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Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{(x-2)(x+2)}-3\sqrt{x-2}=0$
$\Leftrightarrow \sqrt{x-2}(\sqrt{x+2}-3)=0$
\(\Rightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x+2}-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=7\end{matrix}\right.\) (thỏa mãn)
Vậy..........
b) ĐK: $x\geq 0$
PT $\Leftrightarrow (\sqrt{x}-3)^2=0$
$\Leftrightarrow \sqrt{x}-3=0$
$\Leftrightarrow x=9$ (thỏa mãn)
c) ĐK: $x\geq 3$
PT $\Leftrightarrow \sqrt{9(x-3)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4(x-3)}=7$
$\Leftrightarrow 3\sqrt{x-3}+\sqrt{x-3}-\sqrt{x-3}=7$
$\Leftrightarrow 3\sqrt{x-3}=7$
$\Leftrightarrow x-3=(\frac{7}{3})^2$
$\Rightarrow x=\frac{76}{9}$
d)
ĐK: $x\geq \frac{-1}{2}$
PT $\Leftrightarrow 3\sqrt{4(2x+1)}-\frac{1}{3}\sqrt{9(2x+1)}-\frac{1}{2}\sqrt{25(2x+1)}+\sqrt{\frac{1}{4}(2x+1)}=6$
$\Leftrightarrow 6\sqrt{2x+1}-\sqrt{2x+1}-\frac{5}{2}\sqrt{2x+1}+\frac{1}{2}\sqrt{2x+1}=6$
$\Leftrightarrow 3\sqrt{2x+1}=6$
$\Leftrightarrow \sqrt{2x+1}=2$
$\Rightarrow x=\frac{3}{2}$ (thỏa mãn)
a, ĐK :a >= 3
\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)
\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)
b, \(ĐK:x\ge-\frac{1}{2}\)
\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow x=4\left(tm\right)\)
a) đk: \(a\ge3\)
pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
Lời giải:
a) ĐK: \(x>0; x\neq 25; x\neq 36\)
PT \(\Rightarrow (\sqrt{x}-2)(\sqrt{x}-6)=(\sqrt{x}-5)(\sqrt{x}-4)\)
\(\Leftrightarrow x-8\sqrt{x}+12=x-9\sqrt{x}+20\)
\(\Leftrightarrow \sqrt{x}=8\Rightarrow x=64\) (thỏa mãn)
Vậy.......
b)
ĐK: \(x\geq \frac{-1}{2}\)
PT \(\Leftrightarrow \sqrt{9(2x+1)}-\sqrt{4(2x+1)}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow 3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow \frac{4}{3}\sqrt{2x+1}=4\Leftrightarrow \sqrt{2x+1}=3\)
\(\Rightarrow x=\frac{3^2-1}{2}=4\) (thỏa mãn)
c)
ĐK: \(x\geq 2\)
PT \(\Leftrightarrow \sqrt{4(x-2)}-\frac{1}{2}\sqrt{x-2}+\sqrt{9(x-2)}=9\)
\(\Leftrightarrow 2\sqrt{x-2}-\frac{1}{2}\sqrt{x-2}+3\sqrt{x-2}=9\)
\(\Leftrightarrow \frac{9}{2}\sqrt{x-2}=9\Leftrightarrow \sqrt{x-2}=2\Rightarrow x=2^2+2=6\) (thỏa mãn)
a.\(\sqrt{x-2}=\sqrt{4-x}\)
đk: \(\left\{{}\begin{matrix}x-2\ge0\\4-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le4\end{matrix}\right.\Leftrightarrow2\le x\le4\)
pt đã cho tương đương với
\(x-2=4-x\)
\(\Leftrightarrow2x=6\Rightarrow x=3\left(TM\right)\)
b.\(\sqrt{x^2-8x+6}=x+2\)
đk: \(x+2\ge0\Rightarrow x\ge-2\)
pt đã cho tương đương với
\(x^2-8x+6=\left(x+2\right)^2\)
\(\Leftrightarrow x^2-8x+6=x^2+4x+4\)
\(\Leftrightarrow-12x=-2\Rightarrow x=\frac{1}{6}\left(TM\right)\)
c.\(\sqrt{2x-1}+5=\sqrt{8x-4}\)
\(\Leftrightarrow\sqrt{2x-1}+5=\sqrt{4\left(2x-1\right)}\)
\(\Leftrightarrow\sqrt{2x-1}+5=2\sqrt{2x-1}\)
\(\Leftrightarrow\sqrt{2x-1}=5\)
đk: \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)
pt tương đương: \(2x-1=25\)
\(\Leftrightarrow2x=26\Rightarrow x=13\left(TM\right)\)
d.\(\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)
\(\Leftrightarrow\sqrt{16\left(1-2x\right)}-\sqrt{4.3x}=\sqrt{3x}+\sqrt{9\left(1-2x\right)}\)
\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}+3\sqrt{1-2x}\)
\(\Leftrightarrow\sqrt{1-2x}=3\sqrt{3x}\)
đk: \(\left\{{}\begin{matrix}1-2x\ge0\\3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{1}{2}\\x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le\frac{1}{2}\)
pt tương đương: \(1-2x=9.3x\)
\(\Leftrightarrow29x=1\Rightarrow x=\frac{1}{29}\left(TM\right)\)
e. \(\sqrt{x^2-9}-\sqrt{4x-12}=0\)
đk: \(\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ge0\\4x-12\ge0\end{matrix}\right.\Leftrightarrow x\ge3\)
pt đã cho tương đương với
\(\sqrt{\left(x-3\right)\left(x+3\right)}-\sqrt{4\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-3}.\sqrt{x+3}-2\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}.\left(\sqrt{x+3}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\left(TM\right)\\\sqrt{x+3}=2\Leftrightarrow x+3=4\Rightarrow x=1\left(KTM\right)\end{matrix}\right.\)
a) ĐKXĐ : \(x\ge0\)
Ta có : \(\sqrt{3x}-\sqrt{27}+\sqrt{75x}=3\Leftrightarrow\sqrt{x}\left(\sqrt{3}+\sqrt{75}\right)=3+\sqrt{27}\)
\(\Leftrightarrow\sqrt{x}=\frac{3+\sqrt{27}}{\sqrt{3}+\sqrt{75}}=\frac{\sqrt{3}+3}{6}\)
\(\Leftrightarrow x=\frac{\left(3+\sqrt{3}\right)^2}{36}\)
b) ĐKXĐ : \(x\ge1\)
\(\sqrt{x-1}-\sqrt{4x-4}+\sqrt{9x-9}=10\)
\(\Leftrightarrow\sqrt{x-1}-\sqrt{4.\left(x-1\right)}+\sqrt{9.\left(x-1\right)}=10\)
\(\Leftrightarrow\sqrt{x-1}-2\sqrt{x-1}+3\sqrt{x-1}=10\)
\(\Leftrightarrow\sqrt{x-1}=5\Leftrightarrow x=26\) (TMĐK)
c) ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\sqrt{2x+1}+\sqrt{18x+9}-\sqrt{50x+25}=-3\)
\(\Leftrightarrow\sqrt{2x+1}+\sqrt{9\left(2x+1\right)}-\sqrt{25\left(2x+1\right)}=-3\)
\(\Leftrightarrow\sqrt{2x+1}+3\sqrt{2x+1}-5\sqrt{2x+1}=-3\)
\(\Leftrightarrow0=-3\) (Vô lí - loại)
Vậy pt vô nghiệm.
\(\sqrt{x-1}=5\)
\(\Leftrightarrow x-1=25\) (bình phương 2 vế)
\(\Leftrightarrow x=26\)
a)\(2\sqrt{3}-\sqrt{4+x^2}=0\)
\(\Leftrightarrow\sqrt{12}-\sqrt{4+x^2}=0\)
\(\Leftrightarrow\sqrt{4+x^2}=\sqrt{12}\)
\(\Leftrightarrow4+x^2=12\Leftrightarrow x^2=8\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\end{matrix}\right.\)
vậy ....
b)\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18x}=0\) điều kiện xác định x\(\ge0\)
\(\Leftrightarrow3\sqrt{2x}+5\sqrt{4}\sqrt{2x}-\sqrt{9}\sqrt{2x}=20\)
\(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}=20\)
\(\Leftrightarrow10\sqrt{2x}=20\Leftrightarrow\sqrt{2x}=2\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\) (tm)
Vậy ....
c)\(\sqrt{4\left(x+2\right)^2}=8\Leftrightarrow4\left(x+2\right)^2=64\)
\(\Leftrightarrow\left(x+2\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy ...
a) pt <=> \(\sqrt{4+x^2}=2\sqrt{3}\)
<=> x2 + 4 = 12
<=> x2 = 8
<=> x = \(\pm2\sqrt{2}\)
b) ĐKXĐ: x ≥ 0
pt <=> \(3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}=20\)
<=> \(10\sqrt{2x}\) = 20
<=> \(\sqrt{2x}=2\)
<=> x = 2 (TM)
c) pt <=> 2|x + 2| = 8
<=> |x + 2| = 4
<=> \(\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
d) ĐKXĐ: x ≥ 2
pt <=> \(\sqrt{x-2}=3\sqrt{x^2-4}\)
<=> 9x2 - 12 = x - 2
<=> 9x2 - x - 10 = 0
<=> 9(x + 1)(x - \(\dfrac{10}{9}\)) = 0
<=> \(\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{9}\end{matrix}\right.\)(KTM)
e) pt <=> 4x + 1 = -7
<=> 4x = -8
<=> x = -2
\(a,\Leftrightarrow\left|x-2\right|=2x-5\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{7}{3}\left(l\right)\end{matrix}\right.\)
b\(\left|x+3\right|=x+3\Rightarrow\left[{}\begin{matrix}\forall x\ge-3\\x=-3\left(l\right)\end{matrix}\right.\)
\(c,\sqrt{2x}\left(\frac{1}{3}-2+3\right)=12\Rightarrow\sqrt{2x}=9\Rightarrow2x=81\Rightarrow x=\frac{81}{2}\)
Vũ Minh TuấnguPhương Anyen thi vaHoàng Tử HàngttNguyHISINOMA KINguy@Nk>↑@ễNguyễn Huy Tún Thị Diễm QuỳTrần ThaVõ Đông Anh Tuấnnh PhươngnhNHuyềNguHồnTrần Việt Linhg PNguyễn Thanh Hằnghúc Nguyễnyễn Văn ĐạtnIAce LegonaMADOễMystesoyeon_Tiểubàng giảirious Personn HoànLê Thị Thục Hiềng Nhihn
Mn giúp em với ạ
e/ \(\sqrt{x-2}+\sqrt{6-x}=\sqrt{x^2-8x+24}\)
\(\Leftrightarrow4+2\sqrt{\left(x-2\right)\left(6-x\right)}=x^2-8x+24\)
\(\Leftrightarrow2\sqrt{-x^2+8x-12}=x^2-8x+20\)
Đặt \(\sqrt{-x^2+8x-12}=a\left(a\ge0\right)\)thì pt thành
\(2a=-a^2+8\)
\(\Leftrightarrow a^2+2a-8=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-4\left(l\right)\\a=2\end{cases}}\)
\(\Leftrightarrow\sqrt{-x^2+8x-12}=2\)
\(\Leftrightarrow-x^2+8x-12=4\)
\(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x=4\)
a/ \(4x^2+3x+3-4x\sqrt{x+3}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{x+3}+x+3\right)+\left(2x-1-2\sqrt{2x-1}+1\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x+3}\right)^2+\left(1-\sqrt{2x-1}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=\sqrt{x+3}\\1=\sqrt{2x-1}\end{cases}\Leftrightarrow}x=1\)
a, đk x >= 1/2
\(3\sqrt{2x-1}+2\sqrt{2x-1}-\sqrt{2x-1}=12\)
\(\Leftrightarrow4\sqrt{2x-1}=12\Leftrightarrow2x-1=9\Leftrightarrow x=5\left(tmđk\right)\)
b, bạn xem lại đề nhé