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a) ĐK : \(x\ge1\)
pt <=> \(\sqrt{3^2\left(x-1\right)}-\frac{1}{2}\sqrt{2^2\left(x-1\right)}=2\)
<=> \(\left|3\right|\sqrt{x-1}-\frac{1}{2}\cdot\left|2\right|\sqrt{x-1}=2\)
<=> \(3\sqrt{x-1}-1\sqrt{x-1}=2\)
<=> \(2\sqrt{x-1}=2\)
<=> \(\sqrt{x-1}=1\)
<=> \(x-1=1\)=> \(x=2\)( tm )
b) \(3x-\sqrt{49-14x+x^2}=15\)
<=> \(\sqrt{x^2-14x+49}=3x-15\)
<=> \(\sqrt{\left(x-7\right)^2}=3x-15\)
<=> \(\left|x-7\right|=3x-15\)(1)
Với x < 7
(1) <=> 7 - x = 3x - 15
<=> -x - 3x = -15 - 7
<=> -4x = -22
<=> x = 11/2 ( tm )
Với x ≥ 7
(1) <=> x - 7 = 3x - 15
<=> x - 3x = -15 + 7
<=> -2x = -8
<=> x = 4 ( ktm )
Vậy x = 11/2
a) \(ĐKXĐ:x\ge1\)
\(\sqrt{9x-9}-\frac{1}{2}\sqrt{4x-4}=2\)
\(\Leftrightarrow\sqrt{9.\left(x-1\right)}-\frac{1}{2}.\sqrt{4\left(x-1\right)}=2\)
\(\Leftrightarrow3\sqrt{x-1}-\frac{1}{2}.2\sqrt{x-1}=2\)
\(\Leftrightarrow3\sqrt{x-1}-\sqrt{x-1}=2\)
\(\Leftrightarrow2\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )
Vậy phương trình có nghiệm là \(x=2\)
b) \(3x-\sqrt{49-14x+x^2}=15\)
\(\Leftrightarrow3x-\sqrt{\left(7-x\right)^2}=15\)
\(\Leftrightarrow3x-\left|7-x\right|=15\)
+) TH1: Nếu \(7-x< 0\)\(\Leftrightarrow x>7\)
thì \(3x-\left(x-7\right)=15\)
\(\Leftrightarrow3x-x+7=15\)\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)( không thỏa mãn )
+) TH2: Nếu \(7-x\ge0\)\(\Leftrightarrow x\le7\)
thì \(3x-\left(7-x\right)=15\)
\(\Leftrightarrow3x-7+x=15\)
\(\Leftrightarrow4x=22\)\(\Leftrightarrow x=\frac{22}{4}\)( thỏa mãn ĐKXĐ )
Vậy nghiệm của phương trình là \(x=\frac{22}{4}\)
a.
\(DK:49-28x-4x^2\ge0\)
PT\(\Leftrightarrow\sqrt{49-28x-4x^2}=5\)
\(\Leftrightarrow49-28x-4x^2=25\)
\(\Leftrightarrow4x^2+28x-24=0\)
\(\Leftrightarrow x^2+7x-6=0\)
Ta co:
\(\Delta=7^2-4.1.\left(-6\right)=73>0\)
\(\Rightarrow\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\left(n\right)\\x_2=\frac{-7-\sqrt{73}}{2}\left(n\right)\end{cases}}\)
Vay nghiem cua PT la \(\hept{\begin{cases}x_1=\frac{-7+\sqrt{73}}{2}\\x_2=\frac{-7-\sqrt{73}}{2}\end{cases}}\)
1. \(\sqrt{x^2-4}-x^2+4=0\)( ĐK: \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\))
\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)
\(\Leftrightarrow\left(x^2-4\right)^2=x^2-4\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\left(tm\right)\\x=\pm\sqrt{5}\left(tm\right)\end{cases}}\)
Vậy pt có tập no \(S=\left\{2;-2;\sqrt{5};-\sqrt{5}\right\}\)
2. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)ĐK: \(\hept{\begin{cases}x^2-4x+5\ge0\\x^2-4x+8\ge0\\x^2-4x+9\ge0\end{cases}}\)
\(\Leftrightarrow\sqrt{x^2-4x+5}-1+\sqrt{x^2-4x+8}-2+\sqrt{x^2-4x+9}-\sqrt{5}=0\)
\(\Leftrightarrow\frac{x^2-4x+4}{\sqrt{x^2-4x+5}+1}+\frac{x^2-4x+4}{\sqrt{x^2-4x+8}+2}+\frac{x^2-4x+4}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}\right)=0\)
Từ Đk đề bài \(\Rightarrow\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}>0\)
\(\Rightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy pt có no x=2
\(a.2\sqrt{x-2}=16\left(ĐK:x\ge2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)
\(b.\sqrt{x-1}>3\left(ĐK:x\ge1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)
\(c.-5\sqrt{2x+4}\le-10\left(ĐK:x\ge2\right)\\ \Leftrightarrow\sqrt{2x+4}\ge2\\ \Leftrightarrow2x+4\ge4\\ \Leftrightarrow2x\ge0\Leftrightarrow x\ge0\)
\(a.2\sqrt{x-2}=16\left(ĐK:x>2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)
b.\(\sqrt{x-1}>3\left(ĐK:x>1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)
\(c.-5\sqrt{2x+4}< -10\left(ĐK:x>-2\right)\\ \Leftrightarrow\sqrt{2x+4}>2\\ \Leftrightarrow2x+4>4\\ \Leftrightarrow2x>0\Leftrightarrow x>0\)
a)\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)
\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)
\(\Rightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+x-3=0\)
Đặt \(x-3=t\) pt thành
\(\sqrt{t\left(t-6\right)}-t=0\)
\(\Leftrightarrow t^2-6t=t^2\)
\(\Leftrightarrow t=0\)\(\Rightarrow x-3=0\Leftrightarrow x=3\)
b)\(\sqrt{x^2-4}-x^2+4=0\)
\(\Leftrightarrow\sqrt{x^2-4}=x^2-4\)
Đặt \(\sqrt{x^2-4}=t\) pt thành
\(t=t^2\Rightarrow t\left(1-t\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}t=1\\t=0\end{array}\right.\).
Với \(t=0\Rightarrow\sqrt{x^2-4}=0\Rightarrow x=\pm2\)
Với \(t=1\Rightarrow\sqrt{x^2-4}=1\)\(\Rightarrow x=\pm\sqrt{5}\)
a/ \(\sqrt{x^2-14x+49}+4x-7=0\)
\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=7-4x\)
\(\Leftrightarrow\left|x-7\right|=7-4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=7-4x\\x-7=4x-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\left(KTM\right)\\x=0\left(TM\right)\end{matrix}\right.\)
Vậy pt có 1 nghiệm x = 0
b/ đkxđ: x ≥2
\(\sqrt{x+2+4\sqrt{x-2}}=4\sqrt{x-2}-5\)
Đặt \(\sqrt{x-2}\) = t (t ≥ 0)
PT \(\Leftrightarrow\sqrt{t^2+4t+4}=4t-5\)
\(\Leftrightarrow\sqrt{\left(t+2\right)^2}=4t-5\)
\(\Leftrightarrow\left|t+2\right|=4t-5\)
Vì t ≥ 0 => t + 2 > 0
=> \(t+2=4t-5\)
\(\Leftrightarrow-3t=-7\Leftrightarrow t=\dfrac{7}{3}\left(TM\right)\)
\(\Rightarrow\sqrt{x-2}=\dfrac{7}{3}\Rightarrow x-2=\dfrac{49}{9}\)
\(\Leftrightarrow x=\dfrac{67}{9}\)(TM)
Vậy pt có nghiệm \(x=\dfrac{67}{9}\)