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a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
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
Sửa đề: \(\left(x-1\right)^2-\left(3x+2\right)\left(x-12\right)=\left(x^2+1\right)\left(x-2\right)-x^2\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(3x^2-36x+2x-24\right)=x^3-2x^2+x-2-x^2\)
=>\(x^3-3x^2+3x-1-3x^2+34x+24=x^3-3x^2+x-2\)
=>\(x^3-6x^2+37x+23-x^3+3x^2-x+2=0\)
=>\(-3x^2+36x+25=0\)
=>\(x=\dfrac{18\pm\sqrt{399}}{3}\)
\(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-3\left(\frac{2x-4}{x-4}\right)^2=0\)
<=> \(\left(x+1\right)^2.\left(x-2\right)^2.\left(x-4\right)^2+\frac{x+1}{x-4}.\left(x-2\right)^2.\left(x-4\right)^2-\frac{3\left(2x-4\right)^2}{\left(x-4\right)^2}.\left(x-2\right)^2.\left(x-4\right)^2\)\(=0.\left(x-2\right)^2.\left(x-4\right)^2\)
<=> \(\left(x+1\right)^2.\left(x-4\right)^2+\left(x+1\right).\left(x-2\right)^2.\left(x-4\right)^2-3\left(2x-4\right)^2.\left(x-2\right)^2=0\)
<=> \(-\left(x-3\right)\left(5x-4\right)\left(2x^2-9x+16\right)=0\)
<=> \(\orbr{\begin{cases}x-3=0\\5x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
Mà vì: \(2x^2-9x+16\ne0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{4}{5}\end{cases}}\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)
Vậy phương trình vô nghiệm
Đặt BT trên là A.
TH1: \(x< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|=-x\\\left|x-1\right|=-\left(x-1\right)=1-x\\\left|x-2\right|=-\left(x-2\right)=2-x\end{cases}}\)
\(\Rightarrow A=-x-2\left(1-x\right)+3\left(2-x\right)\)
\(=-x-2+2x+6-3x\)
\(=-2x+4=4\Rightarrow x=0\)( Không thỏa mãn trường hợp x < 0 )
Tương tự với các trường hợp 2 ; 3 ; 4 :
TH2 : \(0\le x< 1\)
\(\Rightarrow A=x-2\left(1-x\right)+3\left(2-x\right)\)
\(=x-2+2x+6-3x\Rightarrow4=4\)( Thỏa mãn )
Bởi vậy với mọi x thỏa mãn \(0\le x< 1\) thì biểu thức luôn đúng.
TH3 : \(1\le x< 2\)
\(A=x-2\left(x-1\right)+3\left(2-x\right)\)
\(=2-2x+2+6-3x=10-5x=4\Rightarrow5x=6\Rightarrow x=\frac{6}{5}\)( thỏa mãn trường hợp \(1\le x< 2\))
TH4: \(x\ge2\)
\(\Rightarrow A=x-2\left(x-1\right)+3\left(x-2\right)\)
\(=x-2x+2+3x-6=2x-4=4\Rightarrow x=4\)( Thỏa mãn )
Vậy ...
TH1: x < 0
pt <=> - x - 2 + 2x + 6 - 3x = 4<=> - 2x + 4 = 4 <=> - 2x = 0 <=> x=0 (loại)
TH2: \(0\le x\le1\)
pt <=> x - 2 + 2x + 6 - 3x = 4 <=> 4 = 4 luôn đúng!
TH3: 1 < x < 2
pt <=> x - 2x + 2 + 6 - 3x = 4 <=> - 4x - 8 = 4 <=> -4x = 12 <=> x=-3 (loại)
TH4: \(x\ge2\)
pt <=> x - 2x + 2 + 3x - 6 =4 <=> 2x - 4 =4 <=> 2x = 8 <=> x=4 (nhận)
Vậy .........