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a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
b: Để phương trình có hai nghiệm trái dấu thì (m+2)(m-4)<0
=>-2<m<4
\(a,\) Đặt \(x^2+2x=a\), pt trở thành:
\(a^2-3a+2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=0\left(1\right)\\x^2+2x-2=0\left(2\right)\end{matrix}\right.\)
\(\left[{}\begin{matrix}\Delta\left(1\right)=4+4=8\\\Delta\left(2\right)=4+8=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{8}}{2}\\x=\dfrac{-2+\sqrt{8}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{12}}{2}\\x=\dfrac{-2+\sqrt{12}}{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{2}\\x=-1+\sqrt{2}\\x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{matrix}\right.\)
\(b,\) Đặt \(x^2+x=b\), pt trở thành:
\(b\left(b+1\right)-6=0\\ \Leftrightarrow b^2+b-6=0\\ \Leftrightarrow\left[{}\begin{matrix}b=2\\b=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\x\in\varnothing\left[x^2+x+3=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(d,x^4-2x^3+x=2\\ \Leftrightarrow x^4-2x^3+x-2=0\\\Leftrightarrow\left(x^3+1\right)\left(x-2\right)=0 \\ \Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x\in\varnothing\left[x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Lời giải:
a.
PT $\Leftrightarrow (x^2+2x)^2-(x^2+2x)-2[(x^2+2x)-1]=0$
$\Leftrightarrow (x^2+2x)(x^2+2x-1)-2(x^2+2x-1)=0$
$\Leftrightarrow (x^2+2x-1)(x^2+2x-2)=0$
$\Leftrightarrow x^2+2x-1=0$ hoặc $x^2+2x-2=0$
$\Leftrightarrow x=-1\pm \sqrt{2}$ hoặc $x=-1\pm \sqrt{3}$
b.
PT $\Leftrightarrow (x^2+x)^2+(x^2+x)-6=0$
$\Leftrightarrow (x^2+x)^2-2(x^2+x)+3(x^2+x)-6=0$
$\Leftrightarrow (x^2+x)(x^2+x-2)+3(x^2+x-2)=0$
$\Leftrightarrow (x^2+x-2)(x^2+x+3)=0$
$\Leftrightarrow x^2+x-2=0$ (chọn) hoặc $x^2+x+3=0$ (loại do $x^2+x+3=(x+0,5)^2+2,75>0$)
$\Leftrightarrow x=-1\pm \sqrt{3}$
c. Nghiệm khá xấu. Bạn coi lại đề.
d.
PT $\Leftrightarrow x^3(x-2)+(x-2)=0$
$\Leftrightarrow (x^3+1)(x-2)=0$
$\Leftrightarrow x^3+1=0$ hoặc $x-2=0$
$\Leftrightarrow x=-1$ hoặc $x=2$
b: \(\Leftrightarrow\left(x^2-2x+1-1\right)^2-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left[\left(x-1\right)^2-1\right]^2-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left(x-1\right)^4-2\left(x-1\right)^2+1-2\left(x-1\right)^2-1=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-3\right)\left(x+1\right)=0\)
hay \(x\in\left\{1;3;-1\right\}\)
a: \(\Leftrightarrow2x^3-3x-10=-2\left(8-12x+6x^2-x^3\right)\)
\(\Leftrightarrow2x^3-3x-10=-16+24x-12x^2+2x^3\)
\(\Leftrightarrow-3x-10+16-24x+12x^2=0\)
=>\(12x^2-27x+6=0\)
hay \(x\in\left\{2;\dfrac{1}{4}\right\}\)
\(x^4+\left(x-1\right)\left(x^2-2x+2\right)=0\)
\(\Leftrightarrow x^4+x^3-3x^2+4x-2=0\)
\(\Leftrightarrow x^2\left(x^2-x+1\right)+2x\left(x^2-x+1\right)-2\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+2x-2\right)=0\)
đến đây tự giải nhé
Bài 2:
a: \(\Leftrightarrow\left(x^2-3x+2\right)\left(x^2-3x+3\right)=0\)
=>x^2-3x+2=0
=>x=2 hoặc x=1
b: \(\Leftrightarrow\left(\left|x\right|\right)^2-\left|x\right|+m=0\)
Để phương trình có nghiệm thì \(\text{Δ}>=0\)
=>1-4m>=0
=>m<=1/4
Để phương trình vô nghiệm thì Δ<0
=>m>1/4
c: TH1: m=1
=>-2x+2=0
=>x=1
TH2: m<>1
\(\text{Δ}=\left(-2\right)^2-4\left(1-m\right)\cdot2m\)
\(=4+8m\left(m-1\right)\)
\(=8m^2-8m+4\)
Để phương trình có nghiệm thì Δ>=0
=>\(m\in R\)
1/ \(\left(x-1\right)\left(x^2-2x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\left(1\right)\\x^2-2x-2=0\left(2\right)\end{cases}}\)
+ Từ (1) => x = 1
+ Từ (2) . Ta có: \(\Delta=\left(-2\right)^2-4\left(-2\right)=12\Rightarrow\sqrt{\Delta}=\sqrt{12}=2\sqrt{3}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2+2\sqrt{3}}{2}=1+\sqrt{3}\\x=\frac{2-2\sqrt{3}}{2}=1-\sqrt{3}\end{cases}}\)
Vậy \(x=\left\{1+\sqrt{3};1-\sqrt{3};1\right\}\)
2/ \(\left(x-1\right)^2\left(2x^2-x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\left(1\right)\\2x^2-x+2=0\left(2\right)\end{cases}}\)
+ Từ (1) => x = 1
+ Từ (2). Ta có: \(2x^2-x+2=2\left(x^2-\frac{1}{2}x+1\right)\)
\(=2\left(x^2-2.\frac{1}{4}x+\frac{1}{16}-\frac{1}{16}+1\right)\)
\(=2\left[\left(x-\frac{1}{4}\right)^2+\frac{15}{16}\right]=2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}>0\)
=> pt (2) vô nghiệm
Vậy x = 1
a)(x-1)(x2-2x-2)=0
=>x-1=0 hoặc x2-2x-2=0
=>x1,2=(2±căn 12)/2=1- căn 3 hoặc căn 3+1
b)(x-1)2(2x2-x+2)=0
=>(x-1)2=0 hoặc 2x2-x+2=0
Với Denta<0 =>vô nghiệm
Vậy x=1