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lời giải
a)
\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)
\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)
\(\Leftrightarrow2x\le4\Rightarrow x\le2\)
\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)
\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)
c)Đkxđ: x≥0
x+√x>(2√x+3)(√x−1)
⇔x+√x>2x+√x−3
⇔x−3>0
⇔x>3. (tmđk).
\(\frac{2x-5}{!x-3!}+1>0\Leftrightarrow\frac{2x-5+!x-3!}{!x-3}>0\)
do !x-3!>0 mọi x khác 3=> Bất phương trình tương đương
\(2x-5+!x-3!>0\Leftrightarrow!x-3!>5-2x\)
TH(1) x<3 <=>3-x>5-2x=> x>2
Kết luận(1) \(2< x< 3\)
TH(2) \(x\ge3\Leftrightarrow x-3>5-2x\Rightarrow3x>8\Rightarrow x>\frac{8}{3}\)
Kết luận(2) \(x\ge3\)
(1)và(2) nghiệm của Bpt là: x>2
1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)
\(t^3+2=2t-2\)
\(\Leftrightarrow t^3-2t+4=0\)
\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)
\(\Rightarrow t=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)
\(\Leftrightarrow x^2+5x-2=-8\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)
a/ ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow x+1-\sqrt{2x+2}+\sqrt{2x-1}-1=0\)
\(\Leftrightarrow\frac{x^2+2x+1-2x-2}{x+1+\sqrt{2x+2}}+\frac{2x-1-1}{\sqrt{2x-1}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{x+1+\sqrt{2x+2}}+\frac{2}{\sqrt{2x-1}+1}\right)=0\)
\(\Rightarrow x=1\)
2/ ĐKXĐ:\(\left[{}\begin{matrix}x=0\\x\ge2\\x\le-3\end{matrix}\right.\)
- Nhận thấy \(x=0\) là 1 nghiệm
- Với \(x\ge2\):
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x-2}=2\sqrt{x+3}=\sqrt{4x+12}\)
Ta có \(VT\le\sqrt{2\left(x-1+x-2\right)}=\sqrt{4x-6}< \sqrt{4x+12}\)
\(\Rightarrow VT< VP\Rightarrow\) pt vô nghiệm
- Với \(x\le-3\)
\(\Leftrightarrow\sqrt{1-x}+\sqrt{2-x}=2\sqrt{-x-3}\)
\(\Leftrightarrow3-2x+2\sqrt{x^2-3x+2}=-4x-12\)
\(\Leftrightarrow2\sqrt{x^2-3x+2}=-2x-15\) (\(x\le-\frac{15}{2}\))
\(\Leftrightarrow4x^2-12x+8=4x^2+60x+225\)
\(\Rightarrow x=-\frac{217}{72}\left(l\right)\)
Vậy pt có nghiệm duy nhất \(x=0\)
Bài 3: ĐKXĐ: \(-3\le x\le6\)
Đặt \(\sqrt{3+x}+\sqrt{6-x}=t\) \(\Rightarrow3\le t\le3\sqrt{2}\)
\(t^2=9+2\sqrt{\left(3+x\right)\left(6-x\right)}\Rightarrow-\sqrt{\left(3+x\right)\left(6-x\right)}=\frac{9-t^2}{2}\)
Phương trình trở thành:
\(t+\frac{9-t^2}{2}=m\Leftrightarrow-t^2+2t+9=2m\) (2)
a/ Với \(m=3\Rightarrow t^2-2t-3=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=3\end{matrix}\right.\)
\(\Rightarrow\sqrt{3+x}+\sqrt{6-x}=3\)
\(\Leftrightarrow2\sqrt{\left(3+x\right)\left(6-x\right)}=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)
b/ Xét hàm \(f\left(t\right)=-t^2+2t+9\) trên \(\left[3;3\sqrt{2}\right]\)
\(-\frac{b}{2a}=1< 3\Rightarrow\) hàm số nghịch biến trên \(\left[3;3\sqrt{2}\right]\)
\(f\left(3\right)=6\) ; \(f\left(3\sqrt{2}\right)=6\sqrt{2}-9\)
\(\Rightarrow6\sqrt{2}-9\le2m\le6\Rightarrow\frac{6\sqrt{2}-9}{2}\le m\le3\)
Bài 4 làm tương tự bài 3
\(1\))\(x^2+5x+8=3\sqrt{x^3+5x^2+7x+6}\left(1\right)\\ĐK:x\ge-\dfrac{3}{2} \\ \left(1\right)\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\left(2\right)\)
Đặt \(b=\sqrt{2x+3};a=\sqrt{x^2+x+2}\)
\(\left(2\right)\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)\(\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1\pm\sqrt{5}}{2}\\x=\dfrac{7\pm\sqrt{89}}{2}\end{matrix}\right.\)
4)\(ĐK:x\ge-\dfrac{1}{3}\)
\(x^2-7x+2+2\sqrt{3x+1}=0\\ \Leftrightarrow x^2-7x+6+2\sqrt{3x+1}-4=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)+\dfrac{12\left(x-1\right)}{2\sqrt{3x+1}+4}=0\\ \Leftrightarrow\left(x-1\right)\left(x-6+\dfrac{12}{2\sqrt{3x+1}+4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x-6+\dfrac{12}{2\sqrt{3x+1}+4}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x-5\right)+\dfrac{6}{\sqrt{3x+1}+2}-1=0\\ \Leftrightarrow\left(x-5\right)+\dfrac{4-\sqrt{3x+1}}{\sqrt{3x+1}+2}=0\\ \Leftrightarrow\left(x-5\right)-\dfrac{3\left(x-5\right)}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}=0\\ \Leftrightarrow\left(x-5\right)\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)=3\\ \Leftrightarrow3x+1+6\sqrt{3x+1}+8=3\\ \Leftrightarrow x+2\sqrt{3x+1}+2=0\\ \Leftrightarrow2\sqrt{3x+1}=-x-2\ge0\Leftrightarrow x\le-2\)
Vậy pt có 2 nghiệm là x=1 và x=5