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d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
\(B=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)
\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)
Ta có: \(x=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=1\)
Thay x=1 vào B, ta được:
\(B=-\sqrt{1}\cdot\left(\sqrt{1}-1\right)=0\)
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
Bài 3:
\(A=\dfrac{2\sqrt{x}-4}{3\sqrt{x}-4}+\dfrac{x+22\sqrt{x}-32}{3x-10\sqrt{x}+8}+\dfrac{4+2\sqrt{x}}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-4}{3\sqrt{x}-4}+\dfrac{x+22\sqrt{x}-32}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+4}{\sqrt{x}-2}\)
\(=\dfrac{\left(2\sqrt{x}-4\right)\left(\sqrt{x}-2\right)+x+22\sqrt{x}-32+\left(2\sqrt{x}+4\right)\left(3\sqrt{x}-4\right)}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2x-8\sqrt{x}+8+x+22\sqrt{x}-32+6x-8\sqrt{x}+12\sqrt{x}-16}{\left(3\sqrt{x}-4\right)\cdot\left(\sqrt{x}-2\right)}\)
\(=\dfrac{9x+18\sqrt{x}-40}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{9x-12\sqrt{x}+30\sqrt{x}-40}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(3\sqrt{x}-4\right)\left(3\sqrt{x}+10\right)}{\left(3\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}+10}{\sqrt{x}-2}\)
Bài 2:
b: Tọa độ A là:
\(\left\{{}\begin{matrix}y=0\\-\dfrac{1}{2}x+\dfrac{3}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\3-x=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=0\end{matrix}\right.\)
=>A(3;0)
Tọa độ B là:
\(\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}x+\dfrac{3}{2}=-\dfrac{1}{2}\cdot0+\dfrac{3}{2}=1,5\end{matrix}\right.\)
=>B(0;1,5)
\(OA=\sqrt{\left(3-0\right)^2+\left(0-0\right)^2}=\sqrt{3^2+0^2}=3\)
\(OB=\sqrt{\left(0-0\right)^2+\left(1,5-0\right)^2}=1,5\)
Ox\(\perp\)Oy nên OA\(\perp\)OB
=>ΔOAB vuông tại O
=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=2.25\)
Bài 1:
a: ĐKXĐ: \(x\in R\)
\(\sqrt{x^2+4x+4}=2\)
=>\(\sqrt{\left(x+2\right)^2}=2\)
=>|x+2|=2
=>\(\left[{}\begin{matrix}x+2=2\\x+2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
b: ĐKXĐ: x>=2
\(\sqrt{4x-8}-7\cdot\sqrt{\dfrac{x-2}{49}}=5\)
=>\(2\sqrt{x-2}-7\cdot\dfrac{\sqrt{x-2}}{7}=5\)
=>\(\sqrt{x-2}=5\)
=>x-2=25
=>x=27(nhận)
\(\sqrt{4x-8}-2\sqrt{\dfrac{x-2}{4}}=3\left(x\ge2\right)\\ \Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x-2=9\\ \Leftrightarrow x=11\left(tm\right)\)
ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow2\sqrt{x-2}-\sqrt{x-2}=3\)
\(\Leftrightarrow\sqrt{x-2}=3\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)
a)
Có:
\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)
Vì \(\sqrt{117}>\sqrt{116}\) nên \(3\sqrt{13}>2\sqrt{29}\)
b)
Có:
\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)
\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)
Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\) nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)
c)
Có:
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)
Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)
d)
Có:
\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)
\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)
lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)
\(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)
`a, <=> 5/3 . 3sqrt(x^2+2) + 3/2.2sqrt(x^2+2)-7sqrt6=sqrt(x^2+2)`
`= (5+3-1)sqrt(x^2+2)=7sqrt6`
`<=> 7sqrt(x^2+2)=7sqrt6`.
`<=> x^2+2=36`.
`<=> x^2=34`.
`<=> x=+-sqrt(34)`.
Vậy...
`b, sqrt(4x^2-12x+9)-6=0`
`<=> |2x-3|=6`.
`@ x >=3/2 <=> 2x-3=6.`
`<=> x=9/2 (tm)`.
`@x <3/2 <=> 3-2x=6`
`<=> 2x=-3`
`<=> x=-3/2.`
Vậy...
\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{17-12\sqrt{2}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{3^2-2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot\sqrt{7}+\left(\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{\left(2\sqrt{2}-3\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{\left(3-2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}+\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{2\sqrt{2}-3}+\dfrac{3-2\sqrt{5}}{3-2\sqrt{5}}-\dfrac{1}{\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\left(2\sqrt{2}+3\right)^2}{\left(2\sqrt{2}+3\right)\left(2\sqrt{2}-3\right)}+1-\dfrac{\sqrt{5}-\sqrt{7}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)
\(=17-12\sqrt{2}+1-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{2\cdot\left(18-12\sqrt{2}\right)}{2}-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{36-24\sqrt{2}-\sqrt{5}+\sqrt{7}}{2}\)
Lời giải:
ĐKXĐ: $x\geq 5$
$2x^2-8x-6=2\sqrt{x-5}\leq (x-5)+1$ theo BĐT Cô-si
$\Leftrightarrow 2x^2-9x-2\leq 0$
$\Leftrightarrow 2x(x-5)+(x-2)\leq 0$
Điều này vô lý do $2x(x-5)\geq 0; x-2\geq 3>0$ với mọi $x\geq 5$
Vậy pt vô nghiệm nên không có đáp án nào đúng.