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a/ ĐKXĐ: \(x\ge-\frac{5}{2}\)
\(\sqrt{2x+5}=5\Rightarrow2x+5=25\Rightarrow x=10\)
b/ \(\sqrt{x-7}+3=0\)
Do \(\sqrt{x-7}\ge0\Rightarrow\sqrt{x-7}+3>0\Rightarrow ptvn\)
c/ ĐKXĐ: \(x\ge0\)
\(\sqrt{3x}=\sqrt{10}-1\Rightarrow3x=11-2\sqrt{10}\Rightarrow x=\frac{11-2\sqrt{10}}{3}\)
d/ \(4-7x=11\Rightarrow-7x=7\Rightarrow x=-1\)
d)
ĐK: $x\leq \frac{16}{7}$
PT $\Rightarrow 16-7x=11^2=121$
$\Rightarrow 7x=16-121=-105$
$\Leftrightarrow x=-15$ (thỏa mãn)
e) ĐK: $x\geq 3$
PT $\Rightarrow 10(x-3)=30$ (bình phương 2 vế)
$\Leftrightarrow x-3=3\Leftrightarrow x=6$
(thỏa mãn)
f)
ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{x-2}=6$
$\Rightarrow x-2=6^2=36\Leftrightarrow x=38$ (thỏa mãn)
a)
ĐK: $x\geq \frac{-5}{2}$
PT $\Rightarrow 2x+5=25$ (bình phương 2 vế)
$\Leftrightarrow 2x=10\Leftrightarrow x=5$ (thỏa mãn)
b) ĐK: $x\geq \frac{-1}{3}$
PT $\Rightarrow 3x+1=10$ (bình phương 2 vế)
$\Leftrightarrow 3x=9\Leftrightarrow x=3$ (thỏa mãn)
c)
ĐK: $x\geq 7$
PT $\Leftrightarrow \sqrt{x-7}=3+0=3$
$\Rightarrow x-7=3^2$
$\Leftrightarrow x=16$ (thỏa mãn)
a) + \(VT=\sqrt{x^2+2x+10}+x^2+2x+1+7\)
\(=\sqrt{x^2+2x+1}+\left(x+1\right)^2+7>0\forall x\)
=> ptvn
d) ĐK : \(x^2+7x+7\ge0\)
Đặt \(t=\sqrt{x^2+7x+7}\ge0\) \(\Rightarrow t^2=x^2+7x+7\)
\(pt\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)
\(\Leftrightarrow3t^2+2t-5=0\Leftrightarrow\left(3t+5\right)\left(t-1\right)=0\)
\(\Leftrightarrow t=1\) ( do \(3t+5>0\forall t\ge0\) )
\(\Leftrightarrow x^2+7x+1=0\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\) ( TM )
f) ĐK : \(x\ge1\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\ge0\\b=\sqrt{x+3}\ge0\end{matrix}\right.\) thì pt trở thành :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
Lời giải:
a) ĐK: $x\geq \frac{1}{2}$
PT $\Rightarrow 2x-1=(\sqrt{2}-1)^2=3-2\sqrt{2}$
$\Leftrightarrow 2x=4-2\sqrt{2}$
$\Leftrightarrow x=2-\sqrt{2}$ (thỏa mãn)
Vậy.........
b) ĐK: $x\geq \frac{-11}{3}$
PT $\Rightarrow 3x+11=(3+\sqrt{2})^2=11+6\sqrt{2}$
$\Leftrightarrow x=2\sqrt{2}$ (thỏa mãn)
Vậy.........
c)
ĐK: $x\geq -5$
Ta thấy: $\sqrt{x+5}\geq 0$ với mọi $x\geq -5$, mà $\sqrt{3}-2< 0$ nên PT vô nghiệm.
d)
ĐK: $x\geq -38$
PT $\Rightarrow x+38=(3+\sqrt{5})^2=14+6\sqrt{5}$
$\Leftrightarrow x=6\sqrt{5}-24$ (thỏa mãn)
Vậy........
a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
ĐKXĐ:....
a/ \(\sqrt{2x+3}=1+\sqrt{2}\Leftrightarrow2x+3=3+2\sqrt{2}\Rightarrow x=\sqrt{2}\)
b/ \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\Leftrightarrow10+\sqrt{3x}=10+4\sqrt{6}\)
\(\Rightarrow\sqrt{3x}=4\sqrt{6}\Rightarrow3x=96\Rightarrow x=32\)
c/ \(\sqrt{3x-2}=2-\sqrt{3}\Rightarrow3x-2=7-4\sqrt{3}\)
\(\Rightarrow3x=9-4\sqrt{3}\Rightarrow x=\frac{9-4\sqrt{3}}{3}\)
d/ \(\sqrt{x-1}=\sqrt{5}-3\)
Do \(\left\{{}\begin{matrix}\sqrt{x-1}\ge0\\\sqrt{5}-3< 0\end{matrix}\right.\) \(\Rightarrow\) phương trình vô nghiệm
a)<=>2x +5=25
<=>2x=20
<=>x=10
b)<=>\(\sqrt{x-7}\)=-3
<=>x-7=9
<=>x=16
c)<=>3x+1=10
<=>3x=9
<=>x=3
d)16-7x=121
<=>7x=-105
<=>x=-15
mk ko viết lại đề nha <3