Giải pt X(x-5)-(x+5)(x-2)=4
Bài 1:
a) Giải PT sau: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
b) Giải PT sau: |2x+6|-x=3
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}
Giải hộ mình với: Giải pt
18(x+1)(x+2)(x+5)(2x+5)=(19/4)x^2
Giải pt: x.(x+5).(x-5)-(x+2).(x2-2x+4)=3
=> x( x^ - 5) -(x^3 - 8) = 0
=> x^ 3 - 5x -x^3 +8 = 0
=> 5x = 8
=> x = 8/5
giải pt:
1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)+1/(x-4)(x-5)+1/(x-5)(x-6)=1/10
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)
Vậy phương trình vô nghiệm
ĐKXĐ: \(x\notin\left\{1;2;3;4;5;6\right\}\)
Ta có: \(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}+\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{10\left(x-1\right)}{10\left(x-6\right)\left(x-1\right)}-\dfrac{10\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}=\dfrac{\left(x-1\right)\left(x-6\right)}{10\left(x-1\right)\left(x-6\right)}\)
Suy ra: \(x^2-7x+6=10x-10-10x+60\)
\(\Leftrightarrow x^2-7x+6=50\)
\(\Leftrightarrow x^2-7x-44=0\)
\(\Leftrightarrow x^2-11x+4x-44=0\)
\(\Leftrightarrow x\left(x-11\right)+4\left(x-11\right)=0\)
\(\Leftrightarrow\left(x-11\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-11=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
Vậy: S={11;-4}
ĐKXĐ : \(x\notin\left\{1;2;...;6\right\}\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x-2\right)-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{\left(x-5\right)-\left(x-6\right)}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{5}{\left(x-1\right)\left(x-6\right)}=\dfrac{5}{50}\\ \Rightarrow\left(x-1\right)\left(x-6\right)=50\\ \Leftrightarrow x^2-7x-44=0\\ \Leftrightarrow\left(x-11\right)\left(x+4\right)=0\\ \Leftrightarrow\begin{matrix}x=-4\\x=11\end{matrix}\left(t.m\right)\)
Giải pt:
(1-x)/(2-x)+(5)/(x+2)=(12)/(x^(2)-4)+1
\(\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow x^2+x-2+5x-10=12+x^2-4\)
\(\Leftrightarrow6x-12=8\)
=>6x=20
hay x=10/3(nhận)
ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{1-x}{2-x}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{12}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{x^2+x-2+5x-10-12-x^2+4}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow6x-20=0\\ \Leftrightarrow x=\dfrac{10}{3}\left(tm\right)\)
giải pt: x^5 + 2x^4 +3x^3 + 3x^2 + 2x +1=0
giải pt: x^4 + 3x^3 - 2x^2 +x - 3=0
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
giải pt
`(x-3)^4 + (x-5)^4 =2`
Ta có:
\(\left(x-3\right)^4+\left(x-5\right)^4=2\)
\(\Leftrightarrow\left(x-4+1\right)^4+\left(x-4-1\right)^4=2\)
Đặt: \(y=x-4\) ta có:
\(\Leftrightarrow\left(y+1\right)^4+\left(y-1\right)^4=2\)
\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)
\(\Leftrightarrow2y^4+12y^2+2=2\)
\(\Leftrightarrow2y^4+12y^2=2-2\)
\(\Leftrightarrow2y^4+12y^2=0\)
\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)
Mà: \(y^2+6\ge6>0\forall x\)
\(\Leftrightarrow2y^2=0\)
\(\Leftrightarrow y^2=0\)
\(\Leftrightarrow y=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
Giải PT
a) | x + 5 | + 3| x - 2 | = | x + 4 |
b) | x + 2 | + | x + 5 | + | 6 - 2x | = 20
a) / x + 5 / +3/ x - 2/ = / x + 4/ ( 1)
Lập bảng xét dấu , ta có :
*) Với : x < - 5 , ta có:
( 1 ) ⇔ - x - 5 + 3( 2 - x) = - x - 4
⇔ - x - 5 + 6 - 3x = - x - 4
⇔ 1 - 4x = -x - 4
⇔ 3x = 5
⇔ x = \(\dfrac{5}{3}\) ( không thỏa mãn )
*) Với : - 5 ≤ x < - 4 , ta có :
( 1) ⇔ x + 5 + 3( 2 - x ) = - x - 4
⇔ x + 5 + 6 - 3x = -x - 4
⇔ 11 - 2x = - x - 4
⇔ x = 15 ( không thỏa mãn )
*) Với : - 4 ≤ x < 2 , ta có :
( 1) ⇔ x + 5 + 3( 2 - x) = x + 4
⇔ x + 5 + 6 - 3x = x + 4
⇔ 11 - 2x = x + 4
⇔ 3x = 7
⇔ x = \(\dfrac{7}{3}\) ( không thỏa mãn )
*) Với : x ≥ 2 , ta có :
( 1) ⇔ x + 5 + 3( x - 2) = x + 4
⇔ x + 5 + 3x - 6 = x + 4
⇔ 4x - 1 = x + 4
⇔3x = 5
⇔ x = \(\dfrac{5}{3}\) ( không thỏa mãn )
Vậy , PT trên vô nghiệm
Giải Pt
(2x+4)(x-3)-(x+2)(x-4)=x(x-5)
(x-2)*2 = (2x-4)(x+5)
* là dấu mũ nha
\(\left(2x+4\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)
\(2\left(x+2\right)\left(x-3\right)-\left(x+2\right)\left(x-4\right)=x\left(x+5\right)\)
\(\left(x+2\right)\left(2x-6-x+4\right)=x\left(x+5\right)\)
\(\left(x+2\right)\left(x-2\right)-x^2-5x=0\)
\(x^2-2x+2x-4-x^2-5x=0\)
\(-5x-4=0\)
\(-5x=4\)
\(\Rightarrow\)\(x=\frac{-4}{5}\)
\(\left(x-2\right)^2=\left(2x-4\right)\left(x+5\right)\)
\(\left(x-2\right)^2-2\left(x-2\right)\left(x+5\right)=0\)
\(\left(x-2\right)\left(x-2-2x-10\right)=0\)
\(\left(x-2\right)\left(-x-12\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\-x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}}\)
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