giải pt
`(x-3)^4 + (x-5)^4 =2`
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 (3-x)^4+(2-x)^4=(5-2x)^4
Đặt 3-x = a ; 2-x = b
=> 5-2x = a+b
pt <=> a^4+b^4 = (a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4
<=> a^4+4a^3b+6a^2b^2+4ab^3+b^4-a^4-b^4 = 0
<=> 4a^3b+6a^2b^2+4ab^3 = 0
<=> 2a^3b+3a^2b^2+2ab^3 = 0
<=> ab.(2a^2+3ab+2b^2) = 0
<=> ab=0 ( vì 2a^2+3ab+2b^2 > 0 )
<=> a=0 hoặc b=0
<=> 3-x=0 hoặc 2-x=0
<=> x=3 hoặc x=2
Vậy .............
Tk mk nha
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 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 : \(x\notin\left\{1;2;...;6\right\}\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{\left(x-1\right)-\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x-2\right)-\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+...+\dfrac{\left(x-5\right)-\left(x-6\right)}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-2}+...+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-1}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{5}{\left(x-1\right)\left(x-6\right)}=\dfrac{5}{50}\\ \Rightarrow\left(x-1\right)\left(x-6\right)=50\\ \Leftrightarrow x^2-7x-44=0\\ \Leftrightarrow\left(x-11\right)\left(x+4\right)=0\\ \Leftrightarrow\begin{matrix}x=-4\\x=11\end{matrix}\left(t.m\right)\)
|x+3|/4-|x-4|/9=1/2-(x+5)/36
giải pt
Giải PT; (x+3)4 + (x+5)4 =2
Giải phương trình:
\(\left(x+3\right)^4+\left(x+5\right)^4=2\) \(\left(\text{1}\right)\)
Đặt \(y=x+4\), khi đó phương trình \(\left(\text{1}\right)\) trở thành:
\(\left(y-1\right)^4+\left(y+1\right)^4=2\)
\(\Leftrightarrow\) \(y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1=2\)
\(\Leftrightarrow\) \(2y^4+12y^2+2=2\)
\(\Leftrightarrow\) \(y^4+6y^2+1=1\)
\(\Leftrightarrow\) \(y^4+6y^2+9-9=0\)
\(\Leftrightarrow\) \(\left(y^2+3\right)^2-3^2=0\)
\(\Leftrightarrow\) \(y^2\left(y^2+6\right)=0\) \(\left(\text{1'}\right)\)
Vì \(y^2\ge0\) nên \(y^2+6\ge6>0\) nên từ \(\left(\text{1'}\right)\) suy ra \(y^2=0\), tức là \(\left(x+4\right)^2=0\) \(\Leftrightarrow\) \(x+4=0\) \(\Leftrightarrow\) \(x=-4\)
Vậy, tập nghiệm của pt là \(S=\left\{-4\right\}\)
Giải pt: (x + 5)4 + (x + 3)4 = 2
đặt x+4=a
pt trở thành (a+1)4+(a-1)4=2
\(\leftrightarrow a^4+4a^3+6a^2+4a+1+a^4-4a^3+6a^2-4a+1=2\)
\(\leftrightarrow2a^4+12a^2+2=2\leftrightarrow a^2\left(a^2+12\right)=0\)
a2+12>0 vs mọi a=> a=0 =>x+4=0 <=> x=-4
vậy...
Giải pt sau:
(x+3).(x+5)=(x+4).(2+x)
Giúp tôi với
=>x^2+8x+15=x^2+6x+8
=>8x+15=6x+8
=>2x=-7
=>x=-7/2
Đề:
`=> x^2 + 3x + 5x + 15 = 2x + 8 + x^2 + 4x`
`=> x^2 + 8x + 15 = x^2 + 6x + 8`
`=> x^2 - x^2 + 8x - 6x + 15 - 8 = 0`
`=> 2x + 7 = 0`
`=> 2x = -7`
`=> x = -7/2`
Vậy `x = -7/2`
\(\left(x+3\right)\left(x+5\right)=\left(x+4\right)\left(2+x\right)\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)\left(4+x\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\\4+x=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\\x=-4\\x=-2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-3;-5;-4;-2\right\}\)