Phân tích:
a. 4x - 3\(\sqrt{x}\) - 1
b. 11 - 2\(\sqrt{45}\)
c. 12 - 2\(\sqrt{27}\)
d. 6x - \(\sqrt{x}-1\)
e. 11 + 2\(\sqrt{30}\)
f. 10 + 2\(\sqrt{21}\)
K
Khách
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10 tháng 7 2018
\(a.11+2\sqrt{30}=6+2.\sqrt{6}.\sqrt{5}+5=\left(\sqrt{6}+\sqrt{5}\right)^2\)
\(b.10+2\sqrt{21}=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)
\(c.6x-\sqrt{x}-1=6x+2\sqrt{x}-3\sqrt{x}-1=2\sqrt{x}\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}+1\right)=\left(3\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\) \(d.14-2\sqrt{45}=14-6\sqrt{5}=9-2.3\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)
\(e.4x-3\sqrt{x}-1=4x-4\sqrt{x}+\sqrt{x}-1=4\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{x}-1\right)\left(4\sqrt{x}+1\right)\) \(j.12-2\sqrt{27}=9-2.3.\sqrt{3}+3=\left(3-\sqrt{3}\right)^2\)
9 tháng 7 2021
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
12 tháng 9 2023
a: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>2x=-2 hoặc 2x=4
=>x=2 hoặc x=-1
c: \(\Leftrightarrow\left|x-3\right|=11-x\)
=>x<=11 và (x-3)^2=(11-x)^2
=>x<=11 và x^2-6x+9=x^2-22x+121
=>x<=11 và 16x=112
=>x=7
d:
ĐKXĐ: 3x+19>=0
=>x>=-19/3
PT =>x>=-3 và (3x+19)=(x+3)^2=x^2+6x+9
=>x>=-3 và x^2+6x+9-3x-19=0
=>x>=-3 và (x+5)(x-2)=0
=>x=2
e: =>\(\sqrt{x^2+x+5}=x+1\)
=>x>=-1 và x^2+x+5=x^2+2x+1
=>x>=-1 và 2x+1=x+5
=>x=4
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
a.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$
$\Leftrightarrow \sqrt{2x}=3$
$\Leftrightarrow 2x=9$
$\Leftrightarrow x=\frac{9}{2}$ (tm)
b.
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$
$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$
$\Leftrightarrow 3\sqrt{x+2}=15$
$\Leftrightarrow \sqrt{x+2}=5$
$\Leftrightarrow x+2=25$
$\Leftrightarrow x=23$ (tm)
AH
Akai Haruma
Giáo viên
30 tháng 7 2021
c.
$\sqrt{(x-2)^2}=12$
$\Leftrightarrow |x-2|=12$
$\Leftrightarrow x-2=12$ hoặc $x-2=-12$
$\Leftrightarrow x=14$ hoặc $x=-10$
e.
PT $\Leftrightarrow |2x-1|-x=3$
Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$
$\Leftrightarrow x=4$ (tm)
Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
30 tháng 8 2019
a,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\) (*)(đk \(x\ge-2\))
<=> \(\sqrt{\left(x+2\right)-4\sqrt{x+2}+4}+\sqrt{\left(x+2\right)-6\sqrt{x+2}+9}\)=1
<=> \(\sqrt{\left(\sqrt{x+2}-2\right)^2}+\sqrt{\left(\sqrt{x+2}-3\right)^2}=1\)
<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{x+2}-3\right|\)=1 (1)
TH1: \(0\le\sqrt{x+2}< 2\)
Từ (1) =>\(2-\sqrt{x+2}+3-\sqrt{x+2}=1\)
<=> \(5-2\sqrt{x+2}=1\) <=> \(2\sqrt{x+1}=4\) <=> \(\sqrt{x+1}=2\)
<=> \(x+1=4\) <=> x=3(không t/m \(\sqrt{x+2}\le2\))
TH2 : \(2\le\sqrt{x+2}\le3\)
Từ (1) =>\(\sqrt{x+2}-2+3-\sqrt{x+2}=1\)
<=> \(1=1\) (luôn đúng)
Từ TH2 <=> 4\(\le x+2\le9\) <=> \(2\le x\le7\)
TH3 \(\sqrt{x+2}>3\)
Từ (1) => \(\sqrt{x+2}-2+\sqrt{x+2}-3=1\)
<=> \(2\sqrt{x+2}=6\) <=> \(\sqrt{x+2}=3\) <=> \(x+2=9\) <=> x=7 (không t/m \(\sqrt{x+2}>3\))
Vậy pt (*) có tập nghiệm S=\(\left\{2\le x\le7\right\}\)
b, \(x^2-10x+27=\sqrt{6-x}+\sqrt{x-4}\) (*) (đk :\(4\le x\le6\))
Vs a,b \(\ge0\) ta có \(\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a^2+b^2\right)}\)(tự CM nha)
Dấu "=" xảy ra <=> a=b
Áp dụng bđt trên ta có: \(\sqrt{6-x}+\sqrt{x-4}\le\sqrt{2\left(6-x+x-4\right)}=\sqrt{2.2}=2\)
<=> \(\sqrt{6-x}+\sqrt{x-4}\le2\)(1)
Lại có: \(x^2-10x+27=x^2-10x+25+2=\left(x-5\right)^2+2\ge2\)
<=> \(x^2-10x+27\ge2\) (2)
Từ (1),(2) => Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}6-x=x-4\\x-5=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}6+4=2x\\x=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=5\\x=5\end{matrix}\right.\left(tm\right)\)
Vậy pt (*) có tập nghiệm S=\(\left\{5\right\}\)
c, \(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\)(*) (đk: x\(\ge0\))
<=> \(x\left(x-2\right)-\sqrt{x}\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\left(x-\sqrt{x}\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\sqrt{x}\left(\sqrt{x}-1\right)\left(x-2\right)-4\left(\sqrt{x}-1\right)=0\)
<=> \(\left(\sqrt{x}-1\right)\left[\sqrt{x}\left(x-2\right)-4\right]=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}\left(x-2\right)-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}\left(x-2\right)=4\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x\left(x-2\right)^2=16\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\x\left(x^2-4x+4\right)-16=0\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}x=1\\x^3-4x^2+4x-16=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=1\\x^2\left(x-4\right)+4\left(x-4\right)=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\\left(x^2+4\right)\left(x-4\right)=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\left(tm\right)\)
Vậy pt (*) có tập nghiệm S=\(\left\{1;4\right\}\)
31 tháng 8 2019
d) x2+3x+1=(x+3)\(\sqrt{x^2+1}\)
<=>(\(\sqrt{x^2+1}-3x+3\sqrt{x^2+1}-\left(x^2+1\right)=0\)
<=>\(\left(\sqrt{x^2+1}-3\right)\left(x-\sqrt{x^2+1}\right)=0\)
<=>\(\sqrt{x^2+1}=3\) hoặc \(x=\sqrt{x^2+1}\)
=>x=\(2\sqrt{2}\)
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
CW
3 tháng 8 2018
a) Đk: \(\left[{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
\(\sqrt{x^2-1}-x^2+1=0\)
\(\Leftrightarrow x^2-1-\sqrt{x^2-1}= 0\)
\(\Leftrightarrow\left(\sqrt{x^2-1}-1\right)\sqrt{x^2-1}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}-1=0\\\sqrt{x^2-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=1\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\left(1\right)\\x^2=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=\pm\sqrt{2}\left(N\right)\)
\(\left(2\right)\Leftrightarrow x=\pm1\left(N\right)\)
Kl: \(x=\pm\sqrt{2}\), \(x=\pm1\)
b) Đk: \(\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
\(\sqrt{x^2-4}-x+2=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4=x^2-4x+4\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(N\right)\\x\ge2\end{matrix}\right.\)
kl: x=2
c) \(\sqrt{x^4-8x^2+16}=2-x\)
\(\Leftrightarrow\sqrt{\left(x^2-4\right)^2}=2-x\)
\(\Leftrightarrow\left|x^2-4\right|=2-x\) (*)
Th1: \(x^2-4< 0\Leftrightarrow-2< x< 2\)
(*) \(\Leftrightarrow x^2-4=x-2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=-1\left(N\right)\end{matrix}\right.\)
Th2: \(x^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-2\\x\ge2\end{matrix}\right.\)
(*)\(\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)
Kl: x=-3, x=-1,x=2
d) \(\sqrt{9x^2+6x+1}=\sqrt{11-6\sqrt{2}}\)
\(\Leftrightarrow\sqrt{\left(3x+1\right)^2}=\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(\Leftrightarrow\left|3x+1\right|=3-\sqrt{2}\) (*)
Th1: \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=3-\sqrt{2}\Leftrightarrow x=\dfrac{2-\sqrt{2}}{3}\left(N\right)\)
Th2: \(3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}\)
(*) \(\Leftrightarrow3x+1=-3+\sqrt{2}\Leftrightarrow x=\dfrac{-4+\sqrt{2}}{3}\left(N\right)\)
Kl: \(x=\dfrac{2-\sqrt{2}}{3}\), \(x=\dfrac{-4+\sqrt{2}}{3}\)
e) Đk: \(x\ge-\dfrac{3}{2}\)
\(\sqrt{4^2-9}=2\sqrt{2x+3}\) \(\Leftrightarrow\sqrt{7}=2\sqrt{2x+3}\) \(\Leftrightarrow7=8x+12\)
\(\Leftrightarrow8x=-5\Leftrightarrow x=-\dfrac{5}{8}\left(N\right)\)
kl: \(x=-\dfrac{5}{8}\)
f) Đk: x >/ 5
\(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\left(N\right)\)
kl: x=9
a) 4x-3√x-1=4x-4√x+√x-1=4√x(√x-1)+(√x+1)
=(4√x+1).(√x-1)
c) 12-2√27=9-2.3√3+3=32-2.3√3+(√3)2=(3-√3)2
d) 6x-√x-1=6x-3√x+2√x-1=3√x(2√x-1)+(2√x-1)
=(3√x+1).(2√x-1)
e) 11+2√30=6+2.√6.√5+5=(√6)2+2.√6.√5+(√5)2=(√6+√5)2
f) 10+2√21=7+2.√7.√3+3=(√7)2+2.√7.√3+(√3)2=(√7+√3)2