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a, \(5\sqrt{2x^2+3x+9}=2x^2+3x+3\) (*)
Đặt \(2x^2+3x=a\left(a\ge-9\right)\)
=> \(5\sqrt{a+9}=a+3\)
<=> \(25\left(a+9\right)=a^2+6a+9\)
<=> \(25a+225=a^2+6a+9\)
<=> \(0=a^2+6a+9-25a-225=a^2-19a-216\)
<=> 0= \(a^2-27a+8a-216\)
<=> \(\left(a-27\right)\left(a+8\right)=0\)
=> \(\left[{}\begin{matrix}a=27\\a=-8\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}2x^2+3x=27\\2x^2+3x=-8\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x^2+3x-27=0\\2x^2+3x+8=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left(x-3\right)\left(2x+9\right)=0\\2\left(x^2+2.\frac{3}{4}+\frac{9}{16}\right)+\frac{55}{8}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{9}{2}\left(tm\right)\\2\left(x+\frac{3}{4}\right)^2=-\frac{55}{8}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có tập nghiệm \(S=\left\{3,-\frac{9}{2}\right\}\)
b, \(9-\sqrt{81-7x^3}=\frac{x^3}{2}\left(đk:x\le\sqrt[3]{\frac{81}{7}}\right)\)(*)
<=> \(\sqrt{81-7x^3}=9-\frac{x^3}{2}\)
<=>\(81-7x^3=\left(9-\frac{x^3}{2}\right)^2=81-9x^3+\frac{x^6}{4}\)
<=> \(-7x^3+9x^3-\frac{x^6}{4}=0\) <=> \(2x^3-\frac{x^6}{4}=0\)<=> \(8x^3-x^6=0\)
<=> \(x^3\left(8-x^2\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\8=x^2\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\pm2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có nghiệm x=0
d,\(\sqrt{9x-2x^2}-9x+2x^2+6=0\) (*) (đk: \(0\le x\le\frac{1}{2}\))
<=> \(\sqrt{9x-2x^2}-\left(9x-2x^2\right)+6=0\)
Đặt \(\sqrt{9x-2x^2}=a\left(a\ge0\right)\)
Có \(a-a^2+6=0\)
<=> \(a^2-a-6=0\) <=> \(a^2-3x+2x-6=0\)
<=> \(\left(a-3\right)\left(a+2\right)=0\)
=> \(a-3=0\) (vì a+2>0 vs mọi \(a\ge0\))
<=> a=3 <=>\(\sqrt{9x-2x^2}=3\) <=> \(9x-2x^2=9\)
<=> 0=\(2x^2-9x+9\) <=> \(2x^2-6x-3x+9=0\) <=>\(\left(2x-3\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}2x=3\\x=3\end{matrix}\right.< =>\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)(t/m)
Vậy pt (*) có tập nghiệm \(S=\left\{\frac{3}{2},3\right\}\)
1) \(\sqrt{x-1}=\sqrt{2x+3}\) ĐK: x ≥ 1; x ≥ \(\dfrac{-3}{2}\) => x ≥ 1
=> x - 1 = 2x + 3
=> x - 2x = 3 + 1
=> -x = 4 => x = -4 (ko TMĐK)
Vậy S = ∅
2) \(\sqrt{2x-3}=\sqrt{x-1}\) ĐK: x ≥ \(\dfrac{3}{2}\); x ≥ 1 => x ≥ \(\dfrac{3}{2}\)
=> 2x - 3 = x - 1
=> 2x - x = -1 + 3
=> x = -2 (ko TMĐK)
Vậy S = ∅
3) \(\sqrt{2-x}=\sqrt{3+x}\) ĐK: x ≥ 2; x ≥ -3 => x ≥ 2
=> 2 - x = 3 + x
=> -x - x = 3 - 2
=> -2x = 1 => x = \(\dfrac{-1}{2}\) (ko TMĐK)
Vậy S = ∅
4) \(\sqrt{4x-8}=2\sqrt{x-2}\) ĐK: x ≥ 2
=> 4x - 8 = 2(x - 2)
=> 4x - 8 = 2x - 4
=> 4x - 2x = -4 + 8
=> 2x = 4 => x = 4 : 2 = 2 (TMĐK)
Vậy S = \(\left\{2\right\}\)
5) \(\sqrt{x^2-5}=\sqrt{4x-9}\) ĐK: \(\left|x\right|=\sqrt{5}\) ; x ≥ \(\dfrac{9}{4}\)
<=> x2 - 5 = 4x - 9
<=> x2 - 4x - 5 + 9 = 0
<=> x2 - 4x - 4 = 0 <=> (x - 2)2 =0
=> x = 2 (ko TMĐK)
6) \(\sqrt{x-2}=\sqrt{x^2-2x}\) ĐK: x ≥ 2
=> x - 2 = x2 - 2x
=> x - 2 - x2 + 2x = 0
=> (x - 2) - x(x - 2) = 0
=> (1- x) . (x - 2) = 0
=> \(\left\{{}\begin{matrix}1-x=0\\x-2=0\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-0=1\left(loai\right)\\x=0+2=2\left(TMĐK\right)\end{matrix}\right.\)
Vậy S = \(\left\{2\right\}\)
7) \(\sqrt{x^2-3x}-\sqrt{15-5x}=0\) ĐK: x ≥ 3 hoặc x ≤ 0
<=> \(\sqrt{x^2-3x}=\sqrt{15-5x}\)
<=> x2 - 3x = 15 - 5x
=> x2 - 3x + 5x - 15 = 0
=> x(x -3) + 5(x - 3) = 0
=> (x + 5) . (x - 3) = 0
=> \(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=0-5=-5\\x=0+3=3\end{matrix}\right.\)(TMĐK)
Vậy S = \(\left\{-5;3\right\}\)
8) \(\sqrt{4x^2-9}=\sqrt{-20x-18}\) ĐK: \(\left|x\right|\text{≥}\dfrac{3}{2}\) hoặc x ≤ \(\dfrac{-9}{10}\)
<=> 4x2 - 9 = -20x - 18
<=> 4x2 - 9 + 20x + 18 = 0
<=> 4x2 + 20x + 9 =0
<=> 4x2 + 2x + 18x + 9 =0
<=> 2x(2x + 1) + 9(2x + 1) = 0
<=> (2x + 9) . (2x + 1) = 0
=> \(\left[{}\begin{matrix}2x+9=0\\2x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=-9\\2x=-1\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=\dfrac{-9}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy S = \(\left\{\dfrac{-9}{2};\dfrac{-1}{2}\right\}\)
9) \(\sqrt{x-2}=\sqrt{x-2}\) ĐK: x ≥ 2
<=> x - 2 = x - 2
<=> x - x = 2 - 2
=> 0x = 0 với mọi x TMĐK: x ≥ 2
Kết luận: Phương trình vô nghiệm thoả mãn: x ≥ 2
1,
√(x-1) = √(2x+3)
->(√x-1)^2 = (√2x+3)^2
->x-1=2x+3
->x=-4
2
√(2x−3)=√(x−1) (bình phương lên tiếp)
->2x-3=x-1
->x=2
3->9 tự làm nha tương tự
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
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 :
\(a+b-ab-1=0\)
\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)
\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)
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