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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
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}
a, đk : x >= 1
\(\left[{}\begin{matrix}3x+5=2x-2\\3x+5=2-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\left(ktm\right)\)
vậy pt vô nghiệm
b, đk >= 0
\(\left[{}\begin{matrix}x^2+1=2x\\x^2+1=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
c, \(\left[{}\begin{matrix}2x^2+2x=0\\2x^2+4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x\left(x+1\right)=0\\x^2+2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0;x=-1\\x=-1\end{matrix}\right.\)
\(ĐK:x\in R\)
\(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\) (*)
Đặt \(x^2+x+1=a;a\ge0\)
\(\rightarrow\left\{{}\begin{matrix}x^2+x+4=a+3\\2x^2+2x+9=2a+7\end{matrix}\right.\)
(*) \(\Rightarrow\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow\left(\sqrt{a+3}+\sqrt{a}\right)^2=\left(\sqrt{2a+7}\right)^2\)
\(\Leftrightarrow a+3+a+2\sqrt{a\left(a+3\right)}=2a+7\)
\(\Leftrightarrow2\sqrt{a\left(a+3\right)}=4\)
\(\Leftrightarrow\sqrt{a\left(a+3\right)}=2\)
\(\Leftrightarrow a\left(a+3\right)=4\)
\(\Leftrightarrow a^2+3a-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\left(tm\right)\\a=-4\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) \((tm)\)
Vậy \(S=\left\{0;-1\right\}\)
a.
ĐKXĐ: \(1\le x\le7\)
\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Biến đổi pt đầu:
\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a^2b^2-b^4=b-a\)
\(\Leftrightarrow b^2\left(a+b\right)\left(a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)
Thế vào pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)
\(\Leftrightarrow...\)
Bài 1:
c) |2x - 1| = x + 2
<=> 2x - 1 = +(x + 2) hoặc -(x + 2)
* 2x - 1 = x + 2
<=> 2x - x = 2 + 1
<=> x = 3
* 2x - 1 = -(x + 2)
<=> 2x - 1 = x - 2
<=> 2x - x = -2 + 1
<=> x = -1
Vậy.....
\(ĐKXĐ:x\ge\frac{1}{2}\)
Phương trình đã cho tương đương :
\(4.\left(x^2+1\right)+3.x.\left(x-2\right).\sqrt{2x-1}=2x^3+10x\)
\(\Leftrightarrow3x\left(x-2\right)\sqrt{2x-1}=2x^3-8x^2+10x-4\)
\(\Leftrightarrow3x.\left(x-2\right).\sqrt{2x-1}=2.\left(x-2\right).\left(x-1\right)^2\) (1)
Dễ thấy \(x=2\) là một nghiệm của (1). Xét \(x\ne2\). Khi đó ta có :
\(3x.\sqrt{2x-1}=2.\left(x-1\right)^2\)(*)
Đặt \(\sqrt{2x-1}=a\left(a\ge0\right)\Rightarrow-a^2=1-2x\)
Khi đó pt (*) có dạng :
\(3x.a=2.\left(x^2-a^2\right)\)
\(\Leftrightarrow2x^2-3xa-2a^2=0\)
\(\Leftrightarrow2x^2-4ax+xa-2a^2=0\)
\(\Leftrightarrow2x.\left(x-2a\right)+a.\left(x-2a\right)=0\)
\(\Leftrightarrow\left(x-2a\right)\left(a+2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2a=x\\a=-2x\end{cases}}\)
+) Với \(2a=x\Rightarrow2\sqrt{2x-1}=x\left(x\ge0\right)\)
\(\Leftrightarrow x^2=4.\left(2x-1\right)\)
\(\Leftrightarrow x^2-8x+4=0\)
\(\Leftrightarrow x=4\pm2\sqrt{3}\) ( Thỏa mãn )
+) Với \(a=-2x\Rightarrow\sqrt{2x-1}=-2x\left(x\le0\right)\)
\(\Leftrightarrow4x^2=2x-1\)
\(\Leftrightarrow4x^2-2x+1=0\) ( Vô nghiệm )
Vậy phương trình đã cho có tập nghiệm \(S=\left\{4\pm2\sqrt{3},2\right\}\)
Đặt \(2x^2-2x+2=a\)
\(\Leftrightarrow\dfrac{a-3x}{x-1}=\dfrac{a+3x+15}{x-3}\)
\(\Leftrightarrow6x^2+3x+2a-15=0\)
\(\Leftrightarrow10x^2-x-11=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=-1\end{matrix}\right.\)
\(x+\left(x-2\right)\left(2x-1\right)=2\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=2-x\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)+\left(x-2\right)=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow x=0;2\)
Vậy tập nghiệm phương trình là S = { 0 ; 2 }
x + ( x - 2 )( 2x + 1 ) = 2
<=> x + 2x2 - 3x - 2 - 2 = 0
<=> 2x2 - 2x - 4 = 0
<=> x2 - x - 2 = 0
<=> x2 - 2x + x - 2 = 0
<=> x( x - 2 ) + ( x - 2 ) = 0
<=> ( x - 2 )( x + 1 ) = 0
<=> x = 2 hoặc x = -1
Vậy phương trình có tập nghiệm S = { 2 ; -1 }