\(\sqrt{2x+7}-\sqrt{5-x}\ge\sqrt{3x-2}\)
Giải BPT
\(\sqrt{2x+7}-\sqrt{5-x}\ge\sqrt{3x-2}\)
ĐKXĐ: \(\frac{2}{3}\le x\le5\)
\(\Leftrightarrow\sqrt{2x+7}\ge\sqrt{5-x}+\sqrt{3x-2}\)
\(\Leftrightarrow2x+7\ge2x+3+2\sqrt{-3x^2+17x-10}\)
\(\Leftrightarrow\sqrt{-3x^2+17x-10}\le2\)
\(\Leftrightarrow-3x^2+17x-10\le4\)
\(\Leftrightarrow3x^2-17x+14\ge0\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge\frac{14}{3}\end{matrix}\right.\)
Kết hợp ĐKXĐ: \(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}\le x\le1\\\frac{14}{3}\le x\le5\end{matrix}\right.\)
Ôn Tập Cơ Bản
1) Tìm điều kiện để các biểu thức sau có nghĩa:
a) \(\sqrt{11-2x}\)
b) \(\sqrt{9x-18}\)
c) \(\sqrt{\dfrac{3}{x^2}}\)
d) \(\sqrt{\dfrac{5}{x-7}}\)
2) Rút gọn:
a) \(\sqrt{16x^2}-2x^2\) với x \(\ge\) 0
b) \(\sqrt{9\left(x+5\right)^2}+2-3x\) với x
c) \(\sqrt{\left(x-5\right)^2}-4x\) với x < 5
\(1,\\ a,ĐK:11-2x\ge0\Leftrightarrow x\le\dfrac{11}{2}\\ b,ĐK:9x-18\ge0\Leftrightarrow x\ge2\\ c,ĐK:x\ne0;\dfrac{3}{x^2}\ge0\left(luôn.đúng.do.3>0;x^2>0\right)\Leftrightarrow x\in R\backslash\left\{0\right\}\\ d,ĐK:\dfrac{5}{x-7}\ge0\Leftrightarrow x-7>0\left(5>0;x-7\ne0\right)\Leftrightarrow x>7\\ 2,\\ a,=\left|4x\right|-2x^2=4x-2x^2\\ b,bạn.thiếu.điều.kiện.nhé\\ c,=\left|x-5\right|-4x=5-x-4x=5-5x\)
Bài 2:
a: \(\sqrt{16x^2}-2x^2=4x-2x^2\)
c: \(\sqrt{\left(x-5\right)^2}-4x=5-4x-x=5-5x\)
1) \(\frac{\sqrt{2\left(X^2-16\right)}}{\sqrt{X-3}}+\sqrt{X-3}>\frac{7-X}{\sqrt{X-3}}\)
2) \(\frac{1}{\sqrt{2X^2+3X-5}}\ge\frac{1}{2X-1}\)
3) \(\frac{1-\sqrt{1-4X^2}}{X}< 3\)
4) \(\frac{\sqrt{3X+1}-X}{2X-1}< 1\)
giải các phương trình
a \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\)
b \(\sqrt{3x^2-4x}=2x-3\)
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}=2\)
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
rút gọn các biểu thức sau với x\(\ge\)0
a. \(2\sqrt{3}-4\sqrt{3x}+27-3\sqrt{3x}\)
b.\(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)
a) \(2\sqrt{3}-4\sqrt{3x}+27-3\sqrt{3x}\)
= \(\left(2\sqrt{3}+27\right)-\left(4\sqrt{3x}+3\sqrt{3x}\right)\)
=\(\sqrt{3}\left(2+3\right)-\sqrt{3x}\left(4-3\right)\)
=\(5\sqrt{3}-\sqrt{3x}\)
=\(\sqrt{3}\left(5-\sqrt{x}\right)\)
b)\(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)
=\(3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)
=\(\sqrt{2x}\left(3-10+21\right)+28\)
=\(14\sqrt{2x}+28\)
=\(14\sqrt{2}\left(\sqrt{x}+\sqrt{2}\right)\)
cho \(x\ge-\dfrac{1}{3}\). tìm GTNN của \(E=5x-6\sqrt{2x+7}-4\sqrt{3x-1}+2\)
Bạn xem lại ĐKĐB. Nếu $x\geq \frac{-1}{3}$ thì mình nghi ngờ $\sqrt{3x-1}$ của bạn viết là $\sqrt{3x+1}$Còn nếu đúng là $\sqrt{3x-1}$ thì ĐK cần là $x\geq \frac{1}{3}$.
giải BPT :
a. \(\sqrt[3]{x+6}+\sqrt{x-1}\ge x^2-1\)
b.2\(\sqrt[3]{x+4}+\sqrt{2x+7}+x^2+8x+13\)
c.\(4x^3+5x^2+1\ge\sqrt{3x+1}-3x\)
giúp với ạ
giải bpt:
1. \(\frac{\sqrt{-3x^2+x+4}+2}{x}< 2\)
2. \(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
3. \(\sqrt{x^2-8x+15}+\sqrt{x^2+2x-15}\le\sqrt{4x^2-18x=18}\)
4. 4(x+1)2 \(\ge\) (2x +10)( 1- \(\sqrt{3+2x}\))2
5. \(\sqrt{1+x}-\sqrt{1-x}\ge x\)
BT1: Tính
a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
BT2: Rút gọn
\(3x-\sqrt{27}+\frac{\sqrt{x^3+3x^2}}{\sqrt{x+3}}\) ( x ≥ 0 )
Bài 1:
a) Ta có: \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}\)
\(=\frac{\sqrt{5+2\cdot\sqrt{5}\cdot1+1}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}-2}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|-2}{\sqrt{2}}\)
\(=\frac{\sqrt{5}+1-\left(\sqrt{5}-1\right)-2}{\sqrt{2}}\)(Vì \(\sqrt{5}>1>0\))
\(=\frac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}=\frac{2-2}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0\)
b) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
\(=\sqrt{\frac{7}{2}-2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}-\sqrt{\frac{7}{2}+2\cdot\sqrt{\frac{7}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}+\sqrt{7}\)
\(=\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)^2}+\sqrt{7}\)
\(=\left|\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}\right|-\left|\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right|+\sqrt{7}\)
\(=\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}-\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{1}{2}}\right)+\sqrt{7}\)(Vì \(\sqrt{\frac{7}{2}}>\sqrt{\frac{1}{2}}>0\))
\(=\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}-\sqrt{\frac{7}{2}}-\sqrt{\frac{1}{2}}+\sqrt{7}\)
\(=-2\sqrt{\frac{1}{2}}+\sqrt{7}\)
\(=-\sqrt{2}+\sqrt{7}\)