(x+1)\(\sqrt{x+3}\sqrt{x+5}\le0\)
1. \(\sqrt{2x^2+5x-6}>2-x\)x
2.\(\sqrt{x^2+2}\le x-1\)
3.\(\sqrt{x^2-2x-15}>2x+5\)
4.\(\left(16-x^2\right)\sqrt{x-3}\le0\)
5.\(\sqrt{x^2+2017}\le\sqrt{2018}x\)
6.\(\hept{\begin{cases}\frac{x+3}{2x-3}-\frac{x}{2x-1}\le0\\\sqrt{x^2+3}+3x< 1\end{cases}}\)
Câu 6:
\(\hept{\begin{cases}\frac{x+3}{2x-3}-\frac{x}{2x-1}\le0\\\sqrt{x^2+3}+3< 1\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2x^2-x+6x-3-2x^2+3x}{\left(2x-3\right)\left(2x-1\right)}\le0\\x^2+3< \left(1-3x\right)^2\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}8x-3\le0\\x^2+3< 1-6x+9x^2\end{cases}\Leftrightarrow\hept{\begin{cases}8x-3\le0\\8x^2-6x-2< 0\end{cases}\Leftrightarrow}\hept{\begin{cases}x< \frac{3}{8}\\\frac{-1}{4}x< x< \frac{1}{4}\end{cases}\Rightarrow}S\left(\frac{-1}{4};\frac{3}{8}\right)}\)
Giải bpt
a) \(\frac{3}{\sqrt{x-2}-1}\ge\frac{5}{\sqrt{x-2}-3}\)
b) \(x\sqrt{x-3}-\frac{\sqrt{x-3}}{2-x}\le0\)
c) \(\frac{2\sqrt{x-1}-4}{\sqrt{4-x^2}-1}\ge2-\sqrt{x-1}\)
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\x\ne\left\{3;11\right\}\end{matrix}\right.\)
Đặt \(\sqrt{x-2}=t\ge0\)
\(\Rightarrow\frac{3}{t-1}\ge\frac{5}{t-3}\)
\(\Leftrightarrow\frac{3}{t-1}-\frac{5}{t-3}\ge0\)
\(\Leftrightarrow\frac{3t-9-5t+5}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{-2t-4}{\left(t-1\right)\left(t-3\right)}\ge0\)
\(\Leftrightarrow\frac{t+2}{\left(t-1\right)\left(t-3\right)}\le0\)
\(\Leftrightarrow1< t< 3\)
\(\Rightarrow1< \sqrt{x-2}< 3\)
\(\Leftrightarrow1< x-2< 9\Rightarrow3< x< 11\)
b/
ĐKXĐ: \(x\ge3\)
- Với \(x=3\) BPT thỏa mãn
- Với \(x>3\Rightarrow\sqrt{x-3}>0\) BPT tương đương
\(x-\frac{1}{2-x}\le0\Leftrightarrow x+\frac{1}{x-2}\le0\)
\(\Leftrightarrow\frac{x^2-2x+1}{x-2}\le0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}\le0\Rightarrow\) không tồn tại x thỏa mãn
Vậy BPT có nghiệm duy nhất \(x=3\)
c/
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\4-x^2\ge0\\\sqrt{4-x^2}\ne1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\-2\le x\le2\\x\ne\pm\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le2\\x\ne\sqrt{3}\end{matrix}\right.\)
BPT tương đương:
\(\frac{2\left(\sqrt{x-1}-2\right)}{\sqrt{4-x^2}-1}+\sqrt{x-1}-2\ge0\)
\(\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(\frac{2}{\sqrt{4-x^2}-1}+1\right)\ge0\)
Do \(x\le2\Rightarrow\sqrt{x-1}\le1\Rightarrow\sqrt{x-1}-2< 0\)
BPt tương đương:
\(\frac{2}{\sqrt{4-x^2}-1}+1\le0\)
\(\Leftrightarrow\frac{1+\sqrt{4-x^2}}{\sqrt{4-x^2}-1}\le0\)
\(\Leftrightarrow\sqrt{4-x^2}-1< 0\) (do \(1+\sqrt{4-x^2}>0\) \(\forall x\))
\(\Leftrightarrow\sqrt{4-x^2}< 1\Leftrightarrow x^2>3\Rightarrow x>\sqrt{3}\)
Vậy nghiệm của BPT đã cho là: \(\sqrt{3}< x\le2\)
Khi \(1\le x\le0,\) tìm \(maxE=\dfrac{5}{x+\sqrt{x}+1}\).
Lời giải:
Với mọi $1\geq x\geq 0$ thì $x+\sqrt{x}+1\geq 1$
$\Rightarrow E=\frac{5}{x+\sqrt{x}+1}\leq \frac{5}{1}=5$
Vậy $E_{\max}=5$ khi $x=0$
Giải bất phương trình
1) \(\frac{x^4-1}{x^2+3x}+x^2\ge1\)
2) \(\left(x^4-5x^2+4\right)\left(\frac{x-2}{x}-3\right)\le0\)
3) \(\left(\frac{4}{x}-\frac{2}{x-1}\right)\left(\frac{x^2+1}{x}-2\right)\le0\)
4) \(\left(\sqrt{x^3-4x}-\sqrt{15}\right)\sqrt{\frac{1+x}{x}-2}\le0\)
a/
\(\Leftrightarrow\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+3x}+x^2-1\ge0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{x^2+1}{x^2+3x}+1\right)\ge0\)
\(\Leftrightarrow\left(x^2-1\right)\left(\frac{2x^2+3x+1}{x^2+3x}\right)\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(2x+1\right)}{x\left(x+3\right)}\ge0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(2x+1\right)\left(x+1\right)^2}{x\left(x+3\right)}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x< -3\\x=-1\\-\frac{1}{2}\le x< 0\\x\ge1\end{matrix}\right.\)
b/
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)\left(\frac{-2-2x}{x}\right)\le0\)
\(\Leftrightarrow\frac{-2.\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+1\right)}{x}\le0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)\left(x-2\right)\left(x+1\right)^2}{x}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\le-2\\x=-1\\0< x\le1\\x\ge2\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(\frac{4\left(x-1\right)-2x}{x\left(x-1\right)}\right)\left(\frac{x^2+1-2x}{x}\right)\le0\)
\(\Leftrightarrow\frac{\left(2x-4\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)^2}{x^2\left(x-1\right)}\le0\)
\(\Rightarrow1< x\le2\)
d/
ĐKXĐ: \(\left\{{}\begin{matrix}x^3-4x\ge0\\\frac{1+x}{x}-2\ge0\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x-2\right)\left(x+2\right)\ge0\\\frac{1-x}{x}\ge0\\x\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}-2\le x\le0\\x\ge2\end{matrix}\right.\\0< x\le1\\x\ne0\end{matrix}\right.\)
\(\Rightarrow\) Không tồn tại x thỏa mãn ĐKXĐ
Vậy BPT đã cho vô nghiệm
Bài 1: Giải phương trình
1) \(\sqrt{4x^2+12x+9}=2-x\left(vớix\le0\right)\)
2) \(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ( với \(x^2+5x+4>0\))
3) \(\sqrt{5x+1}=4\)
4) \(\sqrt{3-x}=7\)
Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ĐK:\(x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x^2+1\right)^2}=x^2+5x+4\)
\(\Leftrightarrow\left|x^2+1\right|=x^2+5x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)
\(\left(\sqrt{8-\sqrt{X-3}}-\sqrt{5-\sqrt{X-3}}\right)^2=5^2\)=\(5^2\)
\(\Leftrightarrow-2\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=25-3=22\)
\(\Leftrightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=11\)
Do \(\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\ge0\Rightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\le0\)
\(\Rightarrow\)PT vô nghiệm
\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+4}\). Tìm x để \(A\le0\)
A\(A\le0< =>\dfrac{\sqrt{x}-1}{\sqrt{x}+4}\le0\)
\(< =>\sqrt{x}-1\le0\left(do\sqrt{x}+4\ge0\right)\)
\(< =>\sqrt{x}\le1< =>x\le1\)
Với \(x\ge0\)
\(A\le0< =>\dfrac{\sqrt{x}-1}{\sqrt{x}+4}\le0\)
\(< =>\sqrt{x-1}\le0\) (vì \(\sqrt{x}+4\ge0\))
\(< =>x-1\le0< =>x\le1\)
Kết hợp với ĐKXĐ ta được \(0\le x\le1\)
\(\left(x+5\right)\left(2-x\right)+3\sqrt{x^2+3x}\le0\)
cho biểu thức: P=\(\left[1-\frac{x-3\sqrt{x}}{x-9}\right]:\left[\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9x}{x+\sqrt{x}-6}\right]\) \(\left(x\le0;x\ne9;x\ne4\right)\)
a) Rút gọn P
b) Tìm giá trị của x để P=1
x-9=(cănx-3)(cănx+3)
x+cănx-6=(cănx-2)(cănx+3)=-(2-cănx)(cănx+3)
x-3cănx=x(căn-3)
tự quy đồng rút gọn nha