§1. Đại cương về phương trình

a) \(\sqrt{x^2-4x-12}\le x-4\)

\(ĐK:\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-2\\x\ge6\end{matrix}\right.\\x\ge4\end{matrix}\right.\Leftrightarrow x\ge6}\)

\(x^2-4x-12\le x^2-8x+16\\ \Leftrightarrow4x\le28\Leftrightarrow x\le7\)

Vậy \(6\le x\le7\)

b) \(\left(x-2\right)\sqrt{x^2+4}\le x^2-4\Leftrightarrow\left(x-2\right)\left(x+2-\sqrt{x^2+4}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x+2\ge\sqrt{x^2+4}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x+2\le\sqrt{x^2+4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)

Vd1

3x=6<=>x=2

Vd2

4x-1=11<=>4x=12<=>X=3

VD3

2x^2+3-5=0

<=>2x^2=2<=>x^2=1<=>x=\(\pm\)1

a: |x-2|=x+1

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x-2-x-1\right)\left(x-2+x+1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(2x-1\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\)

b: |x+1|=x-2

\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(x-2-x-1\right)\left(x-2+x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\2x-1=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\left(loại\right)\)

c: =>|2x-2|=x+2

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(2x-2-x-2\right)\left(2x-2+x+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x-4\right)\cdot3x=0\end{matrix}\right.\Leftrightarrow x\in\left\{0;\dfrac{4}{3}\right\}\)

d: |x-2|=2x-1

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x-1-x+2\right)\left(2x-1+x-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(x+1\right)\left(3x-3\right)=0\end{matrix}\right.\Leftrightarrow x=1\)

a, ĐKXĐ: \(x\ge3\)

\(pt\Leftrightarrow\sqrt{x-3}\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-1=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\\x=2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

b, ĐKXĐ: \(x\ge-1\)

\(pt\Leftrightarrow\sqrt{x+1}\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\x+1=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ: \(x>2\)

\(pt\Leftrightarrow\frac{x}{\sqrt{x-2}}=\frac{3-x}{\sqrt{x-2}}\)

\(\Leftrightarrow x=3-x\)

\(\Leftrightarrow x=\frac{3}{2}\left(l\right)\)

\(\Rightarrow\) Phương trình vô số nghiệm

d, ĐKXĐ: \(x>-1\)

\(pt\Leftrightarrow\frac{x^2-4}{\sqrt{x+1}}=\frac{x+3+x+1}{\sqrt{x+1}}\)

\(\Leftrightarrow x^2-4=2x+4\)

\(\Leftrightarrow x^2-2x-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=4\)

b: ĐKXĐ: x>=-1

\(\sqrt{x+1}=x+1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-1\\\left(x+1\right)^2=x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\cdot x=0\\x>=-1\end{matrix}\right.\Leftrightarrow x\in\left\{0;-1\right\}\)

c: \(\sqrt{x-1}=1-x\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1>=0\\1-x< =0\end{matrix}\right.\Leftrightarrow x=1\)

Do đó: x=1 là nghiệm của phương trình

d: \(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)(ĐKXĐ: x<>1)

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)

\(\Leftrightarrow2x^2-2x+3x-3+4-x^2-3=0\)

\(\Leftrightarrow x^2+x-2=0\)

=>(x+2)(x-1)=0

=>x=-2(nhận) hoặc x=1(loại)

a: Ta có: \(\left(x+1\right)^2=0\)

=>x+1=0

hay x=-1

Thay x=-1 vào \(mx^2-\left(2m+1\right)x+m=0\), ta được:

m+2m+1+m=0

=>3m=-1

hay m=-1/3

b:x+2=0

nên x=-2

Thay x=-2 vào \(\dfrac{mx}{x+3}+3m-1=0\), ta được:

\(\dfrac{-2m}{-2+3}+3m-1=0\)

=>-2m+3m-1=0

=>m=1

d: 3x-2=0

=>x=2/3

Thay x=2/3 vào (m+3)x-m+4=0, ta được:

\(\dfrac{2}{3}\left(m+3\right)-m+4=0\)

\(\Leftrightarrow\dfrac{2}{3}m+2-m+4=0\)

=>6-1/3m=0

=>1/3m=6

hay m=18

Câu 1: a)

Đặt \(t=x^2+x+1\) Phương trình trở thành :

\(\sqrt{t+3}+\sqrt{t}=\sqrt{2t+7}\) Bình phương hai vế ta được:

\(2t+3+2\sqrt{\left(t+3\right)t}=2t+7\)\(\Leftrightarrow\sqrt{t\left(t+3\right)}=2\)

\(\Leftrightarrow t^{^2}+3t-4=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-4\left(l\right)\end{matrix}\right.\)

Khi \(t=1\)thì \(x^{^2}+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Câu b)

\(\left\{\begin{matrix} x^2-xy-2=0\\ x^2+y^2+2x+2y-2=0\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} 2=x^2-xy\\ x^2+y^2+2x+2y-2=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+2x+2y-(x^2-xy)=0\)

\(\Leftrightarrow y^2+2x+2y+xy=0\)

\(\Leftrightarrow y(y+2)+x(y+2)=0\Leftrightarrow (y+2)(x+y)=0\)

Xét 2 TH:

TH1: \(y+2=0\Leftrightarrow y=-2\)

Thay vào pt đầu tiên suy ra \(x^2+2x-2=0\Leftrightarrow x=-1\pm \sqrt{3}\)

TH2: \(x+y=0\Leftrightarrow x=-y\)

Thay vào pt đầu tiên suy ra \(x^2+x^2-2=0\Leftrightarrow x=\pm 1\Rightarrow y=\mp 1\)

Vậy \((x,y)=(-1\pm \sqrt{3};-2);(1;-1);(-1;1)\)

\(\Leftrightarrow\left\{{}\begin{matrix}A=0\\B=0\end{matrix}\right.\)

tầm hơn 10h thui bn ak

uk giống mk nhỉ

Chắc là bây giờ mình off