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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) Thay a = -1 vào phương trình
\(\dfrac{x-1}{x+3}+\dfrac{x-3}{x+1}=2\)
\(\Rightarrow\dfrac{x^2-1+x^2-9}{\left(x+3\right)\left(x+1\right)}=2\)
\(\Rightarrow2x^2-10=2\left(x+3\right)\left(x+1\right)=2x^2+8x+6\)
\(\Rightarrow2x^2+8x+6-2x^{10}+10=0\)
\(\Rightarrow8x+16=0\Rightarrow x=-2\)
b, c Làm tương tự như câu a
d)
Phương trình nhận x = 1 làm nghiệm
=> \(\dfrac{1+a}{1+3}+\dfrac{1-3}{1-a}=2\)
\(\Rightarrow\dfrac{a+1}{4}+\dfrac{2}{a-1}=2\)
\(\Rightarrow\dfrac{a^2-1+8}{4\left(a-1\right)}=2\)
\(\Rightarrow a^2+7=2\left(4a-1\right)=8a-2\)
\(\Rightarrow a^2-8x+9=0\)
\(\Rightarrow\left[{}\begin{matrix}a=4+\sqrt{7}\\a=4-\sqrt{7}\end{matrix}\right.\)
Bài 1 :
a )Thế \(m=1\) vào phương trình ta được :
\(2x^2-3x-2=0\)
\(\Leftrightarrow2x^2+x-4x-2=0\)
\(\Leftrightarrow x\left(2x+1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-\frac{1}{2};2\right\}\)
b ) Theo hệ thức vi-et ta có :
\(\left\{{}\begin{matrix}x_1+x_2=\frac{6m-3}{2}\\x_1x_2=\frac{-3m+1}{2}\end{matrix}\right.\)
\(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(\frac{6m-3}{2}\right)^2-\frac{2\left(-3m+1\right)}{2}\)
\(=\frac{36m^2-36m+9}{4}+3m-1\)
\(=\frac{36m^2-36m+9+12m-4}{4}\)
\(=\frac{36m^2-24m+5}{4}\)
\(=\frac{36m^2-24m+4+1}{4}\)
\(=\frac{\left(6m-2\right)^2+1}{4}\ge\frac{1}{4}\)
Vậy GTNN của A là \(\frac{1}{4}\) . Dấu bằng xảy ra khi \(x=\frac{1}{3}\)
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}
\(x^3-x^2-x=\dfrac{1}{3}\)
\(\Leftrightarrow3\left(x^3-x^2-x\right)=1\)
\(\Leftrightarrow3x^3-3x^2-3x=1\)
\(\Leftrightarrow4x^3-x^3-3x^2-3x=1\)
\(\Leftrightarrow4x^3=x^3+3x^2+3x+1\)
\(\Leftrightarrow4x^3=\left(x+1\right)^3\)
\(\Leftrightarrow x=\dfrac{x+1}{\sqrt[3]{4}}\)
\(\Leftrightarrow x=\dfrac{1}{\sqrt[3]{4}-1}\)
Câu 1. thiếu đề đó bạn ạ
Câu 2:
Ta có: x^3+15x^2+74x+120
=(x^3+4x^2) + (11x^2+44x) + (30x+120)
=(x+4)(x^2+11x+30)
=(x+4)(x+5)(x+6)
Ta có bảng xét dấu
| x | -6 | -5 | -4 | ||||
| x+4 | - | | | - | | | - | | | + |
| x+5 | - | | | - | | | + | | | + |
| x+6 | - | | | + | | | + | | | + |
Để (x+4)(x+5)(x+6)<0
Khi có chỉ 1 số âm hoặc cả 3 số âm
<=> x<-6 hoặc -5<x<-4
ĐKXĐ:\(x\ge-3\)
\(x^2+\sqrt{x+3}=3\\ \Leftrightarrow\sqrt{x+3}=3-x^2\left(-\sqrt{3}\le x\le3\right)\\ \Leftrightarrow x+3=x^4-6x^2+9\\ \Leftrightarrow x^4-6x^2-x+6=0\\ \Leftrightarrow\left(x^4-x^3\right)+\left(x^3-x^2\right)-\left(5x^2-5x\right)-\left(6x-6\right)=0\\ \Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)-5x\left(x-1\right)-6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3+x^2-5x-6\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^3+2x^2\right)-\left(x^2+2x\right)-\left(3x+6\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
x=1/2-căn bậc hai(13)/2, x=1
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