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b. Tự đặt đk
\(x^{^2}+5\sqrt{x-3}=21\\\Leftrightarrow x^{^2}-9+5\sqrt{x-3}=12 \)
Đặt \(a=\sqrt{x-3}\) \(\left(a\ge0\right)\) Phương trình trở thành:
\(a^{^2}\left(a^{^2}+6\right)+5a=12\\ \Leftrightarrow a^{^4}+6a^{^2}+5a-12=0\\ \Leftrightarrow a^{^4}-a^{^3}+a^{^3}-a^{^2}+7a^{^2}-7a+12a-12=0\\ \Leftrightarrow\left(a-1\right)\left(a^{^3}+a^{^2}+7a+12\right)=0\\ \Leftrightarrow a=1\left(tmdk\right)\)
Ta có: vì \(a\ge0\) nên \(a^{^3}+a^{^2}+7a+12\ne0\)
Với a = 1 ta có x=4 (tmdk)
`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`
1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)
\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)
\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)
\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)
a) => 5x^2 - 3 = 2 hoặc 5x^2 - 3 = -2
=> 5x^2 = 5 hoặc 5x^2 = 1
b) pt <=> l(x-1)^2l = x + 2
VÌ ( x - 1 )^2 >= 0 => l( x - 1 )^2 l = ( x- 1 )^2
pt <=> x^2 - 2x + 1 = x + 2 <=>
x^2 - 3x - 1 = 0
c) l2x-5l - l2x^2 - 7x + 5 l = 0
<=> l2x-5l - l ( 2x-5)(x-1) l = 0
<=> l2x-5l ( 1 - l x - 1 l = 0
<=> l 2x - 5 l = 0 hoặc 1 - l x - 1 l = 0
d); e lập bảng xét dấu sau đó xét ba trường hợ p ra
a: Ta có: \(\sqrt{x^2-x+3}+7=10\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b: Ta có: \(\sqrt{x^2-4x+8}-7=-5\)
\(\Leftrightarrow x^2-4x+8=4\)
\(\Leftrightarrow x-2=0\)
hay x=2
\(a,ĐK:\dfrac{3}{x+7}\ge0\Leftrightarrow x+7>0\left(3>0;x+7\ne0\right)\Leftrightarrow x>-7\\ b,ĐK:\dfrac{-2}{5-x}\ge0\Leftrightarrow5-x< 0\left(2-< 0;5-x\ne0\right)\Leftrightarrow x>5\\ c,ĐK:x^2-7x+10\ge0\Leftrightarrow\left(x-5\right)\left(x-2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le2\end{matrix}\right.\)
\(d,ĐK:x^2-8x+10\ge0\Leftrightarrow\left(x-4-\sqrt{6}\right)\left(x-4+\sqrt{6}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4-\sqrt{6}\ge0\\x-4+\sqrt{6}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4-\sqrt{6}\le0\\x-4+\sqrt{6}\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4+\sqrt{6}\\x\ge4-\sqrt{6}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4+\sqrt{6}\\x\le4-\sqrt{6}\end{matrix}\right.\)
\(e,ĐK:9x^2+1\ge0\Leftrightarrow x\in R\left(9x^2+1\ge1>0\right)\)
a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)
\(\Rightarrow3x-2\sqrt{x-1}-4=0\)
\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)
\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)
\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)
*TH1: x = 2 (t/m)
*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)
\(\Rightarrow3\sqrt{x-1}+3=2\)
\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)
Vậy S = {2}
b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )
\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)
\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)
\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)
=> x = 2
\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)
\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)
Đáp án A