Bài 7: Phương trình quy về phương trình bậc hai

1: \(\Leftrightarrow\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)

=>(x+y)^2>=4xy

=>(x-y)^2>=0(luôn đúng)

2: \(\Leftrightarrow a^3+b^3-a^2b-ab^2>=0\)

=>a^2(a-b)-b^2(a-b)>=0

=>(a-b)^2(a+b)>=0(luôn đúng)

................?

Ta có các hệ số: a=1;b=-m,c=21

Ta có △=b²-4ac=(-m)²-4.1.21=\(m^2-84\)

Để phương trình có nghiệm kép thì: △=0⇒\(m^2-84=0\Rightarrow m^2=84\Rightarrow m=\pm2\sqrt{21}\)

Ta có phương trình có nghiệm kép ⇒x=\(\dfrac{-b}{2a}=\dfrac{-b}{2.1}=\dfrac{-b}{2}\)

Nếu b=-m=\(-\left(2\sqrt{21}\right)\)thì

\(x=\dfrac{-b}{2}=\dfrac{2\sqrt{21}}{2}=\sqrt{21}\)

Nếu b=-m=\(-\left(-2\sqrt{21}\right)=2\sqrt{21}\)thì

\(x=\dfrac{-b}{2}=\dfrac{-2\sqrt{21}}{2}=-\sqrt{21}\)

Vậy nghiệm kép là: \(\left[{}\begin{matrix}\sqrt{21}\\-\sqrt{21}\end{matrix}\right.\)

Phương trình có nghiện kép \(\Leftrightarrow\Delta=m^2-4.21=0\Leftrightarrow m^2=84\Leftrightarrow m=\pm2\sqrt{21}\)

Lời giải:
Vì \(x\in\mathbb{Z}\Rightarrow 2x+1\neq 0\)

Ta có: \((2x+1)y=x+1\Rightarrow y=\frac{x+1}{2x+1}\)

Vì \(y\in\mathbb{Z}\Rightarrow \frac{x+1}{2x+1}\in\mathbb{Z}\)

\(\Leftrightarrow x+1\vdots 2x+1\)

\(\Rightarrow 2(x+1)\vdots 2x+1\)

\(\Rightarrow 2x+1+1\vdots 2x+1\Rightarrow 1\vdots 2x+1\)

Vậy \(2x+1\in\left\{\pm 1\right\}\Rightarrow x\in\left\{0;-1\right\}\)

+) \(x=0\Rightarrow y=1\)

+) \(x=-1\Rightarrow y=0\)

Vậy.................

Lời giải:

Áp dụng BĐT Bunhiacopxky với $x>0; 1-x> 0$ ta có:

\(\left(\frac{2}{1-x}+\frac{1}{x}\right)[(1-x)+x]\geq (\sqrt{2}+1)^2\)

\(\Rightarrow \frac{2}{1-x}+\frac{1}{x}\geq \frac{(\sqrt{2}+1)^2}{1-x+x}=(\sqrt{2}+1)^2\)

Vậy \(y_{\min}=(\sqrt{2}+1)^2\)

Dấu bằng xảy ra khi \(\frac{\sqrt{2}}{1-x}=\frac{1}{x}\Rightarrow x=\sqrt{2}-1\)

ĐK: \(-4\le x\le1\)

<=>\(\sqrt{1-x}=3-\sqrt{4+x}\)

<=>\(1-x=9+4+x-6\sqrt{4+x}\)

<=>\(0=12+2x-6\sqrt{4+x}\)

<=>\(6+x-3\sqrt{4+x}=0\)

<=>\(6+x=3\sqrt{4+x}\)

<=>\(36+12x+x^2=9\left(4+x\right)\)

<=>\(36+12x+x^2=36+9x\)

<=>\(x^2+3x=0\)

<=>\(x\left(x+3\right)=0\)

<=>\(\left\{{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)(nhận)

Vậy S={-3;0}

C₂: ĐK:-4\(\le x\le\)1

ĐKXĐ : \(x\ge-\dfrac{7}{3}\)

\(x^2+7x+12=2\sqrt{3x+7}\)

\(\Leftrightarrow x^2+7x+12-2\sqrt{3x+7}=0\)

\(\Leftrightarrow\left(x^2+4x+4\right)+\left(3x+7-2\sqrt{3x+7}+1\right)=0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(\sqrt{3x+7}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=0\\\left(\sqrt{3x+7}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=-2\left(TMĐK\right)\)

Vậy \(S=\left\{-2\right\}\)

Chúc bạn học tốt

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

\(\Leftrightarrow\left(2x^2-x-1\right)^2-2\left(2x^2-x-1\right)-7=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-x-1=1+2\sqrt{2}\\2x^2-x-1=1-2\sqrt{2}\end{matrix}\right.\Leftrightarrow x\in\left\{1.82;-1.32\right\}\)

b) x⁴ - 3 = (x + 1)(x - 1)

\(\Rightarrow\) x⁴ - 3 = x² - 1

\(\Rightarrow\) x⁴ = x² + 2

\(\Rightarrow\) 2 = x⁴ - x²

\(\Rightarrow\) 2,25 = (x² - 0,5)²

\(\Rightarrow\) \(\left[{}\begin{matrix}x^2-0,5=1,5\\x^2-0,5=-1,5\end{matrix}\right.\) mà x² - 0,5 \(\ge\) -0,5 nên x² - 0,5 = 1,5

\(\Rightarrow x^2=2\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

ĐK:x\(\ge-1\)

\(x^3+\left(x+1\right)\sqrt{x+1}+2\sqrt{2}=\left(x+\sqrt{x+1}+\sqrt{2}\right)^3\)(*)

Đặt \(a=x;b=\sqrt{x+1};c=\sqrt{2}\)

Vậy (*)\(\Leftrightarrow a^3+b^3+c^3=\left(a+b+c\right)^3\Leftrightarrow3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

_a=-b\(\Leftrightarrow x=-\sqrt{x+1}\)\(\left(x\le0\right)\Leftrightarrow x^2=x+1\Leftrightarrow x^2-x-1=0\Leftrightarrow x=\dfrac{1-\sqrt{5}}{2}\)

_b=-c\(\Leftrightarrow\sqrt{x+1}=-\sqrt{2}\)(vô nghiệm)

_c=-a\(\Leftrightarrow\sqrt{2}=-x\Leftrightarrow x=-\sqrt{2}\left(ktm\right)\)

Vậy S={\(\dfrac{1-\sqrt{5}}{2}\)}