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a) \(x-\sqrt{2x+3}=0\)
⇔ \(x=\sqrt{2x+3}\left(x\ge0\right)\)
⇔ \(x^2=2x+3\)
⇔ \(x^2-2x-3=0\)
⇔ x2 + x - 3x - 3 = 0
⇔ x(x+1) - 3(x+1) = 0
⇔ (x-3)(x+1) = 0
⇔\(\left\{{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\Leftrightarrow x=3\)
a: \(\Leftrightarrow x=\sqrt{2x+3}\)
=>x^2=2x+3 và x>=0
=>x^2-2x-3=0 và x>=0
=>x=3
b: \(\Leftrightarrow\left\{{}\begin{matrix}x< =8\\x^2+x+12=x^2-16x+64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =8\\17x=52\end{matrix}\right.\)
=>x=52/17
a) \(x=\sqrt{2x+3}\) (đk \(x\ge-\dfrac{2}{3}\) )
\(\Leftrightarrow x^2-2x-3=0\Leftrightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=3\left(nh\right)\end{matrix}\right.\)
b)ĐK \(x\le8\)
\(\Leftrightarrow x^2+x+12=\left(8-x\right)^2\)
\(\Leftrightarrow x^2+x+12=x^2-16x+64\)
\(\Leftrightarrow17x=52\Rightarrow x=\dfrac{52}{17}\)
c) ĐK \(x\ge1\)
\(\Leftrightarrow\left(2\sqrt{x-1}+\sqrt{x+2}\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow4\left(x-1\right)+x+2+4\sqrt{\left(x-1\right)\left(x+2\right)}=x^2+6x+9\)
\(\Leftrightarrow4\sqrt{x^2+x-2}=x^2+x+11\)
\(\Leftrightarrow=x^2+x+2-4\sqrt{x^2+x-2}+9=0\)( vô lí)
suy ra pt vô nghiệm
d) ĐK \(x\ge3\)
\(\Leftrightarrow x^2-6x+9-\left(x-3\right)-2\sqrt{x-3}+2=0\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)-2\sqrt{x-3}+2=0\)
Đặt \(t=\sqrt{x-3}\)
\(\Leftrightarrow t^4-t^2-2t+2=0\)
\(\Leftrightarrow t^2\left(t^2-1\right)-2\left(t-1\right)=0\)
\(\Leftrightarrow t^2\left(t-1\right)\left(t+1\right)-2\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(t^3+t^2-2\right)=0\)
\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t^2+2t+2=0\left(vl\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x-3}=1\Leftrightarrow x-3=1\Rightarrow x=4\)
a) điều kiện xác định : \(x\ge0\)
ta có : \(pt\Leftrightarrow x=\sqrt{2x+3}\Leftrightarrow x^2-2x-3=0\) --> ...
b) điều kiện xác định : \(x\le8\)
ta có : \(pt\Leftrightarrow x^2+x+12=x^2-16x+64\) --> ...
c) điều kiện xác định : \(x\ge1\)
ta có : \(pt\Leftrightarrow5x-2+4\sqrt{x^2+x-2}=x^2+6x+9\)
\(\Leftrightarrow x^2+x-2-4\sqrt{x^2+x-2}+13\) (1)
đặc \(x^2+x-2=t\left(t\ge0\right)\) \(\Rightarrow\left(1\right)\Leftrightarrow t^2-4t+13\) ( vô nghiệm) --> ...
d) điều kiện xác định : \(x\ge3\)
ta có : \(pt\Leftrightarrow\left(x-3\right)\left(x-4\right)-2\sqrt{x-3}+2=0\) (1)
đặc \(\sqrt{x-3}=t\left(t\ge0\right)\)
\(\Rightarrow\) (1) \(\Leftrightarrow t^2\left(t^2-1\right)-2t+2=0\Leftrightarrow t^2\left(t+1\right)\left(t-1\right)-2\left(t-1\right)=0\)
\(\Leftrightarrow\left(t^3+t^2-2\right)\left(t-1\right)=0\Leftrightarrow\left(t^3-t^2+2t^2-2t+2t-2\right)\left(t-1\right)\)
\(\Leftrightarrow\left(t^2\left(t-1\right)+2t\left(t-1\right)+2\left(t-1\right)\right)\left(t-1\right)=0\)
\(\Leftrightarrow\left(t^2+2x+2\right)\left(t-1\right)^2=0\) --> ...
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) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
a.
$x^2-11=0$
$\Leftrightarrow x^2=11$
$\Leftrightarrow x=\pm \sqrt{11}$
b. $x^2-12x+52=0$
$\Leftrightarrow (x^2-12x+36)+16=0$
$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)
Vậy pt vô nghiệm.
c.
$x^2-3x-28=0$
$\Leftrightarrow x^2+4x-7x-28=0$
$\Leftrightarrow x(x+4)-7(x+4)=0$
$\Leftrightarrow (x+4)(x-7)=0$
$\Leftrightarrow x+4=0$ hoặc $x-7=0$
$\Leftrightarrow x=-4$ hoặc $x=7$
d.
$x^2-11x+38=0$
$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$
$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)
Vậy pt vô nghiệm
e.
$6x^2+71x+175=0$
$\Leftrightarrow 6x^2+21x+50x+175=0$
$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$
$\Leftrightarrow (3x+25)(2x+7)=0$
$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$
$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
b)đk:\(x\ge\dfrac{1}{2}\)
Có: \(\sqrt{2x^2-1}\le\dfrac{2x^2-1+1}{2}=x^2\)
\(x\sqrt{2x-1}=\sqrt{\left(2x^2-x\right)x}\le\dfrac{2x^2-x+x}{2}=x^2\)
=>\(\sqrt{2x^2-1}+x\sqrt{2x-1}\le2x^2\)
Dấu = xảy ra\(\Leftrightarrow x=1\)
Vậy....
c) đk: \(x\ge0\)
\(\Leftrightarrow\sqrt{x}=\sqrt{x+9}-\dfrac{2\sqrt{2}}{\sqrt{x+1}}\)
\(\Rightarrow x=x+9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
\(\Leftrightarrow0=9+\dfrac{8}{x+1}-4\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\)
Đặt \(a=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\left(a>0\right)\)
\(\Leftrightarrow\dfrac{a^2-2}{2}=\dfrac{8}{x+1}\)
pttt \(9+\dfrac{a^2-2}{2}-4a=0\) \(\Leftrightarrow a=4\) (TM)
\(\Rightarrow4=\sqrt{\dfrac{2\left(x+9\right)}{x+1}}\) \(\Leftrightarrow16=\dfrac{2\left(x+9\right)}{x+1}\) \(\Leftrightarrow x=\dfrac{1}{7}\) (TM)
Vậy ...
a)ĐKXĐ: x≥-1/3; x≤6
<=>\(\dfrac{3x-15}{\sqrt{3x+1}+4}+\dfrac{x-5}{\sqrt{x-6}+1}+\left(x-5\right)\cdot\left(3x+1\right)=0\Leftrightarrow\left(x-5\right)\cdot\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{\sqrt{x-6}+1}+3x+1\right)=0\Leftrightarrow x-5=0\Leftrightarrow x=5\)(nhận)
(vì x≥-1/3 nên3x+1≥0 )
a: \(\Leftrightarrow x=\sqrt{2x+3}\)
=>x^2=2x+3 và x>=0
=>x^2-2x-3=0 và x>=0
=>x=3
b: \(\Leftrightarrow\left\{{}\begin{matrix}x< =8\\x^2+x+12=x^2-16x+64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =8\\17x=52\end{matrix}\right.\)
=>x=52/17