightarrow\left[2x+2-\left(3x-3\right)\right].\left[2x+2+\left(3x-3\right)\right]=0\)
\(\Leftrightarrow\left(2x+2-3x+3\right).\left(2x+2+3x-3\right)=0\)
\(\Leftrightarrow\left(5-x\right).\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5-x=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{5;\frac{1}{5}\right\}.\)
Chúc bạn học tốt!
\(4\left(x+1\right)^2-9\left(x-1\right)^2=0\)
\(\Leftrightarrow4\left(x^2+2x+1\right)-9\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow4x^2+8x+4-9x^2-18x-9=0\)
\(\Leftrightarrow-5x^2-10x-5=0\)
\(\Leftrightarrow-5\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow-5\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy S = {1}