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Đk \(x\ge\dfrac{3}{2}\)
\(\Leftrightarrow3\sqrt{2x-3}-4\sqrt{2x-3}=1-2\sqrt{2x-3}\)
\(\Leftrightarrow\sqrt{2x-3}=1\)
\(\Leftrightarrow2x-3=1\)
\(\Leftrightarrow x=2\)
Vậy S=\(\left\{2\right\}\)
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: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
a) ĐK: \(x\ge0\)
PT \(\Leftrightarrow\sqrt{4x}\left(\dfrac{3}{4}-1-\dfrac{1}{4}\right)+5=0\)
\(\Leftrightarrow2\sqrt{x}.\left(-\dfrac{1}{2}\right)+5=0\)
\(\Leftrightarrow x=25\) (thỏa)
Vậy \(x=25\)
b) Đk: \(x\le3\)
PT \(\Leftrightarrow\sqrt{3-x}-\sqrt{9\left(3-x\right)}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}=6\)
\(\Leftrightarrow\sqrt{3-x}\left(1-\sqrt{9}+\dfrac{5}{4}.\sqrt{16}\right)=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow x=-1\) (thỏa)
Vậy \(x=-1\)
2:
a:
Sửa đề: \(P=\left(\dfrac{2}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{2}{\sqrt{1-a^2}}+1\right)\)
\(P=\dfrac{2+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}:\dfrac{2+\sqrt{1-a^2}}{\sqrt{1-a^2}}\)
\(=\dfrac{2+\sqrt{1-a^2}}{\sqrt{1+a}}\cdot\dfrac{\sqrt{1-a^2}}{2+\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}\)
\(=\sqrt{1-a}\)
b: Khi a=24/49 thì \(P=\sqrt{1-\dfrac{24}{49}}=\sqrt{\dfrac{25}{49}}=\dfrac{5}{7}\)
c: P=2
=>1-a=4
=>a=-3
a . \(3\sqrt{2x}-\dfrac{1}{3}\sqrt{18x}=\sqrt{24}\) ( ĐK : \(x\ge0\) )
\(\Leftrightarrow3\sqrt{2x}-\sqrt{2x}=\sqrt{24}\)
\(\Leftrightarrow2\sqrt{2x}=\sqrt{24}\)
\(\Leftrightarrow\sqrt{2x}=\sqrt{6}\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
làm mốt câu còn lại nha .
b) ta có : \(\sqrt{x^2+10\left|x\right|+25}=2\left|x\right|+1\Leftrightarrow\sqrt{\left(\left|x\right|+5\right)^2}=2\left|x\right|+1\)
\(\Leftrightarrow\left|x\right|+5=2\left|x\right|+1\Leftrightarrow\left|x\right|=4\Leftrightarrow x=\pm4\)
vậy \(x=\pm4\)
a: =>3 căn 2x-1/3x3 căn 2x=2 căn 6
=>2 căn 2x=2 căn 6
=>2x=6
=>x=3
b: =>||x|+5|=2|x|+1
\(\Leftrightarrow\left(2\left|x\right|+1-\left|x\right|-5\right)\left(2\left|x\right|+1+\left|x\right|+5\right)=0\)
=>|x|-4=0
=>x=4 hoặc x=-4
\(ĐKXĐ:x\ge-1\)
Ta có : \(\sqrt{x+1}=32x^3+48x^2+18x+1\)
\(\Leftrightarrow\sqrt{x+1}-1=32x^3+48x^2+18x\)
\(\Leftrightarrow\frac{\left(x+1\right)-1^2}{\sqrt{x+1}+1}=2x.\left(16x^2+24x+9\right)\)
\(\Leftrightarrow\frac{x}{\sqrt{x+1}+1}-2x\left(4x+3\right)^2=0\)
\(\Leftrightarrow x.\left[\frac{1}{\sqrt{x+1}+1}-2.\left(4x+3\right)^2\right]=0\) (*)
Với mọi \(x\inĐKXD\) thì \(2.\left(4x+3\right)^2>\frac{1}{\sqrt{x+1}+1}\) nên từ (*) suy ra :
\(x=0\) ( Thỏa mãn ĐKXĐ )
Vậy pt có nghiệm duy nhất \(x=0\)
\(\Leftrightarrow\sqrt{3-x}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}-\sqrt{9\left(3-x\right)}=6\)
\(ĐKXĐ:x\le3\)
\(\Leftrightarrow\sqrt{3-x}+5\sqrt{3-x}-3\sqrt{3-x}=0\)
\(\Leftrightarrow3\sqrt{3-x}=6\)
\(\Leftrightarrow\sqrt{3-x}=2\)
\(\Leftrightarrow x=-1\)
\(\sqrt{3-x}+\dfrac{5}{4}\sqrt{48-16x}-\sqrt{27-9x}=6\) (ĐKXĐ :x\(\ge\)3) \(\Leftrightarrow\sqrt{3-x}+\dfrac{5}{4}\sqrt{16\left(3-x\right)}-\sqrt{9\left(3-x\right)}=6\Leftrightarrow\sqrt{3-x}+\dfrac{5}{4}.4\sqrt{3-x}-3\sqrt{3-x}=6\Leftrightarrow\sqrt{3-x}+5\sqrt{3-x}-3\sqrt{3-x}=6\Leftrightarrow3\sqrt{3-x}=6\Leftrightarrow\sqrt{3-x}=2\Leftrightarrow\left(\sqrt{3-x}\right)^2=4\Leftrightarrow3-x=4\Leftrightarrow x=-1\)(loại vì không thỏa mãn ĐKXĐ)
Vậy phương trình đã cho có tập nghiệm là \(S=\left\{\varnothing\right\}\)
\(ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow3\sqrt{2x-3}-2\sqrt{2x-3}+6\sqrt{2x-3}=1\\ \Leftrightarrow7\sqrt{2x-3}=1\\ \Leftrightarrow\sqrt{2x-3}=\dfrac{1}{7}\\ \Leftrightarrow2x-3=\dfrac{1}{49}\Leftrightarrow x=\dfrac{74}{49}\left(tm\right)\)