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ĐKXĐ: \(\left|x\right|\ge\frac{1}{2}\)
\(3x^2+4x+10=2\sqrt{14x^2-7}\)
<=> \(2x^2-1-2\sqrt{7\left(2x^2-1\right)}+7+\left(x^2+4x+4\right)=0\)
<=> \(\left(\sqrt{2x^2-1}-\sqrt{7}\right)^2+\left(x+2\right)^2=0\)
Nhận thấy: \(\left(\sqrt{2x^2-1}-\sqrt{7}\right)^2\ge0\) \(\forall x\)t/m ĐKXĐ
\(\left(x+2\right)^2\ge0\) \(\forall x\)
suy ra: \(\left(\sqrt{2x^2-1}-\sqrt{7}\right)^2+\left(x+2\right)^2\ge0\)
Từ đó, dấu "=" phải xảy ra
Khi đó: \(\hept{\begin{cases}\sqrt{2x^2-1}-\sqrt{7}=0\\x+2=0\end{cases}}\) <=> \(x=-2\) (t/m)
Vậy...
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 \(x=\frac{22}{4}\)
ĐKXĐ: \(\begin{cases}3x+1\ge0\\ 6-x\ge0\end{cases}\Rightarrow\begin{cases}x\ge-\frac13\\ x\le6\end{cases}\Rightarrow-\frac13\le x\le6\)
Ta có: \(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
=>\(\sqrt{3x+1}-4+1-\sqrt{6-x}+3x^2-14x-8+4-1=0\)
=>\(\frac{3x+1-16}{\sqrt{3x+1}+4}+\frac{1-6+x}{1+\sqrt{6-x}}+3x^2-14x-5=0\)
=>\(\frac{3x-15}{\sqrt{3x+1}+4}+\frac{x-5}{\sqrt{6-x}+1}+\left(x-5\right)\left(3x+1\right)=0\)
=>\(\left(x-5\right)\left(\frac{3}{\sqrt{3x+1}+4}+\frac{1}{\sqrt{6-x}+1}+3x+1\right)=0\)
=>x-5=0
=>x=5(nhận)
\(3x^2+4x+10=2\sqrt{14x^2-7}\)
\(\Leftrightarrow2\sqrt{14x^2-7}=3x^2+4x+10\)
\(\Leftrightarrow\left(2\sqrt{14x^2-7}\right)^2=\left(3x^2+4x+10\right)^2\)
\(\Leftrightarrow56x^2-28=9x^4+76x^2+10+24x^3+80x\)\(\Leftrightarrow56x^2-28-9x^4-76x^2-100-24x^3-80x=0\)\(\Leftrightarrow-20x^2-128-9x^4-24x^3-80x=0\)
\(\Leftrightarrow-9x^4-18x^3-6x^3-12x^2-8x^2-16x-64x-128=0\)\(\Leftrightarrow-9x^3\cdot\left(x+2\right)-6x^2\cdot\left(x+2\right)-8x\cdot\left(x+2\right)-64\left(x+2\right)=0\)\(\Leftrightarrow-\left(x+2\right)\cdot\left(9x^3+18x^2+12x^2-24x+32x+64\right)=0\)\(\Leftrightarrow-\left(x+2\right)\cdot\left(9x^2\left(x+2\right)-12x\cdot\left(x+2\right)+32\left(x+2\right)\right)=0\)
\(\Leftrightarrow-\left(x+2\right)^2\cdot\left(9x^2-12x+32\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-\left(x+2\right)^2=0\\9x^2-12x+32=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x\notin R\end{matrix}\right.\)
\(\Leftrightarrow x=-2\)